Solutions to Example_problems

Solutions to ExampleProblems in EngineeringNoise Control, 2nd Edn.

A companion to"Engineering Noise Control", 3rd Edn

Colin H. HansenDepartment of Mechanical EngineeringUniversity of AdelaideSouth Australia 5005AUSTRALIA

FAX: +61-8-8303-4367e-mail:[email protected]

Published by Colin H Hansen, Adelaide, South Australia

Apart from fair dealing for the purposes of research or private study, orcriticism or review, as permitted under the Australian copyright Act, 1968,this publication may not be reproduced, stored, or transmitted, in any form, orby any means, without the prior permission in writing of the author.

The author makes no representation, express or implied, with regard to theaccuracy of the information contained in this book and cannot accept any legalresponsibility or liability for any errors or omissions that may be made. Acatalogue record for this book is available from the Australian NationalLibrary.

© 2003 Colin H Hansen

Published by Colin H Hansen

ISBN 0-9751704-0-6

This version updated August 4, 2006

ContentsAll page, equation and table references are to the third edition of the textbook,Engineering Noise Control by DA Bies and CH Hansen

1. Solutions to problems in fundamentals 1

2. Solutions to problems relating to the human ear 65

3. Solutions to problems relating to noise measurement andinstrumentation 71

4. Solutions to problems relating to criteria 83

5. Solutions to problems related to sound sources and outdoor soundpropagation 99

6. Solutions to problems related to sound power, its use andmeasurement 130

7. Solutions to problems related to sound in enclosed spaces 143

8. Solutions to problems related to sound transmission loss, acousticenclosures and barriers 199

9. Solutions to problems related to muffling devices 259

10. Solutions to problems in vibration isolation 298

11. Solutions to problems in active noise control 312

12. Errata in Third Edition of Engineering Noise Control 314

Preface

This book provides detailed and instructive solutions to the book of problemson acoustics and noise control which is intended as a companion to the 3rd

edition of the book "Engineering Noise Control" by David A. Bies and ColinH. Hansen. The problems and solutions cover chapters 1 to 10 and 12 in thattext. Some of the problems and solutions are formulated to illustrate thephysics underlying the acoustical concepts and others are based on actualpractical problems. Many of the solutions extend the discussion in the textand illustrate the more difficult concepts by example, thus acting as a valuablesource and understanding for the consultant and student alike.

Although most of the problems and solutions have been tested on students,it is highly likely that there exist errors of which I am unaware. I would dearlylike to hear from any readers who may discover any errors, no matter howminor.

C.H.H., September, 2003e-mail:[email protected]

Acknowledgments

The author would like to thank the many people (too numerous to mention)who were responsible for providing ideas for problems. The author would alsolike to thank his undergraduate and graduate students who provided anexcellent opportunity for fine tuning the solutions to the problems in thisbook.

Finally, the author would like to express his deep appreciation to hisfamily, particularly his wife Susan and daughters Kristy and Laura for theirpatience and support during the many nights and weekends needed toassemble the problems and solutions in a form suitable for publication.

This book is dedicatedto Susan,to Kristy

and to Laura.

cair

cHe

'MHe

γHe

×γair

Mair

1/2

'4

5/3

1/2 1.429

1/2

' 0.34

1

Solutions to problemsin FundamentalsUnless otherwise stated an air temperature of 20EC corresponding to anair density of 1.206kg/m3 and a speed of sound of 343m/s has beenassumed.

Problem 1.1

• undertake an assessment of the current environment where there appearsto be a problem, including the preparation of noise level contours whererequired;

• establish the noise control objectives or criteria to be met;• identify noise transmission paths and generation mechanisms;• rank order noise sources contributing to any excessive levels;• formulate a noise control program and implementation schedule;• carry out the program; and• verify the achievement of the objectives of the program.

See pages 8-10 in the third edition of the text for more details.

Problem 1.2

(a) Speed of sound given by equation 1.4 in text. The only variables whichare different for the two gases are γ and M. Thus:

(b) The wavelength of sound emitted from ones mouth is a function of thevocal cord properties which remain unchanged by the presence of helium.As fλ = c and λ is fixed and c is faster in helium, the sound emanatingfrom your mouth will be higher in pitch.

Solutions to problems2

h '4240

101.4×103× 95

100' 0.0397

cdry ' γRT /M ' 1.4×8.314×303.2/0.029 ' 348.8m/s

cwet ' 348.8(1 % 0.0397×0.16) ' 351.0m/s

0PV '0m

MRT

0V '250,000

24 × 3,600' 2.894m 3/s

0m 'P 0VMRT

'101,400 × 0.029 × 2.8935

8.314 × 288.2' 3.550kg/s

Problem 1.3

Using the relation for h given in the question and knowing that atmosphericpressure is 101.4kPa, we can write:

The speed of sound in dry air at 30EC is given by:

The speed of sound in wet air is then:

Problem 1.4

Gas volume flow rate = 250,000 m3 per day at STPGas Law also applies to a moving fluid, so:

P and T are the static pressure and temperature respectively and

(a) The mass flow rate is given by:

(b) Density of gas in pipe is:

ρ '0m0V'

PMRT

'8 × 106 × 0.0298.314 × 393.2

' 71.0kg/m 3

Fundamentals 3

U '4

πd 20V '

4

πd 2

0mρ

'4

π × 0.01× 3.551

71.0' 6.37m/s

c 'γPρ

1/2

' 1.4 × 8 × 106 /71.0 ' 397m/s

PV 'mM

RT or RM

'PVmT

c '1.35 × 101400 × 1273

1.4 × 273

1/2

' 675 m/s

dPdV

'mM

RV

dTdV

&mM

RT

V 2or dp

dV'

PT

dTdV

&PV

(c) Gas flow speed in discharge pipe is:

Speed of sound (relative to fluid) is:

(d) Speed of sound relative to the pipe is thus: 397.2 + 6.4 = 404 m/s.

Problem 1.5

The speed of sound is given by: and for any gas, (R/M) is fixed.c ' γRT /MWe can find (R/M) by using the properties at 0EC and the expression:

As (m/V) = 1.4 kg/m3, and thus:RM

'101400

1.4 × 273

Problem 1.6

The Universal gas law may be written as .PV 'mM

RT or P 'mM

RTV

Differentiating gives:

which may be rewritten as:


dPP

%dVV

'dTT

dVV

' &1γ

dPP

dPP

1 &1γ

'dTT

I 'p 2

2ρc' Iref 10

LI /10and p ' pressure amplitude

pP

1 &1γ

'τT

τ ' 1.617(1 & 1/1.4)101400

× 298 ' 1.4 × 10&3 EC

ρc 'γP

γRT /M'

1.4 × 101400

1.4 × 8.314 × 313.2/0.029' 400.4

Thus, p ' [2 × 1.205 × 343 × 10&12 × 109.5 ]1/2 ' 1.62 Pa

Another gas property associated with the adiabatic expansion and contractionof a sound wave is , which leads to:PV γ ' const

Combining the above two equations gives:

The acoustic pressure associated with a sound wave of intensity 95dB iscalculated as follows:Thus,

We can write the results of the preceding analysis as:

Substituting in given values and rearranging the equation to give τ, we obtain:

which is the amplitude of the temperature fluctuations.

Problem 1.7

Using equations 1.4a and b in the text, it can be shown that:

Fundamentals 5

λ ' c / f ' 789/40 ' 19.7 m

f ' 250 ' c /2L ' c /2.4. Thus, c ' 600 m/s

c 'γRTM

'1.4 × 8.314 × T

0.035

ρ 'γP

c 2'

1.4 × 101.4 × 103

6002' 0.39 kg/m 3

c 'γRTM

'1.4 × 8.314 × 1873

0.035' 789 m/s

ρ 'γP

c 2'

1.4 × 101.4 × 103

7892' 0.23 kg/m 3

T '6002 × 0.0351.4 × 8.314

' 1083EK ' 810EC

Problem 1.8

f = c/4L = 343/(4×4) = 21.4Hz

Problem 1.9

(a) Assume that the tail pipe is effectively open at each end as it follows amuffler.

Thus:

(b)

Assumption is that the gas in the pipe is at atmospheric pressure at sea level.

Problem 1.10

(a)

(b)

(c)


D ' &V MVMp

&1

γdVg

Vg

%dPP

' 0

Lw ' Lp & 10log104(1 & α )

S α&10log10

ρc400

' 120 & 10log104 × 0.98

149 × 0.02& 10log10

789 × 0.23400

' 122.3 dB

W ' 10&12 × 10122.3/10 ' 1.7 W

If we treat the furnace like a closed end tube, , sof ' Nc /2LL ' c /2f ' 789/80 ' 9.9 mSo we can expect one dimension of the furnace to be 9.9 m.

(d) Surface area, S = πdL + 2πr2 = π × 4 × 9.86 + 2π × 4 = 149 m2

(e) Second burner has twice the sound power level, so increase in soundpressure level will be 10 log10(3) or an increase of 4.8 dB. So the newSPL = 124.8 dB.

Problem 1.11

(a) The bulk modulus of the fluid is given by equation 1.2 in the text as:

Consider a unit volume of fluid (V = 1), let the proportion (and volumeVg) of gas be x and the proportion (and volume VL) of liquid is then (1 -x). Assume a pressure increment of Δp, and a corresponding volumeincrement ΔV.

From the above equation, we can write, ΔVL = -Δp(1 - x)/D

Assuming adiabatic compression of gas,. Differentiating P with respect to VPV γ ' const or P ' const × V &γ

and rearranging gives:

Fundamentals 7

ΔVg ' &xΔPγP

ΔV ' &1 & x

D&

xγP

ΔP

Deff ' &ΔVΔP

&1

' &1 & x

D&

xγP

&1

ρeff ' (1 & x)ρL % xρg

c '1

1 & xD

%xγP

(1 & x)ρL % xρg

Substituting x for Vg and rearranging gives:

Thus, the total volume change for a change Δp in pressure =ΔVL + ΔVg = ΔV. Thus:

Thus the effective bulk modulus of the fluid is given by:

The effective density of the fluid is equal to the total mass for unitvolume of fluid and is given by:

The speed of sound is given by . Substituting in the effectivec ' D /ρvalues calculated for D and ρ above we obtain:

where x is the proportion of gas in the fluid.

(b) As x approaches 0, using the result of (a) above gives:

= which is the speed of sound in the liquid.c '1

(1/D)ρL

D /ρL

As x approaches 1, using the result of (a) above gives:

which is the speed of sound in the gas.c '1

(1/γP)ρg

'γPρg


φ 'Ar

e j(ωt & kr)

pu

'jωρ

1/r % jk

pu

' ρc jkr(1 & jkr)

1 % k 2r 2'ρckr(j % kr)

1 % k 2r 2'

ρckr

1 % k 2r 2× j % kr

1 % k 2r 2

pu

' ρccosβ (jsinβ % cosβ) ' ρccosβejβ

u 'p

ρccosβe&jβ

Problem 1.12

Equation 1.40c in the text is the harmonic solution to the spherical waveequation. That is:

For spherical waves, equation 1.6 may be written as and thus theu ' &MφMr

particle velocity may be written as in equation 1.42. Using equations 1.7 and1.40, the acoustic pressure may be written as in equation 1.41b. Usingequations 1.41b and 1.42, we may write:

Using ω = kc and multiplying the numerator and denominator of the aboveequation by r gives equation 1.43 in the text.

Multiplying the numerator and denominator of equation 1.43 by (1 - jkr)gives:

Defining the phase β = tan-1(1/kr) as the phase by which the pressure leads theparticle velocity, the preceding equation may be written as:

which is the same as equation 1.72.

Harmonic intensity is defined as I = 0.5Re where the bar denotes thep u (

complex amplitude. Using equation 1.72, we may write:

Thus, , which is the plane wave0.5 × Rep u ' 0.5 p 2 cosβρccosβ

'¢p 2¦ρc

expression.

Fundamentals 9

u ' &MφMr

'Akρrω

e j(ωt & kr) %A

jρr 2ωe j(ωt & kr)

U 'Aρr0ω

ej(ωt & kr0)

k % 1jr0

W ' 2πω2r 40 ρU

20 /c

' 2π(2π × 100)2 × 0.054 × 1.205 × 22 /344 ' 0.22W

Problem 1.13

(a) At any location, . The velocity potentialp 'Ar

e jω(t & r /c) 'Ar

e j(ωt & kr)

is then and the particle velocity isφ '1ρ mpdt '

Ajρrω

ej(ωt & kr)

(b) When r = r0, u = U and the particle velocity may be written as:

But U = U0ejωt , so . U0 '

Aρr0ω

k % 1jr0

e& jkr0

Thus, A 'ρr0ωU0

k % 1jr0

ejkr0

However, ejkr0 ' coskr0 % jsinkr0 . 1 % jkr0 for small kr0

Thus, A 'ρr0ωU0

k % 1jr0

1 % jkr0 ' jρr 20ωU0

(c) ω = 200π, ρ = 1.206, U0 = 2, r0 = 0.05

Power, W = IS = (p2/ρc)4πr2 = and*p*2

2ρc4πr 2 '

*A*2

2ρcr 24πr 2

. Thus:*A*2 ' ω2 r 40 ρ

2 U 20


p 'Ar

e j(ωt & kr)

φ 'A

jωρre j(ωt & kr)

u ' &MφMr

'jkA

jrωρej(ωt & kr) %

A

jr 2ωρej(ωt & kr)

'A

rρcej(ωt & kr) 1 &

jkr

'pρc

1 &j

kr

*u* '*p*ρc

*1 &j

kr*

Problem 1.14

Spherical wave solution to the wave equation is:

Using equation 1.7 in the text, the velocity potential is:c

Using the one dimensional form of equation 1.6 in the text, the particlevelocity is:

Thus:

Amplitude is twice when or .*p* /ρc *1 & j /kr* ' 2 1 % (kr)&2 ' 4

Thus kr ' 1/ 3 ' 0.58

Problem 1.15

(a) Plane wave: p = ρcu = 1.206 × 343 × 0.2

Lp ' 20log101.206 × 343 × 0.2

2 × 10&5' 132dB

(b) Lp ' 20log10988 × 1486 × 0.2

10&6' 229dB

Non-linear effects are definitely important as the level in air is above 130dB

Fundamentals 11

Lp ' 20log10

prms

10&6' 20log10

106.07 × 103

10&6' 220.5dB

*u* '*p*ρc

*1 &j

kr* ' 150 × 103

988 × 14811 %

14812π × 1000 × 1

2

' 105mm/s

and the level in water is also very high.

Cavitation in water would occur if the sound pressure amplitude exceeds themean pressure which would happen if the water were less than a certain depth.

The depth is calculated by calculating the weight of a column of water, 1 m2

in cross section. Weight = 988 × 9.81 = 9692.3 N, so the water pressure is9692.3 Pa per meter depth. The acoustic pressure amplitude is 1.41 × 988 ×1486 × 0.2 = 0.414 MPa. So the depth of water this corresponds to (allowingfor atmospheric pressure) is:

h = (0.414 × 106 - 101400)/9692.3 = 32.2m deep.

Problem 1.16

(a) The instantaneous total pressure will become negative if the acousticpressure amplitude exceeds the mean pressure. This corresponds to anr.m.s. acoustic pressure of 150//2 = 106.07 kPa. The sound pressurelevel is then:

(b) The particle velocity is given by . Thus the amplitude of theu ' p /ρcparticle velocity is u = 150 × 103/(988 × 1481) = 103 mm/s

(c) Using the analysis of problem 1.14, we can show that for a sphericalwave, the particle velocity amplitude is:


p 'Ar

e j(ωt & kr)

φ 'A

jωρre j(ωt & kr)

u ' &MφMr

'jkA

jrωρe j(ωt & kr) %

A

jr 2ωρe j(ωt & kr)

'A

rρce j(ωt & kr) 1 &

jkr

'pρc

1 &j

kr

*u* '*p*ρc

*1 &j

kr*

*p* ' 2 × pref × 10Lp /20

' 2 × 2 × 10&5 × 10110/20

' 8.94 Pa

*u* '8.94

413.61 % 0.5462 ' 25 mm/s

Problem 1.17

(a) The sound pressure level at 10m will be 20log10(10/1) less than at 1m,which translates to 90dB.

(b) Spherical wave solution to the wave equation is:

Using equation 1.7 in the text, the velocity potential is:


Thus:

ρc = 1.206 × 343 = 413.7; k = 2πf/c = 2π × 100/343 = 1.832At 1m, 1/kr = 0.546; at 10m, 1/kr = 0.0546. The acoustic pressureamplitude at 1m is:

Thus at 1m, the particle velocity amplitude is:

The acoustic pressure amplitude at 10m is:

Fundamentals 13

*p* ' 2 × pref × 10Lp /20

' 2 × 2 × 10&5 × 1090/20

' 0.894 Pa

*u* '0.894413.6

1 % 0.05462 ' 2.2 mm/s

p 'Ar

e j(ωt & kr)

φ 'A

jωρre j(ωt & kr)

u ' &MφMr

'jkA

jrωρe j(ωt & kr) %

A

jr 2ωρe j(ωt & kr)

'A

rρce j(ωt & kr) 1 &

jkr

'pρc

1 &j

kr

Z ' ρc 1 &j

kr

&1

*Z* ' ρckr 1 % k 2 r 2

1 % k 2r 2'

ρckr

1 % k 2 r 2' ρc /2

4k 2 r 2 ' 1 % k 2 r 2 or k 2 r 2 ' 0.3333

And at 10m, the particle velocity amplitude is:

Problem 1.18

(a) Spherical wave solution to the wave equation is:

Using equation 1.7 in the text, the velocity potential is:


(b) Specific acoustic impedance, Z = p/u. Thus:

(c) The modulus of the impedance of the spherical wave is half that of aplane wave (ρc) when

Thus:


Thus, r ' (λ /2π) 0.3333 ' 0.092λ

Problem 1.19

(a) r.m.s. sound pressure

prms ' Iρc 'Wρc

4πr 2'

1 × 1.206 × 343

4π × 0.32' 19.12Pa

(b) SPL = 20log1019.1

2 × 10&5' 119.6dB

(c) r.m.s. particle velocity (see equation 1.43 in text):

*ur* '*pr* 1 % k 2r 2

krρc'

19.025 1 % k 2r 2 1/2

413.6kr

kr '2π × 1000

343× 0.3 ' 5.50

Thus, u '19.125[1 % 5.502 ]1/2

413.7 × 5.50' 0.047m/s

(d) Phase between pressure and particle velocity given by equation 1.73 as:, and the acoustic pressureβ ' tan&1[1/(kr)] ' tan&1(1/5.50) ' 10.3E

leads the particle velocity.

(e) As shown on p35 in the text, spherical wave intensity is .p 2rms / (ρc)

Substituting in the value for prms calculated in (a) above gives:I ' 19.1252 / (1.205 × 343) ' 0.885 W/m 2

(f) From equation 1.74 in the text, the amplitude of the reactive intensity is:

Ir 'p 2

rms

ρckr'

19.1252

1.205 × 343 × 5.50' 0.161 W/m 2

Fundamentals 15

Loct ' 10log10 107.8 % 107.3 % 108 ' 80.8dB

p 2rms '

1.206 × 3434π

10100

%20

200%

15100

' 11.52Pa 2

tanθ '0.1A (1 % 1/ 2)

0.1A (1.5 % 1/ 2' 0.7734

A B

CD

10 m

10 m

(g) The sound intensity level is:

LI ' 10log10 0.885 % 120 ' 119.5dB

(h) 1500Hz is a different frequency to 1000Hz so the mean square pressuresadd.

Thus, .prms ' 2 × 19.125 ' 27.0Pa

Problem 1.20

Problem 1.21

(a) For a spherical source:

.p 2rms ' ρcI ' ρcW /S ' ρcW /4πr 2

As the sources are uncorrelated we mayadd p2 for each. For source A, r = 10.For source B, r = 10/2. For source C, r = 10.

Thus the total p2 at location D is:

Thus, the sound pressure level is:

Lp = 10log1011.52

(2 × 10&5)2' 104.6 dB

(b) Intensity, I % p2. Thus, I % W/r2 = AW/r2. The resultant intensity can thusbe calculated using the figure on the next page.


45o

45o

45o

0.1A

0.1A

0.1A

0.1A0.15A

2

2

Thus the direction of theintensity vector is 37.7Ebelow the horizontal.

(c) Sound intensity is ameasure of the net flow ofenergy in an acousticdisturbance. For a singlefrequency field the soundintensity is the product ofthe acoustic pressure withthe in-phase component ofthe particle velocity. For abroadband field it is the time average product of the acoustic pressureand particle velocity. Thus its measurement requires a knowledge of theacoustic pressure and particle velocity. Equations 1.6 and 1.7 in the textindicate that the particle velocity is proportional to the pressure gradientwhich can be approximated by subtracting the measurements from twomicrophones and dividing the result by the separation distance. In the farfield of the source, the pressure and particle velocity are in phase andrelated by p = ρcu. Thus, in the far field, only one microphone isnecessary as the pressure gradient need not be calculated.

(d) For the determination of sound power, sound intensity measurements canstill give accurate results in the presence of reflecting surfaces or othernearby noisy equipment, whereas results obtained using sound pressuremeasurements are likely to be seriously in error.

Problem 1.22

(a) Specific acoustic impedance is the ratio of acoustic pressure to particlevelocity at any location in the medium containing the acousticdisturbance. It is a function of the type of disturbance as well as of theacoustic medium.

(b) Characteristic impedance is a material or acoustic medium property, isequal to ρc and is the specific acoustic impedance of a plane wave in aninfinitely extending medium.

Fundamentals 17

10log10 1098/10 & 1095/10 ' 95.0dB

10log10 10102/10 & 1098/10 ' 99.8dB

10log10 1096/10 & 1094/10 ' 91.7dB

(c) Interference describes the interaction between two or more sound wavesof the same frequency such that regions of reinforcement (increasedsound pressure) and regions of cancellation (reduced sound pressure) areformed.

(d) Phase speed is the speed at which a single frequency sound wavepropagates and is proportional to the rate of change of phase experiencedby a stationary observer as the sound wave propagates past.

(e) Sound power is a measure of the total rate of energy emission by anacoustic source and has the units of watts.

(f) Particle velocity describe the oscillatory motion of particles in an acousticmedium during propagation of an acoustic disturbance.

Problem 1.23

A flat spectrum level implies equal energy in each 1Hz wide frequency band.As the bandwidth of an octave band doubles from one band to the next, theenergy level will increase by 3dB each time the octave band centre frequencyis doubled.

Similarly, the level will increase by 1dB for one third octave bands, each timethe band centre frequency is stepped up.

Problem 1.24

(a) The noise levels due to the machine only at each of the three locationsare:

The Leq at 500Hz is:

10log10 1095/10 % 1099.8/10 % 1091.7/10 ' 101.5dB


p 21 ' p 2

2 ' p 2ref 10

Lp/10and p 2

t ' 4 × p 21

p1 ' p2 ' pref 1085/20

p 2t ' p 2

ref 108.5 % 108.5 % 2×108.5× cos30

Lpt' 10log10

p 2t

p 2ref

' 90.7dB

dB reduction ' &10log1013

10&1.5 % 10&2 % 10&2.3 ' 18.1 dB

(b)

Problem 1.25

Use equation 1.90 in text, 3rd edn. as the two signals will be coherent.

If the two speakers operate together, the phase difference between the twosignals is zero as the speakers are identical and driven by the same amplifier.Thus cosθ = 1.

Thus, Lpt' Lp1

% 10log10 4 ' 85 % 6 ' 91dB

If a 45E phase shift were introduced, then cosθ = 0.707 and

.p 2t ' (2 % 2 × 0.707)p 2

1 ' 3.414p 21

Thus: .Lpt' Lp1

% 10log10 3.414 ' 85 % 5.3 ' 90.3dB

Problem 1.26

Use equation 1.90 in text, as the signals to be added are coherent.

For signals 180E out of phase, cos180E = -1 and thus pt2 = 0 which means that

Lpt = - 4. In practice the measured level would be greater than this due toelectronic instrumentation noise.

Fundamentals 19

prms ' A 2 /2 % A 2 /2 % A 2 cosφ ' A (1 % cosφ)

ΔL ' 10log10

prms

p )

rms

2

' 10log10 (2 % 2cosφ) ' 4.8dB

β2 & β1 '2πfc

c×

c2fc

' π

¢p 2t ¦ ' ¢p 2

1 ¦ % ¢p 22 ¦ & 2 ¢p1 p2 ¦ ' ¢ (p1 & p2 )2 ¦, p1 > p2

Problem 1.27

(a) Each signal has an amplitude of A, an r.m.s. value of A//2 and therelative phase between them is φ radians.

(b) As the signals are at the same frequency and are shifted in phase by aconstant amount, they are coherent, so equation 1.90 in the text is usedto add them together. Therefore the r.m.s. value of the combined signalis given by:

With just a single source, . The difference in dB betweenp )

rms ' A / 2

the two is thus:

Thus for a value of φ = 60E, ΔL = 4.8dB

Problem 1.28

(a) As the waves are from the same source, one may be described by

and the other by , where x1 and x2 arep1 ' P1ej (ωt & kx1)

p2 ' P2ej (ωt & kx2)

the path lengths of the two waves. The phase difference is thus

, where fc is the centre frequency ofβ2 & β1 ' k(x1 & x2) '2πfc

c(x1 & x2)

the band of noise.

(b) If (x1 - x2) = λ/2 = c/2fc, the phase difference is:

Substituting this result into equation 1.90 in the text gives:


β2 & β1 '2πfc

c×

cfc

' 2π

¢p 2t ¦ ' ¢p 2

1 ¦ % ¢p 22 ¦ % 2 ¢p1 p2 ¦ ' ¢ (p1 % p2 )2 ¦

¢p 2t ¦ ' ¢p 2

1 ¦ % ¢p 22 ¦ % 2 ¢p1 p2 ¦ 1

2πm2π

0

cosα dα

' ¢p 21 ¦ % ¢p 2

2 ¦ % 2 ¢p1 p2 ¦ 12π

sinα 2π0

' ¢p 21 ¦ % ¢p 2

2 ¦

¢p 2t ¦ ' ¢p 2

1 ¦ % ¢p 22 ¦ % 2 ¢p1 p2 ¦cos(β1 & β2)

¢p 2t ¦ ' ¢p 2

1 ¦ % ¢p 22 ¦ % 2 ¢p1 p2 ¦

If (x1 - x2) = λ = c/fc, the phase difference is:

Substituting this result into equation 1.90 in the text gives:

(c) If all phases are present the total pressure is given by equation 1.90averaged over all phases. Thus:

which is the result for the incoherent case.

Problem 1.29

Following example 1.4 on p49 in the text, the level due to the "first" signalalone is given by .10log10 107.5 & 106.9 ' 73.7dB

Problem 1.30

(a) The phenomenon is the superposition of acoustic waves of the same

frequency and fixed phase. Thus the total pressure, , is given by:¢p 2t ¦

If the phase difference between the two waves is zero, then:

Fundamentals 21

¢p 2t ¦ ' ¢p 2

1 ¦ % ¢p 22 ¦ & 2 ¢p1 p2 ¦

20log10

2p1

2 × 10&5' 60

20log10

p1

2 × 10&5' 60 & 6 ' 54dB

10log10 109.7 & 109.4 . 94dB

If p1 = p2, then ¢p 2t ¦ ' 4¢p 2

1 ¦

When the two waves are 180E out of phase:

If p1 = p2, then ¢p 2t ¦ ' 0

The noise level could be reduced substantially by using a control systemwhich ensured that when one pump was turned on, the other was turnedon at such a time that it was 180E out of phase with the first pump.

(b) When the problem is noticed, the sound pressure level at the house is60dB. This would occur when the two pumps are in phase. Thus:

If one pump only were operating, then the level should be:

If the level which is measured with one pump operating is closer to 54dBthan 57dB, then the theory of in-phase addition of sound waves would beverified. However if the level with one pump operating were closer to57dB, then incoherent addition would be suggested and the problemwould need further investigation.

Problem 1.31

Following example 1.4 in the text, the level due to the machine alone is equalto:

Thus the machine is in compliance with specifications.


10log10 109.5 & 109.1 . 92.8dB

NR ' 10log10 [100/10 % 10&5/10]

& 10log10[10&8/10 % 10&13/10 % 10&13/10 % 10&8/10 % 2(10&18/10 % 10&12/10 )]' 1.2 & (&2.4) ' 3.6dB

Problem 1.32

Following example 1.4 in the text, the level due to the machine alone is equalto:

Problem 1.33

See pages 49 and 50 in text, "combining level reductions". The differencelevel with the barrier removed can be calculated by adding the barrier noisereduction to 60dB(A). The noise reduction is calculated using equation 1.97and is:

Thus the level with the barrier removed is 60 + 3.6 = 63.6dB.

Problem 1.34

(a) Level at the receiver due to both waves = 75dB.Reflected signal has suffered a 5dB loss.Let the signal due to the direct wave = xdB. Then:75 ' 10log10 10x /10 % 10(x & 5) /10

or 75 ' 10log10 10x /10 % 10log10 1 % 10(& 5/10)

Thus, x = 75 - 1.2 = 73.8dB

(b) Contributions to the total level from various paths are:Path A: 73.8 - 4 - 7 = 62.8Path B: 73.8 - 5 - 5 = 63.8Path C: 73.8 - 4 = 69.8Sound pressure level at receiver = 10 log10 106.28 % 106.38 % 106.98 ' 71.4dB

(c) The answer can be found by calculating what the direct field contribution

Fundamentals 23

utot ' &1ρ

MMx mp dt ' &

Ajωρ

MMx

e jωt e&jkx % 0.25e jkx

'kAωρ

e jωt e&jkx & 0.25e jkx

I '12

Rep u ( '12

Re A e&jkx % 0.25e jkx kAωρ

e jkx & 0.25e&jkx

I '12

Re kA 2

ωρ1.0 & 0.252 % 0.25e2jkx & 0.25e&2jkx ' 0.47 kA 2

ωρ

is, using the level with the barrier in place (note incoherent addition withthe barrier in place). Let the direct sound field = xdB. The total levelwith the barrier in place is 70dB. Thus:

70 ' 10log10 10(x & 11) /10 % 10(x & 10))10 % 10(x & 4) /10

or

70 ' 10log10 10x /10 % 10log10 10&11/10 % 10&10/10 % 10&4/10

Thus, x = 70.0 + 2.4 = 72.4dBReduction due to destructive interference = 72.4 - 65 = 7.4dB

Problem 1.35

The positive going wave may be represented as and thepi ' Ae j(ωt & kx)

negative going wave as . At x = 0, the phase between the two(A/4)e j(ωt % kx % θ)

waves is 0. Thus, θ = 0. The total pressure field may then be written as:ptot ' Ae j(ωt & kx) % (A/4)e j(ωt % kx) ' Ae jωt e&jkx % 0.25e jkx

.Combining equations 1.6 and 1.7 in the text gives for the particle velocity:

The active acoustic intensity is then:

The preceding equation may be rearranged to give:

Alternatively, the intensity of each of the two waves could have beencalculated separately and combined vectorially. That is, for the positive going


Ii '¢p 2

i ¦ρc

'A 2

2ρc

Ii '¢p 2

r ¦ρc

'A 2

42 ×2ρc

Itot ' Ii & Ir 'A 2

2ρc&

A 2

32ρc' 0.47 A 2

ρc' 0.47 kA 2

ωρ

plane wave:

and for the negative going plane wave:

The total intensity is then:

If the two waves had the same amplitude it is clear from the precedingequations that the active intensity would be zero.

Problem 1.36

(a) Higher order mode cut-on frequency is: fco = 0.586c/d, where d is the tubediameter (see p456 in text).Thus fco = 0.586 × 343/0.05 = 4020HzFrequency range for plane waves = 0 to 4020Hz.

(b) For plane waves, the acoustic power is:

, where S is the duct cross sectional area.W ' I S 'S ¢p 2 ¦ρc

' ρcS ¢u 2 ¦¢u 2 ¦ ' ω2ξ2

0 /2

' (2π × 500 × 0.0001)2 /2 ' 0.04935(m/s)2

ρc = 1.206 × 343 = 413.7, S = (π/4) × 0.052

Thus, W ' 413.7 × 1.964 × 10&3 × 0.04935 ' 0.04watts

(c) As power is proportional to the square of the cone velocity, the conevelocity squared should be kept constant which means that thedisplacement squared of the speaker cone should vary inversely withfrequency. That is, the displacement should vary inversely with thesquare root of the frequency.

Fundamentals 25

u(x , t) ' &1

jωρMp(x, t)Mx

'p(x, t)ρc

ξ(x , t) '1ρc

5500

ej(500t & k1x & π /2)

%3

200e

j(200t & k2x & π /2)

'1

343×1.2065

500e j(500t & 2500/343 & π /2)

%3

200e j(200t & 1000/343 & π /2)

'1

413.71

100e j(500t & 7.29 & π /2)

%3

200e j(200t & 2.91 & π /2)

Problem 1.37

(a) The acoustic pressure is given by:

p(x , t) ' 5ej(500t & k1x)

% 3ej(200t & k2x)

The particle velocity can be obtained using equation 1.7 and the onedimensional form of equation 1.6 as follows:

Using the above expression for acoustic pressure and k1 = 500/343 k2 =200/343 and x = 5, the acoustic particle velocity can be written as:

u(x , t) '1ρc

5ej(500t & k1x)

% 3ej(200t & k2x)

'1

343×1.2065e j(500t & 2500/343) % 3e j(200t & 1000/343)

'1

413.75e j(500t & 7.29) % 3e j(200t & 2.92)

The displacement is given by . Thus: ξ(x , t) ' mu dt


u(x , t) ' &1

jωρ

M(pR(x, t) & pL(x, t))

Mx'

pR(x, t) & pL(x, t)

ρc

(b) r.m.s. values are:

urms '1

2× 1

413.652 % 32 1/2

' 0.010m/s

ξrms '1

2× 1

413.61/100 2 % 3/200 2 1/2

' 3.1 ×10&5 m

(c) Active intensity. As we have a plane wave propagating in only onedirection, the sound intensity is given by equation 1.70 in the text. Alsonote that the mean square pressures for two different frequencies add.Thus:

Ia 'p 2

rms

ρc' 0.5 52 % 32

413.7' 0.041 W/m 2

(d) Reactive intensity. This is undefined as we have more than onefrequency component in the wave.

(e) The acoustic pressure for each wave is given by:

pR(x , t) ' 5ej(500t & k1x)

% 3ej(200t & k2x)

pL(x , t) ' 4ej(500t % k1x)

% 2ej(200t % k2x)

Total acoustic pressure, p = pR + pL. Thus:

p = + 5ej(500t & k1x)

% 3ej(200t & k2x)

4ej(500t % k1x)

% 2ej(200t % k2x)

The total acoustic particle velocity is then:

Thus:

u(x , t) '1

1.206 × 3435e

j(500t & k1x)% 3e

j(200t & k2x)

& 4ej(500t % k1x)

& 2ej(200t % k2x)

Fundamentals 27

I '0.5

1.206 × 343Re 5e

&jk1x% 4e

jk1x5e

jk1x& 4e

&jk1x

%0.5

1.206 × 343Re 3e

&jk2x% 2e

jk2x3e

jk2x& 2e

&jk2x

Z 'p(x, t)u(x, t)

' ρcAe j(ωt & kx)

Ae j(ωt & kx)' ρc

x0

Z = p uT T/

-L

The active intensity is given by I = 0.5Re where the bar denotesp u (

the complex amplitude. The amplitude of the reactive component is notdefined as there is more than one frequency present. The active intensityis calculated at each frequency and the results added together as follows:

Thus, I = 1.20 × 10-3 (9 + 5) = 0.017 W/m2

Problem 1.38

(a) The acoustic pressure may be written as . Usingp(x, t) ' Ae j(ωt & kx)

equations 1.6 and 1.7 in the text, the acoustic particle velocity may be

written as: .u(x, t) 'kAρω

e j(ωt & kx) 'Aρc

e j(ωt & kx)

(b) Particle velocity is the magnitude of the motion of the particles disturbedduring the passage of an acoustic wave, whereas the speed of soundrefers to the speed at which the disturbance propagates. Acoustic particlevelocity is a function of the loudness of the noise, whereas the speed ofsound is independent of loudness.

(c) The specific acoustic impedance is the ratio of acoustic pressure toparticle velocity. Using the preceding equations we obtain:

(d)


pT ' A e j(ωt & kx) % e j(ωt % kx)

uT 'Aρc

e j(ωt & kx) & e j(ωt % kx)

Zρc

'e&j kx % e j kx

e&j kx & e j kx'

cos(kx)&jsin(kx)

' jcot(kx)

pT ' A e j(ωt & kx) % e j(ωt % kx % 2kL)

x0

Z = -p uT T/

-L

rigid terminationspeaker

To simplify the algebra, set the origin of the coordinate system at the rigid endof the tube as shown in the figure. As the tube is terminated non-anechoically,the pressure will include a contribution from the reflected wave. For a rigidtermination, the phase shift on reflection is 0E and the amplitude of thereflected wave is equal to the amplitude of the incident wave. As the origin,x = 0 is at the point of reflection, the phase of the two waves must be the samewhen x = 0. Of course if the origin were elsewhere, this would not be true andthe following expressions would have to include an additional term (equal tothe distance from the origin to the point of reflection) in the exponent of thereflected wave. With the origin at the point of reflection, the total acousticpressure and particle velocity at any point in the tube may be written as:

and

The specific acoustic impedance is then:

Problem 1.39

The coordinate system is asshown in the figure at right.

(a) For a rigid termination,the phase shift onreflection is 0E and theamplitude of thereflected wave is equal to the amplitude of the incident wave. As theorigin, x = 0 is at the loudspeaker location, the phase of the two wavesmust be the same when x = -L. Thus, the total acoustic pressure at anypoint in the tube may be written as:

The velocity potential and acoustic particle velocity may be derived from

Fundamentals 29

φT 'A

jρωe j(ωt & kx) % e j(ωt % kx % 2kL)

uT 'Aρc

e j(ωt & kx) & e j(ωt % kx % 2kL)

pT 'U0 ρc

1 & e j2kLe j(ωt & kx) % e j(ωt % kx % 2kL)

uT 'U0

1 & e j2kLe j(ωt & kx) & e j(ωt % kx % 2kL)

I '12

Re pT u (

T

' ρcU 20 Re

e&j kx % e j(kx % 2kL) × e j kx & e&j(kx % 2kL)

2 1 & e j2kL × 1 & e&j2kL

' ρcU 20 Re

e&j kx % e j(kx % 2kL) × e j kx & e&j(kx % 2kL)

2 2 & e j2kL & e&2jkL

'ρcU 2

0

2 2 & 2cos(2kL)Re 1 & 1 % e 2jk(x % L) & e&2jk(x % L) ' 0

the above expression as:

(b) At x = 0, uT = U0ejωt,

thus and so U0 'Aρc

1 & e j2kL A 'U0ρc

1 & e j2kL

(c) Rewriting the expressions of (a) in terms of U0, we obtain:

and

The real part of the acoustic intensity (where the bar denotes the complexamplitude which is time independent) is:

The amplitude of the imaginary part of the acoustic intensity can bederived in a similar way and from the last line in the above equation, it


I '12

Im pT u (

T

'ρcU 2

0

2 & 2cos(2kL)sin[2k(x % L)]

pi ' Ae j(ωt & kx)

and pr ' Be j(ωt % kx % θ)

pT ' Ae j(ωt & kx) % Be j(ωt % kx % θ)

x

speaker1

speaker2

0

is:

(d) The acoustic intensity is a vector quantity and as the amplitudes of thetwo waves travelling in opposite directions are the same, their intensitieswill vectorially add to zero.

Problem 1.40

(a) Incident wave and reflected wave pressures may be written as:

The total pressure is thus:

The maximum pressure amplitude occurs when the left and right goingwaves are in-phase which is at location x, such that the phase, θ = -2kx,giving a pressure amplitude of (A + B). The minimum pressure amplitudeoccurs when the left and right going waves are π radians out of phase, atthe location x, such that the phase θ = -2kx + π, with a correspondingamplitude of A - B. Thus, the ratio of maximum to minimum pressure is

Fundamentals 31

π & 2kxmin ' &2kxmax

xmin & xmax 'π2k

'λ4

'c4f

f 'c

4(xmin & xmax)'

3434(0.09 &0.03)

' 1430 Hz

uT '1ρc

(pi & pr)

uT '1ρc

Ae j(ωt & kx) & Be j(ωt % kx % θ)

uT '1ρc

A & Be jθ

θ ' &0.06k ' &0.06 π2(xmin & xmax)

' &π /2

* uT* '1

413.7A 2 % B 2 1/2

(A + B)/(A - B) and the standing wave ratio is 20log10[(A %B) / (A & B)]

The location of the minimum closest to the end where x = 0 (left end) is0.09m. Thus θ = -0.18k + π. As θ is a constant:

So:

and thus:

(b) The total particle velocity can be calculated using equations 1.6 and 1.7in the text as:

Thus:

The complex particle velocity amplitude at x = 0 is then:

The phase angle θ is given by part (a) as:

Using the above 2 equations, the particle velocity amplitude can bewritten as:

The standing wave ratio is given by:


* A % BA & B

* ' 10(100 & 96.5) /20 ' 1.496

A % B ' 2 × 10&5 × 1.414 × 10100/20 ' 2.828

Zm ' ρcS A % Be j(2kL % θ)

A & Be j(2kL % θ)

Zm ' 413.7 × 0.001 ×5.032 % ejπ /2

5.032 & ejπ /2' 0.4136 ×

5.032 % j5.032 & j

' 0.01571 × (5.032 % j)2 ' 0.382 % 0.158j

Thus, A = 5.03B. However, the maximum pressure amplitude is A + B.So:

From the preceding two equations, we have A = 2.36 and B = 0.469.Thus the velocity amplitude at x = 0 is:

.* uT* '1

413.72.362 % 0.4692 1/2

' 5.8mm/s

Thus the volume velocity amplitude is 5.9 × 10-6 m3/s (0.001m2 area)which is equivalent to an r.m.s volume velocity of4.2 × 10-6 m3/s.

(c) The mechanical impedance of the second loudspeaker is given by thecross-sectional area multiplied by the ratio of the pressure and particlevelocity at the surface of the loudspeaker. Thus Zm = pS/u. Using therelationships derived in parts (a) and (b), we have:

We previously found that θ = -π/2 and k = -θ/0.06. Also L = 0.3 and 2kL= (2π/0.12)×0.3 = 5π. Thus Zm may be written as:

Problem 1.41

(a) It is sufficient to show that adding two waves of the same frequency butshifted in phase will give a third wave of the same frequency but shifted

Fundamentals 33

p1 ' A1ejωt and p2 ' A2e

jωt % β

p1 ' A1cosωt and p2 ' A2cos(ω % Δω)t

t

p2

p3

p1

Im

Re

in phase. Assuming plane wave propagation, let the two waves to beadded be described as:

and the third wave as . To find A3 and θ in terms of A1, A2p3 ' A3ejωt % θ

and β, it is easiest to express p1, p2 and p3 as rotating vectors and use thecosine rule as shown in the figure.

From the cosine rule:

A 23 ' A 2

1 % A 22 % 2A1A2 cosβ

which is the same as equation 1.90 in the text.

The phase angle θ is:

θ ' tan&1A2 sinβ

A1 % A2 cosβ

(b) It is sufficient to show that adding together two plane waves of slightlydifferent frequency with the same amplitude result in a third wave. Letthe two waves to be added be described as:

The sum of the two pressures written above may be expressed as:


p1 % p2 ' A(cosωt % cos(ω % Δω)t)

' 2A cost2

(ω % ω % Δω) cost2

(ω & ω & Δω)

' 2AcosΔω2

t cos ω %Δω2

t

pi ' Ae j(ωt % kx) and pr ' Be j(ωt & kx % θ)

pT ' Ae j(ωt % kx) % Be j(ωt & kx % θ)

x0

Z = -p /Sua T T

L

which is a sine wave of frequency (ω + Δω) modulated by a frequencyΔω/2

(c) If Δω is small we obtain the familiar beating phenomenon (see page 46,fig 1.9 in text which shows a beating phenomenon where the two wavesare slightly different in amplitude resulting in incomplete cancellation atthe null points).

Problem 1.42

(a)

As the origin is at the left end of the tube, the incident wave will be travellingin the negative x direction. Assuming a phase shift between the incident andreflected waves of θ at x = 0, the incident wave and reflected wave pressuresmay be written as:


The total particle velocity can be calculated using equations 1.6 and 1.7in the text as:

Fundamentals 35

uT '1ρc

(pr & pi)

uT '1ρc

Be j(ωt & kx % θ) & Ae j(ωt % kx)

P0 ' A % Be jθ and ρcU0 ' Be jθ & A

A ' 0.5(P0 & ρcU0 ) and Bejθ ' 0.5(P0 % ρcU0 )

pT ' 0.5(P0 & ρcU0 )e j(ωt % kx) % 0.5(P0 % ρcU0 )e j(ωt & kx)

uT '0.5ρc

(P0 % ρcU0 )e j(ωt & kx) & (P0 & ρcU0 )e j(ωt % kx)

Za ' &pT

SuT

' &ρcS

(P0 & ρcU0 )e jkx % (P0 % ρcU0 )e&jkx

(P0 % ρcU0 )e&jkx & (P0 & ρcU0 )e jkx

Za 'ρcS

jρcU0 sin(kx) & P0cos(kx)

ρcU0 cos(kx) & jP0sin(kx)

Thus:

At x = 0, pT = P0ejωt and uT = U0e

jωt. Thus:

Thus:

and the total acoustic pressure and particle velocity may be written as:

and

The acoustic impedance looking towards the left in the negative x-direction is the negative ratio of the total acoustic pressure to the productof the duct cross-sectional area and the total acoustic particle velocity(see the preceding figure). Thus:

As , theejkx ' cos(kx) % jsin(kx) and e&jkx ' cos(kx) & jsin(kx)impedance may be written as:


Za 'ρcS

P0cos(kx) & jρcU0 sin(kx)


Za 'ρcS

Z0cos(kL) % jρc sin(kL)

ρc cos(kL) % jZ0sin(kL)

2m

(b)

Using a similar analysis to that outlined above, the following expressionis obtained for the acoustic impedance looking to the right in the positivex direction:

The expressions in parts (a) and (b) for the impedance can be shown tobe equal if evaluated at the open end of the tube (x = L in part (a) and x= -L in part (b). In addition, the quantity -P0/U0 is equal to thetermination impedance Z0 in part (a), while in part (b), the quantity P0/U0

is equal to the termination impedance Z0. Making the appropriatesubstitutions, both expressions give the following for the impedance atthe open end of the tube.

Problem 1.43

(a)

Set the origin, x = 0 at the surface of the sample. Then the entrance of thetube is at x = -2.0. As the origin is at the surface of the sample, the incidentwave will be travelling in the positive x direction. Assuming a phase shiftbetween the incident and reflected waves of θ at x = 0, the incident wave and

Fundamentals 37

pi ' Ae j(ωt & kx) and pr ' Be j(ωt % kx % θ)


uT '1ρc

Ae j(ωt & kx) & Be j(ωt % kx % θ)

Z 'pT

uT

' ρcAe& j kx % Be j kx % jθ

Ae& j kx & Be j kx % jθ' ρc

A % Bej(2kx % θ)

A & Bej(2kx % θ)

' ρcA /B % cos(2kx % θ) % jsin(2kx % θ)A /B & cos(2kx % θ) &jsin(2kx % θ)

Rp ' (B /A)ejθ ' 0.5 % 0.5j

reflected wave pressures may be written as:



The pressure reflection coefficient, Rp, is defined as pr /pi. Thus:

The reflection coefficient amplitude is B/A and the phase is θ. Thus, B/A= 0.707 and θ = 45E = 0.7854 radians.

Thus the specific acoustic impedance at any point in the tube may bewritten as:

= 1.832 and x = -2. Thus, 2kx + θ = -6.542.k ' ω /c ' 2π × 100/343

. cos(2kx % θ) ' 0.9667 and sin(2kx % θ) ' &0.25587 and A /B ' 1.414Thus, Z = 414 × (2.3807 - j0.25587)/(0.4475 + j0.25587)

= 1558(1 - j0.7237) = 1560 - j1130

(b) The absorption coefficient is defined as , so α = 0.5α ' 1 & *Rp*2




10L0 /20

'A % BA & B

x0

Z = -p uT T/

-L

speakersample

Problem 1.44

(a) To simplify the algebra, assume that the tube is horizontal with the leftend at x = 0 containing the sample of material whose absorptioncoefficient is to be determined, as shown in the figure. As the origin isat the left end of the tube, the incident wave will be travelling in thenegative x direction. Assuming a phase shift between the incident andreflected waves of θ at x = 0, the incident wave and reflected wavepressures may be written as:


The maximum pressure will occur when θ = 2kx, and the minimum willoccur when θ = 2kx + π. Thus:

and pmax ' e jkx A % B pmin ' e jkx A & B

and the ratio of maximum to minimum pressures is (A + B)/(A - B)The standing wave ratio, L0, is defined as:

Thus the ratio (B/A) is:

Fundamentals 39

BA

'10

L0 /20&1

10L0 /20

%1

*Rp*2 '

10L0 /20

&1

10L0 /20

%1

2

α ' 1 &1015/20 & 1

1015/20 % 1

2

' 0.51

uT '1ρc

(pr & pi)

uT '1ρc


Z ' &pT

uT

' &ρcAe j kx % Be&j kx % jθ

Be&j kx % jθ & Ae j kx' &ρc

A % Be j(&2kx % θ)

Be j(&2kx % θ) & A

Zs

ρc' &

pT

ρcuT

'A % Be jθ

A & Be jθ

Zs 'A /B % cosθ % jsinθA /B & cosθ & jsinθ

'(A /B)2 & 1 % (2A /B) jsinθ

(A /B)2 % 1 & (2A /B)cosθ

The amplitude of the pressure reflection coefficient squared is which can be written in terms of L0 (= 95 - 80) as:*Rp*

2 ' (B /A)2

The absorption coefficient is defined as , so:α ' 1 & *Rp*2

(b) The total particle velocity can be calculated using the equation in part (a)for the acoustic pressure and equations 1.6 and 1.7 in the text as:

Thus:


At x = 0, the specific acoustic impedance is the normal impedance, Zs, ofthe surface of the sample. Thus:

The above impedance equation may be expanded to give:


*Zs* '(A /B)2 & 1 2

% (2A /B)2 sin2θ

(A /B)2 % 1 & (2A /B)cosθ

β ' tan&1 2(A /B) sinθ

(A /B)2 & 1

*Zs*

ρc'

(1.4332 & 1)2 % (2.8662 × 0.9662)

1.4332 % 1 & 2.866 × 0.258' 1.28

β ' tan&1 2 × 1.433 × (&0.966)

1.4332 & 1' &69.2E

αst '8 cosβξ

1 &cosβξ

loge(1 % 2ξ cosβ % ξ2) %

cos(2β)ξ sinβ

tan&1 ξ sinβ1 % ξ cosβ

The modulus of the impedance is then:

and the phase is given by:

Using the previous analysis, .AB

'10

L0 /20% 1

10L0 /20

& 1' 1.433

At the pressure minimum, θ = 2kx - π, where k = 2π/λ. x = 0.2m and f = 250Hz, thus k = 2πf/c = 4.58.Thus, θ = 2 × 4.58 × 0.2 - π = -1.31 radianscosθ = 0.258 and sinθ = -0.966

Thus from the preceding equations:

and the phase is:

(c) The statistical absorption coefficient is given by equation C.37 in the textas

Substituting in the modulus and phase of the impedance, we obtain”

Fundamentals 41

αst '8 × cos(&69.17)

1.281 &

cos(&69.17)1.28

×

× loge 1 % 2 × 1.28 × cos(&69.17) % 1.282

%cos(&138.4)

1.28sin(&69.17)tan&1 1.28 × sin(&69.17)

1 % 1.28 × cos(&69.17)

' 2.22(1 & 0.278 × 1.266 % 0.625 × (&0.688)

' 2.22(1 & 0.352 & 0.430) ' 0.485

W 'S2

Re pT u (

T 'S* uT*

2

2Re Z

x0

Z = -p /uT T

L

Problem 1.45

(a)Assume a horizontal tube with the left end containing the terminationimpedance Z0 at x = 0. As the origin is at the left end of the tube, theincident wave will be travelling in the negative x direction.

The power radiated by a source at the other end of the tube is related tothe specific acoustic impedance, Z, it "sees" as follows:

where the bar represents the complex amplitude and S is the tube cross-sectional area.

Assuming a phase shift between the incident and reflected waves of θ at

x = 0, the incident wave and reflected wave pressures may be written as:




uT '1ρc

(pr & pi)

uT '1ρc


Z ' &pT

uT



A % Be j(&2kx % θ)

Be j(&2kx % θ) & A

Z0

ρc' &

P0

ρcU0

' &A % Be jθ

Be jθ & A

A ' 0.5(P0 & ρcU0 ) and Be jθ ' 0.5(P0 % ρcU0 )


uT '0.5ρc




Thus:


At x = 0, Z = Z0. For convenience also set pT = P0ejωt and uT = U0e

jωt

Thus:

Thus:


and

Fundamentals 43

Z ' &ρc(P0 & ρcU0 )e jkx % (P0 % ρcU0 )e&jkx


Z ' ρcjρcU0 sin(kx) & P0 cos(kx)


Z ' ρcZ0 /ρc % j tankL

1 % j(Z0 /ρc)tankL' ρc

(R0 % jX0 ) /ρc % j tankL

1 % j((R0 % jX0) /ρc)tankL

ReZ ' ρc(R0 /ρc) (1 % tan2kL)

(1 & X0 /ρc)2 tan2kL % (R0 /ρc)2tan2kL

W 'SρcU 2

L

2

(R0 /ρc) (1 % tan2kL)

(1 & X0 /ρc)2 tan2kL % (R0 /ρc)2tan2kL

The specific acoustic impedance looking towards the left in the negativex direction may then be written as:

As , thee jkx ' cos(kx) % jsin(kx) and e&jkx ' cos(kx) & jsin(kx)impedance may be written as:

Dividing through by ρcU0cos(kx), replacing x with L and replacing

gives:&P0

U0

with Z0

Thus:

and the power is then:

(b) It can be seen from the equation derived in part (a) that whenR0 = 0, the power will be zero.

(c) If all losses are zero, the impedance presented to the loudspeaker will begiven by the previously derived expression for Z with R0 = 0. In this case:


Z ' ρcjX0 /ρc % j tankL

1 & (X0 /ρc)tankL' jρc

X0 /ρc % tankL

1 & (X0 /ρc)tankL


Rp 'Be jθ

A

Rp 'Be jθ

A'

(P0 % ρcU0 )

(P0 & ρcU0 )

Rp '&ρc & jX0 % ρc

&ρc & jX0 & ρc'

jX0

2ρc % jX0

*Rp* ' X0 [4ρ2c 2 % X 20 ]&1/2

which is imaginary. Thus there will be no real power generated; onlyimaginary power which represents non-propagating energy stored in thenear field.

(d) As no real power is generated, the acoustic pressure and particle velocitymust be 90E out of phase.

(e) The pressure amplitude reflection coefficient is given by:

From part (a):

Thus:

Dividing numerator and denominator by U0 and putting , we obtain:Z0 ' ρc % jX0 ' &P0 /U0

Thus:

Fundamentals 45

10SWR /20 'A % BA & B



pT ' A e j(ωt % kx) % Re j(ωt & kx)

x0

Z = -p uT T/

L

samplespeaker

Problem 1.46

(a)

As given in the problem, the tube is assumed to be horizontal with the leftend at x = 0 containing the sample of material whose impedance is to bedetermined, as shown in the figure. As the origin is at the left end of thetube, the incident wave will be travelling in the negative x direction.Assuming a phase shift between the incident and reflected waves of θ atx = 0, the incident wave and reflected wave pressures may be written as:


At the surface of the sample, the pressure amplitude reflection coefficientis thus and . Thus the total pressure at anyR ' (B /A)e jθ B ' (RA)e&jθ

location, x, in the tube may be written as:

(b) Returning to the first expression for the total pressure of part (a), themaximum pressure will occur when θ = 2kx, and the minimum will occurwhen θ = 2kx + π. Thus , and thepmax ' e jkx A % B pmin ' e jkx A & B

ratio of maximum to minimum pressures is (A + B)/(A - B)

(c) The standing wave ratio (SWR) which is L0 is defined as:


BA

'10

L0 /20&1

10L0 /20

%1

*Rp*2 '

10L0 /20

&1

10L0 /20

%1

2

uT '1ρc

(pr & pi)

uT '1ρc


Z ' &pT

uT



A % Be j(&2kx % θ)

Be j(&2kx % θ) & A

Zρc

' &pT

ρcuT

'A % Be jθ

A & Be jθ'

1 % Rp

1 & Rp

Thus the ratio (B/A) is:

From part (a), the amplitude of the pressure reflection coefficient squaredis which can be written in terms of L0 as:*Rp*

2 ' (B /A)2

(d) The absorption coefficient is defined as , which onα ' 1 & *Rp*2

substituting the equation derived in part (c) for , gives the required*Rp*2

result.

(e) The total particle velocity can be calculated using equations 1.6 and 1.7in the text as:

Thus:



Fundamentals 47



uT '1ρc

(pr & pi)

uT '1ρc


Z ' &pT

uT


Be&j kx % jθ & Ae j kx

' &ρcA % Be j(&2kx % θ)

Be j(&2kx % θ) & A

x0

Z = -p /uT T

L

Problem 1.47

(a)

Assume a horizontal tube with the left end containing the terminationimpedance ZL at x = 0. As the origin is at the left end of the tube, theincident wave will be travelling in the negative x direction. Assuming aphase shift between the incident and reflected waves of θ at x = 0, theincident wave and reflected wave pressures may be written as:



Thus:



ZL

ρc' &

P0

ρcU0

' &A % Be jθ

Be jθ & A



uT '0.5ρc


Z ' &ρc(P0 & ρcU0 )e jkx % (P0 % ρcU0 )e&jkx


Z ' ρcjρcU0 sin(kx) & P0 cos(kx)


Z ' ρcZL /ρc % j tankL

1 % j(ZL /ρc)tankL

At x = 0, Z = ZL. For convenience also set pT = P0ejωt and

uT = U0ejωt Thus:

Thus:


and:

The specific acoustic impedance looking towards the left in the negativex direction may then be written as:

As , thee jkx ' cos(kx) % jsin(kx) and e&jkx ' cos(kx) & jsin(kx)impedance may be written as:

Dividing through by ρcU0cos(kx), replacing x with L and replacing

gives:&P0

U0

with ZL

The same result can be obtained by putting the open end of the tube tothe left as shown in the figure above.

Fundamentals 49


uT '1ρc

&Be j(ωt % kx % θ) % Ae j(ωt & kx)

Z 'pT

uT

' ρcAe j kx % Be&j kx % jθ

&Be&j kx % jθ % Ae j kx' ρc

A % Be j(&2kx % θ)

A & Be j(&2kx % θ)

Rp 'Be jθ

A

ZL

ρc'

A % Be jθ

A & Be jθ'

1 % Be jθ /A

1 & Be jθ /A'

1 % Rp

1 & Rp

x0

Z = p uT T/

-L

The same result can be obtained by putting the open end of the tube tothe left as shown in the figure below.


and the total particle velocity is thus:

The specific acoustic impedance at any point in the tube is then:

The remaining part of the analysis is the same as before.

(b) The reflection coefficient, Rp, is defined as the ratio of the complexamplitudes of the reflected to incident waves at x = 0. Thus, from part(a):

At x = 0:

Thus the pressure reflection coefficient may be written as:


Rp 'ZL /ρc & 1

ZL /ρc % 1

ZL ' ZH

ST

SH

ZL

ρc'

A % Be jθ

A & Be jθ'

1 % Be jθ /A

1 & Be jθ /A'

1 % Rp

1 & Rp

Rp 'ZL /ρc & 1

ZL /ρc % 1'

j(ST /SH)tankR & 1

j(ST /SH)tankR % 1

ZL ' j(ST /SH)ρc tan[kR(1 & M)] % Ra

which is equal to zero when ZL = ρc.

(c) Assuming continuity of acoustic pressure and volume velocity (uS, whereS is the tube cross sectional area) implies that the acoustic impedance iscontinuous across the hole and thus the specific impedance in the tube atx = 0 is related to the specific acoustic impedance, ZH of the hole by:

where SH is the cross sectional area of the hole and ST is the crosssectional area of the tube.

From equation 9.14 in the text we have ZH = jρctankR, where R is theeffective length of the hole. From part (a), we have at x = 0:

Thus the pressure reflection coefficient may be written as:

(d) Using equation 9.8 in the text and the condition of continuity of acousticpressure and acoustic volume velocity at the hole, the equality ZL/ST =ZH/SH holds. Thus:

(e) For good absorption, Rp = 0, which as we showed in part (b), means thatZL = ρc. For this to be true, the first term in the equation derived in (d)must be large compared to the other terms. Thus, the product kR/M mustbe much less than 1. For a thin plate, the effective length of the hole ismade up of two parts, one corresponding to the side of the hole in thetube and the other corresponding to the side looking into free space.

Fundamentals 51

Z ' ρcZL /ρc % j tankL

1 % j(ZL /ρc)tankL

Z ' ρc 4 /ρc % jkL1 % j(4 /ρc )kL

'ρcjkL

'&jρckL

pξ

'jωpu

'ρcωkL

mξ % Kξ ' f or &ω2mξ % Kξ ' f

Z ' ρc 0 % jkL1 % j0kL

' jρckL

Using equations 9.16 and 9.19 in the text, assuming that the holediameter is very small compared to the tube diameter, it can be shownthat for small M, the effective length of equation 9.7 is R = 0.73dH. Thusthe condition that kR/M << 1 implies that 2πf/c × 0.73dH << M, or fdH <<70M.

(f) The specific acoustic impedance looking into a tube was shown in part(a) to be:

In the limit of small L and large λ (small k), tan(kL) = kL. Also for arigidly terminated tube, ZL = 4. Thus the preceding expression becomes:

The ratio of the pressure to the particle displacement is then:

which indicates that the pressure is in phase with the particledisplacement. Considering the analogy of a single degree of freedomspring-mass system the equation of motion is:

where K is the spring stiffness and m is the mass. It can be seen from theabove equation that for a stiffness only, the exciting force will be inphase with the displacement and for a mass only it will be 180E out ofphase with the displacement. Thus it is clear from this analogy that theprevious case of the pressure and particle displacement in-phaserepresents a stiffness.

For an open ended tube the impedance, ZL = 0 and the impedance of thetube becomes:


pξ

'jωpu

' &ρcωkL

Z0

ρc'

Z0 /ρc % j tankL

1 % j(Z0 /ρc)tankL

Z0

ρc% j

Z0

ρc

2

tankL 'Z0

ρc% j tankL

The ratio of the pressure to the particle displacement is then:

which indicates that the acoustic pressure is 180E out of phase with theparticle displacement. Thus by the preceding argument, this representsa lumped mass.

(g) Using the equation given in the question and using the hint for maximumpower we may substitute Z0 for ZL and Z to give kL at maximum power. Thus:

Rearranging gives:

For the left side to equal the right side, tan(kL) = 0. Thus kL = nπ, andthe optimum length is given by L = nλ/2, wheren = 1, 2, 3, .......

(h) The above result suggests that the frequency response will becharacterised by a number of resonant peaks and so will be very poor andnon-uniform. The frequency response could be made smoother byadding some absorptive porous acoustic material to the tube. This wouldhave the effect of damping the resonances, thus reducing the differencebetween the peaks and troughs in the frequency response.

(i) The resonances are less damped at low frequencies, thus resulting inbigger differences between the peaks and troughs in the frequencyresponse.

Problem 1.48

Using the result derived in the answer to 1.45(a), the impedance seen by theloudspeaker may be written as:

Fundamentals 53

Z ' ρcZr /ρc % j tankL

1 % j(Zr /ρc)tankL

Zρc

'Rr /ρc % j Xr /ρc % tan(kL) 1 & (Xr /ρc)tan(kL) & j (Rr /ρc) tan(kL)

1 & (Xr /ρc)tan(kL) 2 % (Rr /ρc) tan(kL) 2

Rρc

'Rr /ρc 1 % tan2(kL)

1 & (Xr /ρc)tan(kL) 2 % (Rr /ρc) tan(kL) 2

Rρc

'1.04 × 10&3 1 % tan2(kL)

1 & 0.00486tan(kL) 2 % 1.04 × 10&3 tan(kL) 2

which on substituting Zr = Rr + jXr may be rewritten as:

The power output of the loudspeaker will vary with tube length because thepower output of a source is dependent on impedance presented to it and theimpedance it is presented is dependent on the tube length. The maximumpower output will occur when the impedance presented to the loudspeaker isequal to the internal impedance of the loudspeaker which is infinite. Thus themaximum power output will occur when the denominator in the aboveequation is zero. Of course the real power output is only dependent on theresistive impedance while the imaginary power output is dependent on thereactive impedance. Thus the specific resistive impedance is:

At 250Hz, k = 2πf/c = 2π × 250/343 = 4.58As a = 0.075, ka = 4.58 × 0.075 = 0.343, πa2 = 0.0177Thus, Rr /ρc = 0.0177 × 0.3432/2 = 1.04 × 10-3

and Xr /ρc = 0.0177 × 0.343 × 0.8 = 0.00486

Substituting in the expression given for R/ρc, we obtain:

By trial and error it can be shown that the maximum value of the aboveexpression occurs when tan(kL) = 3.636, or kL = 1.262. Thus optimum L formaximum power out = 1.262/4.58 = 276mm.

Problem 1.49

(a) A standing wave tube is used to determine the normal specific impedance




uT '1ρc

(pr & pi)

uT '1ρc


x0

Z = -p /uT T

sample

L

of a solid by placing the sample in one end of the tube which is thenrigidly closed, and then generating a pure tone sound field in the tubewith a loudspeaker placed at the other end of the tube. The ratio ofmaximum to minimum acoustic pressure in the standing wave generatedin the tube is measured, as is the distance of the first minimum from thesurface of the sample.To simplify the algebra, a horizontal tube with the left end at x = 0containing the solid whose impedance is to be determined will beassumed as shown in the figure. As the origin is at the left end of thetube, the incident wave will be travelling in the negative x direction.Assuming a phase shift between the incident and reflected waves of θ atx = 0, the incident wave and reflected wave pressures may be written as:



Thus:


Fundamentals 55

Z ' &pT

uT


Be&j kx % jθ & Ae j kx

' &ρcA % Be j(&2kx % θ)

Be j(&2kx % θ) & A

Zs

ρc' &

pT

ρcuT

'A % Be jθ

A & Be jθ

Zs

ρc'

(A /B) % e jθ

(A /B) & e jθ

10SWR /20 'A % BA & B


To calculate the impedance, it is necessary to evaluate the constants Aand B, and the phase angle, θ. Returning to the above expression for thetotal acoustic pressure, it can be seen that the maximum sound pressurein the tube will occur when θ = 2kx, and the amplitude will be (A + B).The minimum pressure amplitude occurs at the location, whereθ = 2kx - π, with a corresponding amplitude of A - B. Thus, the ratio ofmaximum to minimum pressure is (A + B)/(A - B) and the standing waveratio is ,which is the difference in dB between20log10[(A %B) / (A & B)]

the maximum and minimum sound pressure levels in the tube. The phaseangle θ is determined by the distance, x of the first minimum soundpressure level from the face of the sample by using θ = 2kx - π. This isequivalent to the equation for θ, given on p623 the text, where k = 2π/λ= 2πf/c. The first minimum is used because the effect of any losses dueto non rigid tube walls will be minimised. The minimum rather than themaximum is used because its location is much more sharply defined.

The equation for the impedance may be rewritten as:

The standing wave ratio (SWR) which can be measured is defined as:

Thus, and the impedance may be calculated.AB

'10SWR /20 % 1

10SWR /20 & 1


Zs

ρc'

A /B % cosθ % jsinθA /B & cosθ & jsinθ

'(A /B)2 & 1 % (2A /B)jsinθ

(A /B)2 % 1 & (2A /B)cosθ

*Zs*

ρc'

(A /B)2 & 1 2% (2A /B)2 sin2θ

(A /B)2 % 1 & (2A /B)cosθ

β ' tan&1 2(A /B) sinθ

(A /B)2 & 1

*Zs*

ρc'

(4.2162 & 1)2 % (8.4322 × 0.9512)

4.2162 % 1 % 8.432 × 0.309' 0.87

β ' tan&1 2 × 4.216 × 0.951

4.2162 & 1' 25.5E

This is done by expanding the above impedance equation to give:

The modulus of the impedance is then:

and the phase is given by:

(b) As can be seen in the above derivation, the amplitude of the pressure

reflection coefficient is . The sound power*Rp* 'BA

'10SWR /20 & 1

10SWR /20 % 1

reflection coefficient is defined as and the absorption coefficient*Rp*2

is given by .α ' 1 & *Rp*2

(i) From part (a), = .AB

'10SWR /20 % 1

10SWR /20 & 1

100.21 % 1

100.21 & 1' 4.216

θ = 2kx - π, where k = 2π/λ. x = 0.4λ, thus θ = 4π × 0.4 - π = 0.6πcosθ = -0.309 and sinθ = 0.951

Thus from the preceding equations:

and the phase is:

Fundamentals 57

I '12

Re pT u (

T '1

2ρcRe (A % Be jθ ) × (Be&jθ & A)

'1

2ρcRe B 2 & A 2 & 2jABsinθ '

12ρc

(B 2 & A 2 )

A % B ' 2pref 10Lp /20

' 0.0894

(ii) The normal incidence sound power reflection coefficient is given by:

*Rp*2 ' * 10SWR /20 & 1

10SWR /20 % 1*2 ' 4.216&2 ' 0.056

(iii) The absorption coefficient is given by α ' 1 & *Rp*2 ' 0.94

(iv) The intensity is given by:

The maximum sound pressure level is 70dB. Thus:

Using A = 4.216B, we have 5.216B = 0.0894. Thus B = 0.0171 andA = 0.0723. Thus the sound intensity is:

I = (2×413.7)-1 × (0.01712 - 0.07232) = -5.96×10-6 W/m2.

The negative sign indicates that the net intensity is in the negative x-direction. The intensity will not vary along a lossless tube as the soundintensity is a vector quantity which in this case is the vector sum of theintensities in the left and right going waves which is independent of tubelocation for a lossless tube. This can be verified by using the expressionsfor pT and uT derived in part (a).

(v) Let successive minima be located at x1 and x2. Then using the equationderived in part (a) for the total pressure, setting θ = 2kx, and equating the

pressures at x1 and x2 we obtain . For this to(A % B)ejkx1 ' (A % B)e

jkx2

be true, . Thus , and , whichejk(x2 & x1)

' 1 k(x2 &x1) ' &π x1 & x2 ' λ /2

implies that the minima are separated by half a wavelength. At 200Hz,this is equal to 343/(2 × 200) = 0.86m.

Problem 1.50

(a) Assuming no losses in the tube, the intensity of a single plane wave


I '1ρc

¢p 2%

¦ & ¢p 2&

¦

u ' &1ρ

MMx mp dt

propagating in any one direction is , where the amplitude of p as¢p 2 ¦ /ρcwell as its r.m.s. value is independent of axial location. Thus the

intensity of the positive going wave may be written as and the¢p 2%

¦ /ρc

negative going wave as . As intensity is a vector quantity, the¢p 2&

¦ /ρc

intensities of the positive and negative going waves can be combined byadding the intensity of the positive going wave to the negative value ofthe negative going wave to give:

which is independent of location.

(b) The intensity cannot be measured with a single microphone because itcannot distinguish between the pressures associated with the two wavecomponents travelling in opposite directions.

(c) Two identical microphones can be used because intensity is also the timeaveraged product of the acoustic pressure and particle velocity and fromequations 1.6 and 1.7 in the text we can show that the acoustic particlevelocity is related to the acoustic pressure gradient by:

where p1 and p2 are the pressures measured by microphones 1 and 2respectively. The above equation may be rewritten as equation 3.21 inthe text. Equations 3.21 and 3.22 may then be used to write equation3.23 which is an expression for the intensity as a function of themeasurements made by microphones 1 and 2. Equation 3.23 needs aslight modification as in this case (replacement of n with 1) the intensityis in the positive x-direction down the tube rather than in an arbitrarydirection n.

Fundamentals 59

Rp 'Zs & ρc

Zs % ρc

α ' 1 & *R 2p * ' 1 &

*Zs & ρc*2

*Zs % ρc*2'

*Zs % ρc*2 & *Zs & ρc*2

*Zs % ρc*2

α '[ReZs % ρc ]2 % [ ImZs ]2 & [ReZs & ρc ]2 & [ ImZs ]2

*Zs % ρc*2

'[ReZs ]2 % 2ρcReZs % (ρc)2 & [ReZs ]2% 2ρcReZs & (ρc)2

*Zs % ρc*2

'4ρcReZs

*Zs % ρc*2

Problem 1.51

(a)

and

Rearranging gives:

(b) From the above expression it can be seen that the maximum value of αis 1 which would occur when Zs = ρc.

Problem 1.52

(a) The sound power reflection coefficient is simply the square of themodulus of the pressure amplitude reflection coefficient. Referring toequation 5.129 in the text, the required result can be obtained by allowingθ = 0 (normal incidence) and substituting ρ2c2 for Zm and ρ1c1 for ρc.Note that if θ = 0, then sinθ = 0 and from equation 5.130, cosψ = 1.

(b) Again, referring to equation 5.129 in the text, the required result can beobtained by substituting ρ2c2 for Zm and ρ1c1 for ρc. The angle ψ isdefined in equation 5.130, with the substitutions, k1 = k and k2 = km.


pT ' pI % pR ' AI e jωt e&j(kxx & kyy)

& e&j(kxx % kyy)

' AI e jωt e&jkxx e

jkyy & e&jkyy

' 2jAI sinky e jωt

Rp 'ρwcw & ρc

ρwcw % ρc'

ρwcw

ρc& 1

ρwcw

ρc% 1

Problem 1.53

(a) As a result of the property of zero pressure at the pressure releaseboundary, we have at the boundary (y = 0), pR = -pI. Thus usingequations 5.118 and 5.119 in the text and setting y = 0, we obtain AR = -AI. As the wave is propagating along the y-axis, θ = 0,ky = k and kx = 0. Thus the total pressure (adding the time dependentterm, ) is given by:e jωt

(b) For normally incident sound (and with Zm = ρwcw), equation 5.129 in thetext can be written as:

From the preceding equation it can be seen that if ρwcw >> ρc, then Rp = 1and the reflected wave amplitude will be equal to the incident waveamplitude. For an air/water interface, ρwcw = 1026 × 1500 = 1.54 × 106

which is much greater than ρc = 413.7 thus satisfying the requiredcondition.

(c) Using equations 1.6.and 1.7 in the text and the result of part (a), theacoustic particle velocity amplitude is 2Ai/ρc. Using equations 1.6, 1.7and 5.118, the acoustic particle velocity amplitude (normally incidentwave) is Ai/ρc. Thus the total particle velocity amplitude is twice theincident wave particle velocity.

Problem 1.54

(a) Referring to the analysis on pages 210 – 212 in the text, for normal

Fundamentals 61

Rp '830 & ρc830 % ρc

'830 & 413.6830 % 413.6

' 0.335

Pi ' 2 × 10&5 × 1060/20 ' 0.02Pa

ρ2c2 cosθ & ρccosψ ' ρ2c2 cosθ % ρccosψ

cosψ ' 1 &c 2

2

c 2sin2θ

1/2

ρ2c2 cosθ & ρc 1 &c 2

2

c 2sin2θ

1/2

' ρ2c2 cosθ % ρc 1 &c 2

2

c 2sin2θ

1/2

incidence, θ = 0. Thus, from equation 5.130, ψ = 0, and equation 5.129becomes (with Zm = ρ2c2 = 830):

The r.m.s. amplitude of the incident wave is given by:

Thus the amplitude of the reflected wave is:

Pr ' 0.335Pi ' 0.0067Pa

(b) When all the energy is reflected, , Thus:*Rp* ' 1

Using Snell's Law, . Thus:sinψ 'c2 sinθ

c

Substituting this result into the previous equation gives:

This is true only if (c2/c)sinθ = 1, or θ = sin-1(c/c2) which is the angle abovewhich all incident energy will be reflected.

Problem 1.55

(a) Referring to the analysis on pages 210 – 212 in the text and substituting


pT

pI

'AT

AI

'2cosθ / (ρc)

cosθρc

%cosψρwcw

'2ρwcwcosθ

ρwcwcosθ % ρccosψ

τ '4ρcρwcwcos2θ

(ρwcwcosθ % ρccosψ )2' (1 & R 2

p )cosθ /cosψ

c

w wc

y

x

rRrI

rT

incident wave reflected wave

transmittedwave

air

water

ρwcw for Zm in equation 5.129, the required result is obtained.

(b) The transmission coefficient, τ, is an energy related quantity and isdefined by:

τ '*pT*

2

*pI*2

ρcρwcw

'*AT*

2

*AI*2

ρcρwcw

Using equation 5.121 and 5.127 in the text and substituting Z1 = ρc and Z2 =ρwcw, we obtain:

Thus:

(c) For air, ρc = 413.6 and for sea waterρwcw = 1.026 × 103 × 1500 = 1539 × 103.For θ = 10E, cosθ = 0.9848 and sinθ = 0.1736.Using Snell's Law,

Fundamentals 63

Rp '1.539 × 106 × 0.9848 & 413.6 × 0.6506

1.539 × 106 × 0.9848 % 413.6 × 0.6506' 0.99964

τ ' (1 & R 2p )cosθ /cosψ ' 0.0011

ρwcwcosθ & ρccosψ ' ρwcwcosθ % ρccosψ

cosψ ' 1 &c 2

w

c 2sin2θ

1/2

ρwcwcosθ & ρc 1 &c 2

w

c 2sin2θ

1/2

' ρwcwcosθ % ρc 1 &c 2

w

c 2sin2θ

1/2

. Thus ψ = 49.4E andsinψ 'cwsinθ

c' 1500 × 0.1736/343 ' 0.7594

cosψ = 0.6506.

Substituting these values into the result of part (a) gives:

(d) When all the energy is reflected, , Thus *Rp* ' 1

Using Snell's Law, . Thus,sinψ 'cwsinθ

c

Substituting this result into the previous equation gives

This is true only if (cw/c)sinθ = 1, or θ = 13E, which is the angle abovewhich all incident energy will be reflected.

(e) If the sound source were a point source, the amount of sound powerentering the water would be independent of altitude (except foratmospheric losses) as the same amount of power is contained in anyspecified included angle. However, the power would be distributed overan ever increasing area of ocean surface. A distributed sound source


uI 'AI

ρc( sinθ & cosθ )e

&jk1x x

uR 'AR

ρc( sinθ % cosθ )e

&jk1x x

uR

uI

'AR

AI

( sinθ % cosθ )( sinθ & cosθ )

pu

' ρcjkr

1 % jkr

pu

. ρcjkr

p ' ρc0ukrω

' ρ 0ur

such as a helicopter would exhibit similar behaviour.

(f) Velocity reflection coefficient.Using equations 1.6, 1.7, 5.115, 5.116, 5.118 and 5.119 in the text, theacoustic velocity amplitude of the incident wave at y = 0 is:

and the velocity amplitude of the reflected wave is:

The velocity amplitude reflection coefficient is the ratio of reflected toincident velocity amplitudes. Thus:

Problem 1.56

(a) Assuming that the bubble is a spherical source, the specific acousticimpedance is given by equation 1.43 in the text as:

At the bubble surface, kr << 1, so:

Acceleration is defined as , so:0u ' jωu

Fundamentals 65

γ dVV

%dPP

' 0

pP

' &γdVV

ρ 0urP

' &γdVV

0u ' &γPρr

dVV

0u ' juω

dV '4πr 2 u

jω; V '

43πr 3

juω ' &γP4πr 2 u

ρr jω (4 /3)πr 3' &

3γPu

ρr 2jω

ω2 '3γP

ρr 2'

3c 2

r 2

fres 'ω2π

'c 32πr

(b) Adiabatic compression PV γ ' const or P ' const × V &γ

Differentiating P with respect to V and rearranging gives:

dP . p, where p is the acoustic pressure. Thus:

Substituting the result for p from part (a) gives:

or

(c) Resonance frequency is the frequency at which the bubble prefers tovibrate given the physical parameters.

Substituting for dV and V in the expression of part (b), we obtain:

Rearranging gives:

Thus:

(d) At resonance, the bubble screen should act like a Helmholtz resonator(see Ch. 9) and remove considerable energy from the sound field.

± 28 × 10&9 /9.81 ' ±2.87 × 10&9 m

2

Solutions to problems relatingto the Human Ear

Problem 2.1

(a) Weight of a column of water equivalent to the weight of a column ofatmosphere of cross sectional area 1m2 is equal to 101.4kN, which isequal to 101400/9.81 kg. Density of water = 1000kg/m3,thus volume of water = 101.4/9.81 m3 andthus height of water column = volume/(1×1)= 10.34m.

Minimum audible sound = 0dB = 20 × 10-6 Pa r.m.s = 28.2 × 10-6 Papeak. This corresponds to a variation in water height of

(b) 120dB represents an increase in pressure by a factor of 106 over 0dB, sovariation in column height would be .±2.9mm

(c) Figure 2.9 indicates an increase of 60dB for 31.5Hz sound whichrepresents a factor of 1060/20 = 1000.

(d) See p56, text. Overall mechanical advantage = 15:1 = 23.5dB soundpressure level. Linkage mechanical advantage = 3:1 = 9.5dB.

Problem 2.2

(a) A scaling of physical dimensions by a factor of 10 would mean that themouse's range of hearing is 200Hz to 180kHz.

(b) Figure 3.4 and equations 3.14 and 3.16 indicate that the sensitivity is

proportional to d4, so the mouse's ear should be 4 orders of magnitude (or40dB) less sensitive.

The human ear 67

(c) The mouse's transduction mechanism must be 4 orders of magnitudemore sensitive than the human mechanism.

(d) Yes, because the differences would be necessary to make the mousetransduction mechanism more sensitive.

(e) The differences could take the form of a larger mechanical advantage inthe mouse's middle ear as well as differences in the relative physicaldimensions of the inner ear.

Referring to possible inner ear differences, a critical component of theinner ear in regard to sensitivity is the hinge mechanism of the tectorialmembrane. The sensing of sound by the inner and outer hair cells is bya shearing action imposed on the hair cell stereocelia caused by relativemovement between the tectorial membrane and the rods of Corti. Toincrease the sensitivity, it would be necessary to increase this relativemovement which could be achieved by lengthening the rods of Corti.

Referring to the middle ear differences, a much smaller oval window anda different arrangement of the bone linkage could account for a largesensitivity increase.

Problem 2.3

(a) Sound introduced to the ear using ear muffs will be fairly reverberant,having frontal as well as lateral components; thus, we would expect theMAP to be lower than the MAF. If the earmuffs distort the pinnasufficiently, the MAP may approach the MAF in magnitude.

(b) As the sound field within the ear muffs is similar to a diffuse field, onewould expect the MDF to be similar to the MAP and less than the MAF.


Problem 2.4

The first symptoms of noise induced hearing loss are difficulties withunderstanding conversation in a noisy environment, in focussing on thespeaker and in localising noise sources. These symptoms are caused by thebreaking off of stereocelia on the outer hair cells leaving only the inner haircell stereocelia functional. The loss usually occurs first in the 4kHz range andthen extends to higher and lower frequencies. A conventional hearing aidwhich amplifies all frequencies by the same amount will not be much help asit will only amplify the "noise" experienced and not help with the symptomsmentioned above. A frequency selective hearing aid will do a little better byamplifying the signal at the frequencies most affected but it is difficult to seehow even this type of hearing aid will ameliorate the above mentionedsymptoms of early noise induced hearing loss.

Problem 2.5

Not necessarily. As stated in the text, repeated exposures to noise whichresults in a temporary threshold shift will result in permanent damage. On theother hand, there is evidence that damage is accumulative, so even oneexposure will contribute to the eventual permanent damage.

Problem 2.6

Pitch: Determined by location on the basilar membrane which responds mostto the noise. Also the hair cells are tuned to maximum output at frequenciescorresponding to the resonance frequencies of the parts of the basilarmembrane to which they are attached. However, at low frequencies and loudnoise, some neurons will fire for hair cells all along the basilar membrane anda second mechanism by which neurons fire in locked phase with the acousticsignal (at acoustic signal maxima or once per cycle) dominates the pitchdetermination. As the frequency increases, the localisation of the basilarmembrane excitation (and associated resonant hair cells) becomes increasinglyimportant as the method for determining pitch until at 5000Hz the lockedphase phenomenon of neuron firing ceases altogether and the neurons firerandomly.

The human ear 69

Loudness: Determined by the rate of neuron firing, which is controlled by themotion of the hair cell stereocelia which in turn is controlled by the basilarmembrane. There is also a hair cell feedback mechanism whereby the voltagegenerated by the hair cells causes them to deform, thus increasing themovement of the tectorial membrane. This feedback mechanism effectivelyincreases the dynamic range of the hearing mechanism and also results inimproved pitch resolution.

Problem 2.7

(a) Two signals may have exactly the same spectral content and thus soundthe same, but they may have entirely different phase relationshipsbetween the particular spectral components making up the signal. Thusthe time histories as seen on an oscilloscope could look quite different.Both amplitude and relative phase of the spectral components of a signalare necessary to reconstruct a signal uniquely. However, the ear discardsphase information.

(b) In 0.05 seconds sound will travel a distance of 17.15m. This correspondsto a wavelength at 20Hz. Thus for a 20Hz tone, the peak sound pressureswill occur at intervals of 0.05 seconds. Below this frequency there willbe greater intervals between the peak sound pressures and the sound willbe heard as a sequence of auditory events.

(c) The dimensions of an auditorium must be such that the sound arriving atany location after being reflected from a wall, ceiling or floor must notarrive longer than 0.05 seconds after the direct sound. This means thedifference between direct and reflected paths should be less than 17m.

Problem 2.8

(a) A low frequency warning device would be more effective mainly becauseit is not as easily masked by the 500Hz noise as higher frequencies wouldbe but also because it would diffract more effectively around obstaclesto create a more uniform coverage.

(b) See fig 2.10(a) in text and read off values as accurately as possible.


Octavebandcentre

frequency

63 125 250 500 1k 2k 4k 8k

SPLforward

40 44 49 54 56 55 45 33

SPLside

39 43 47 51 53 49 39 27

SPLrear

38 42 44 48 49 44 27 15

Sonesforward

0 0.6 1.7 2.8 3.8 4.3 2.8 1.6

Sonesside

0.05

0.55

1.4 2.4 3.2 3 2 1.1

Sonesrear

0.02

0.5 1.1 2 2.5 2.2 0.9 0.3

Phonsforward

0 33 48 55 59 61 55 46.8

Phonsside

0 33 45 53 57 55.8 50 41.4

Phonsrear

0 30 41 50 53 51.4 39 22.6

Overall Levels:Sones Forward = 4.3 + 0.3[0.0+0.6+1.7+2.8+3.8+2.8+1.6] = 8.3 Sones Side = 3.2 + 0.3[0.05+0.55+1.4+2.4+3.0+2.0+1.1] = 6.3Sones Rear = 2.5 + 0.3[0.02+0.5+1.1+2.0+2.2+0.9+0.3] = 4.7

Using equation 2.1,Phons = 40 + (10log10S)/(log102).Thus, Phons forward = 70.5Phons side = 66.6Phons rear = 62.3

The human ear 71

(c) Yes, a person with no hearing loss would be able to hear, as the levels inthe bands important for speech recognition are well above the hearingthreshold level or MAF.

(d) If the distance increases to 10m from 2m, the sound pressure level in allbands will decrease by 20log10(10/2) = 14dB (assuming free fieldconditions). The MAF from figure 2.9 and the new sound pressure levelsare in the table below.

Octavebandcentre

frequency(Hz)

63 125 250 500 1k 2k 4k 8k

MAF 39 22 15 9 4 0 0 12

New SPLforward

26 30 35 40 42 41 31 19

New SPLside

25 29 33 37 39 35 25 13

New SPLrear

24 28 30 34 35 30 13 1

Difficulty in hearing sound in the 63Hz and 8kHz bands would beencountered but as these bands are not important for speech, it isexpected that there would be no difficulty in understanding speech forany head orientation.

(e) Speech recognition is just possible for S = 8 Sones. A speaker speakingtwice as loudly will increase the rear level from 4 to 8 sones, makingspeech recognition just possible again, so the situation will be improved.

Problem 2.9

See figure 2.9(a). Masking tone is 800Hz at 60dB. From the figure, thethreshold of detection of an 630Hz tone would be increased by 20dB. Figure2.5 indicates that the MAF at 630Hz is approximately 5dB, so the sound levelwould need to be 25dB in order to be heard.

3

Solutions to problems relating toinstrumentation and measurement

Problem 3.1

(a) Pressure response: microphone response when subjected to a uniformpressure field which is accomplished in practice electrostatically.

Free-field response: microphone response when subjected to a soundwave coming from a specified direction in an otherwise free field (freeof reflected sound).

Random incidence response: microphone response averaged over allpossible angles of sound wave incidence.

Microphones are designed so that the roll-off in pressure response at highfrequencies is just compensated for by the increase in free field pressuredue to diffraction of a normally incident wave (normal incidencemicrophone) or by the increase in pressure due incident waves averagedover all possible directions of incidence (random incidence microphone).

(b) Random incidence microphones are used to measure noise in reverberanttest chambers, reverberation times in auditoria and industrial noise whenthe direction of origin is uncertain or there are a number of sourceslocated in various directions from the observer.

Free field microphones are used in anechoic test chambers and inindustrial noise measurement cases where the origin of the noise is froma single direction or a narrow direction angle.

Instrumentation and measurement 73

Lp ' 20log10 E & S % 94 ' &110 % 26 % 94 ' 10dB

Lp ' 10log10 101.3 & 101.0 ' 10dB

Problem 3.2

(a) Using equation 3.16, the sound level represented by the noise floor on theinstrument is:

If the sound level meter actually reads 13dB, then the contribution dueto the actual noise is:

(b) The ear can hear 0dB of frontally incident sound at 2kHz and it canprobably discern signals just above its noise floor. So the equivalentsensitivity of the sound level meter would be approximately 10dB,implying that the ear is 10dB more sensitive.

Problem 3.3

(a) Sensitivity = -25dB re 1V per Pa. Using equation 3.15, we have: -25 =20log10[E/p] which gives a sensitivity of 10(-25/20) volts/Pa which = 56.2mV/Pa.

(b) The pressure response of the microphone would have to be 4.5dB less at10kHz than at 250Hz for the overall response to remain flat; that is, -29.5dB re 1V per Pa.

(c) The microphone overall sensitivity for a 0E incident field at 10kHz is-29.5 + 5 = -24.5dB.

(d) The microphone overall sensitivity for a reverberant field at 10kHz is-29.5 + 1.5 = -28.0dB.

(e) At 250Hz, it can be seen from figure 3.3 in the text that the correction fora 180E angle of incidence would be 0dB. Thus the overall microphonesensitivity at 250Hz would be equal to -24.5dB, the pressure sensitivity.At 10kHz, the overall sensitivity is: -29.5 - 0.2 = -29.7dB.


Problem 3.4

The MAF is the minimum audible field to frontally incident sound and wouldcorrespond to the free field calibration of a microphone. Similarly the MDFwould correspond to the diffuse field calibration of a microphone. Thepressure calibration of a microphone would correspond to the MAP which isthe minimum pressure audible at the tympanic membrane of the ear.

Problem 3.5

The random incidence microphone is used in cases where one is not sure ofthe direction from which the sound is coming or if it is coming from a numberof directions simultaneously.

To minimise measurement error, the microphone would be held vertically,thus resulting in most sound being incident at angles close to 90E. Thedifference between the microphone diffuse field response and the 90Eresponse is much smaller than the error resulting from pointing a free fieldmicrophone in the wrong direction (see figure 3.3 in text). In both casesmeasured sound pressure levels will be less than actual levels.

Problem 3.6

This would be similar to connecting the microphone to ground through a lowimpedance and this would seriously reduce the microphone sensitivity,resulting in large measurement errors.

Problem 3.7

The A-weighted sound pressure level is related to loudness perception of lowlevel environmental noise as well as to hearing damage (although not toloudness perception of loud industrial noise), and regulations are written interms of this quantity due to its ease of measurement by unskilled people. Itis a reasonably valid measure of noise exposure as there is a directcorrespondence between A-weighted sound level and hearing loss suffered bynoise exposed people.


The advantages of the A-weighted level for characterising equipment andworkplaces are its direct relationship to noise exposure and its single numbersimplicity. However it does not provide the frequency content informationnecessary for effective noise control measures to be specified and this is themain disadvantage.

Problem 3.8

(a) The A-weighted levels are calculated by adding the A-weightingcorrections (most are negative) to each octave band level and thenlogarithmically adding the results as described on page 48 of the text.The overall linear level is calculated by adding the values given in theproblem together logarithmically as described on page 48 of the text.The answers are: A-weighted = 89.4dB(A) and Linear = 98.5dB.

(b) The main source of error is a result of the assumption that all frequenciesin each octave band can be weighted by a single quantity (the weightingcorresponding to the band centre frequency) when in fact the A-weighting is a smoothly varying function of frequency. The maximumpossible error can be estimated by comparing the results using the A-weighting corresponding to the band centre frequency with resultsobtained bu using the A-weighting corresponding to the upper and lowerband limit frequencies.

Problem 3.9

Plot out the values given in Table 3.1 on graph paper using a logarithmicfrequency scale. Join all of the points by a straight line which is as good asa smooth curve for the present purposes. Then plot the upper and loweroctave band frequency limits from table 1.2 and read off from your graph theA-weighting corrections at these frequencies The difference between thesevalues and the octave band centre frequency A-weighted values represent thelargest errors which could occur if all of the energy just happened to be atfrequencies at the edge of each octave band. To find the overall possiblevariation calculate the dB(A) levels for each extreme and compare them to theoverall level calculated using band centre frequency corrections as illustratedin the table on the next page.


Combining the band levels together as described on p 38 in the text gives:

Upper limit = 84.7dB(A)Lower limit = 83.3dB(A)Centre value = 84.1dB(A) (corresponding to A-weighted corrections at bandcentre frequency.

The A-weighted levels peak at high frequencies so the noise would sound alittle "hissy".

Octavebandcentrefreq.(Hz)

63 125 250 500 1000 2000 4000 8000

Upper flimit

88 176 353 707 1,414 2,825 5,650 11,300

Lower flimit

44 88 176 353 707 1,414 2,825 5,650

UpperdB(A)

adj.

-21 -12.2 -5.8 -1.2 1.2 1.3 1.1 0.0

CentredB(A)

adj.

-26.2 -16.1 -8.6 -3.2 0.0 1.2 1.0 -1.1

LowerdB(A)

adj.

-35 -21 -12.2 -5.8 -1.2 1.1 0.0 -3.5

SPL 76 71 68 70 73 76 79 80

UpperdB(A)

55 58.8 62.2 68.8 74.2 77.3 80.1 80.0

CentredB(A)

49.8 54.9 59.4 66.8 73 77.2 80 78.9

LowerdB(A)

41 50 55.8 64.2 71.8 77.1 80.0 76.5


Problem 3.10

First remove the background noise contribution from each octave bandmeasurement as described in example 1.4, p.49 in the text. Thenarithmetically add (some numbers are negative) the A-weighted correctionsto each octave band level as was done for problem 3.9. Then add the A-weighted octave band levels together logarithmically as described on p.48 inthe text. The final answer is 89.0dB(A).

Problem 3.11

(a) A-weighted level = 86.8dB(A)(b) It is not a good way to calculate overall weighted levels because errors

arise from the inherent assumption that the A-weighting is uniform acrossany particular octave band.

Problem 3.12

To begin, assume an arbitrary level of 50dB in the 63Hz octave band,calculate the A-weighted levels in all bands, the overall A-weighted level andthus the amount to add to each band level to reach an overall level of105dB(A). The calculations are summarised in the following table, where itis noted that a constant spectrum level is reflected in octave band levelsincreasing at the rate of 3dB per octave (reflecting a doubling of bandwidthper octave).

Octave band centrefrequency (Hz)

63 125 250 500 1000 2000 4000 8000 total

Un-weightedlevel

50 53 56 59 62 65 68 71

A-weightedlevel

23.8 36.9 47.4 55.8 62 66.2 69 69.9 73.8

Adjustmentneeded

31.2 31.2 31.2 31.2 31.2 31.2 31.2 31.2

Unweightedlevel for

105dB(A)81.2 84.2 87.2 90.2 93.2 96.2 99.2 102.2 105.0


Problem 3.13

Although the signal levels in the experiments to determine equal loudnesscontours varied substantially, the A-weighted scale corresponds approximatelyto the loudness contour of 60dB. As industrial noise is usually much louderthan this and equal loudness contours for high sound levels do not have thesame shape as those at 60dB, it is unlikely that the A-weighted scale willindicate correct loudness levels for most industrial noise.

Problem 3.14

(a) If the observer is too close to the microphone when noise measurementsare being taken, then reflections from the observer can affect the noiselevels being measured. In tonal sound fields, the reflections could resultin an increase in the measured sound level of up to 5dB (and alsodecreases) and in broadband sound fields, the increase measured couldbe up to 2dB. The reasons for the above numbers not being 6 and 3dBrespectively is because the observer will absorb and scatter some of thenoise while reflecting it.

(b) The fast response (0.1s time constant) of the sound level meterapproximates the way the ear hears but the slow response (1.0s timeconstant) is useful for determining LAeq levels (average of the upperswings of the needle on an analog meter) and L90 levels (average of lowermeter swings). However most modern instrumentation allows directdigital readout of these quantities. In addition, most legislation is writtenin terms of measurements taken using the "slow" response as someresearchers suppose that this is more representative of the hearingdamage caused by the noise.

(c) The "frontal/diffuse" control is used for selecting the microphonecharacteristic most suitable for the measurement being undertaken. Notethat the microphone remains unchanged - the electronics in the soundlevel meter effectively change its characteristics to compensate for thedifferent diffraction effects of the microphone grid for each of the twotypes of field.


'18

85

5% 4 × 84 % 24 × 83 % 32 × 82 % 16 × 8 × 10&4 ' 0.4675

Leq ' 10log10p 2

p 2ref

' 10log100.4675

4 × 10&10' 90.7dB

LAeq ' 10log101

1/4 % 2 % 2 % 1/12 % 4×

× 0.25 × 108 % 2 × 107 % 2 × 109 % (1/12) × 109.9 % 4 × 107.5

Problem 3.15

, thusp ' (t 2 % 8t % 4) × 10&2

p 2 ' (t 4 % 16t 3 % 72t 2 % 64t % 16) × 10&4

Average p 2 ' m8

0

(t 4 % 16t 3 % 72t 2 % 64t % 16) × 10&4 dt

A-weighting at 250Hz is -8.6dB, so LAeq = 82.1dB(A).

Problem 3.16

LAeq is generally used to describe noise as it is an A-weighted energy averagewhich seems to be related to loudness perception of low level environmentalnoise as well as to hearing damage (although not to loudness perception ofloud industrial noise), and regulations are written in terms of this quantity dueto its ease of measurement by unskilled people.

= 85.3dB(A).

Problem 3.17

A sound level meter on site is preferable, as a tape recorder is not sufficientlyaccurate for legal disputes (see table 3.3 in the text).

Problem 3.18

Sources of measurement error:


1m

1m

1mpossible wind screen

configuration

shadecloth

Microphone vibration;SLM vibration;background noise;overloading input amplifier when taking octave or 1/3 octave bandmeasurements with an old SLM;too cold or too hot;moisture or dust on microphone diaphragm;reflections from nearby surfaces; andwind noise.

Can minimise effects of wind noise by placing a foam wind shield on themicrophone AND placing the mic in an enclosure made using shade cloth as

shown in the figure below.

Problem 3.19

(a) See text, p114 B 120.

(b) (i) At low frequencies, errors arise because the phasedifference between the two microphone signals (due to thespacing being small compared to a wavelength) is notsufficiently large compared to the phase accuracy of themicrophones.

(ii) At high frequencies, errors arise because the microphonespacing becomes significant compared to a wavelength


prms ' pref 10Lp /20

' 2 × 10&5 × 1095/20 ' 1.12Pa

causing the finite difference approximation for the pressuregradient to be inaccurate.

(iii) In very reactive sound fields, the phase between theacoustic pressure and particle velocity is close to 90E, soany phase errors translate to a large error in the intensity asit is proportional to the cosine of the phase angle.

(iv) In the presence of external noise sources, sound powermeasurements could exhibit significant errors if the soundpressure level of the external noise is sufficiently high(usually about 10dB or more above the level from the noisesource being measured). This is because the powermeasurement relies on averaging normal intensitymeasurements and the result of the external source will beto create a situation where small differences between largenumbers will dominate the result.

Problem 3.20

(a) See part (a) in previous question.

(b) See part (b) in previous question.

(c) Applications include:sound power measurement;localisation and identification of noise sources;sound transmission loss measurement;determining the importance of flanking sound transmission paths in noisecontrol applications;

Problem 3.21

(a) Sound pressure associated with 95dB sound level is

Force on microphone diaphragm is


Frms ' 1.12 × π × (0.012)2 /4 ' 1.27 × 10&4 N

puA

' &jρc 2

Vω

prms ' 2 × 10&5 × 1065/20 ' 3.557 × 10&2

urms 'pωV

ρc 2S; (S is the area of the microphone diaphragm)

'3.557 × 10&2 × 2π × 500 × 0.01 × 0.022 × (π /4)

1.206 × 3432 × 0.0122 × (π /4)

' 2.19 × 10&5 m/s

Vol.displ.ampl '2urms S

ω

'2 × 2.188 × 10&5 × π × 0.0122

4 × 2 × π × 500

' 1.11 × 10&12 m 3

(b) r.m.s. velocity of the diaphragm is the same as the air particle velocity.From equation 9.35 in the text, the ratio of the sound pressure to particlevelocity in a cavity of dimensions much smaller than a wavelength is

The sound pressure measured by the monitoring microphone is 65dBwhich corresponds to an r.m.s.pressure of

Thus the particle velocity is

(c) The volume displacement in the cavity corresponding to a sound pressurelevel of 65dB can be calculated using the same equation as used in part(b). Thus the volume displacement is

(d) Mechanical input impedance, Zm = F/u. Thus


Zm '1.27 × 10&4

2.188 × 10&5' 5.8 N&s /m

(e) The volume displacement of the monitoring microphone will be 30dB(95 - 65) below the displacement of the test microphone. This representsa percentage difference of 100/101.5 = 3% which will not affect the soundpressure sensed by the test microphone significantly.

(f) Upper test frequency is limited by the onset of resonant cavity modes.As the largest dimension is the radius, cross modes will occur beforeaxial modes. This is explained in Chapter 7 in the text.

Ta ' 8 × 2&(99 & 90) /3 ' 1 hour

Ta ' 8 × 2&(99 & 90) /5 ' 2.3 hours

4

Solutions to problemsrelating to criteria

Problem 4.1

(a) Using equation 4.42 in the text, the allowable exposure time usingEuropean criteria is

(b) The allowable exposure using USA criteria is

Problem 4.2

(a) A-weighted SPL is given by:

LpA ' 10log10(10(9.5 & 0.86) % 10(9.7 & 0.32) % 109.9 ) ' 100 dB(A)

(b) Allowed daily exposure time in Australia is:

Ta ' 8 × 2& (100 & 90) /3 ' 0.8 hours

Allowed daily exposure time in USA is: Ta ' 8 × 2& (100 & 90) /5 ' 2 hours

Problem 4.3

Fan noise = 91dB(A)Saw idling noise = 88dB(A)Saw cutting noise = 93dB(A)Let required fan noise for LAeq,8h = 90dB(A) be x dB(A).

Criteria 85

90 ' 10log1018

6.4 1088/10 % 10x /10 % 1.6 1093/10 % 10x /10

10x /10 ' 9.618 × 107

90 ' 16.667log1018

6.4 100.3 × [ 10log10(108.8

% 10x / 10) & 90] /5

% 1.6 100.3 × [ 10log10(109.3

% 10x /10) & 90] /5% 90

(a) European criteriaUsing equation 4.3 or 4.39 with LB = 90 and L = 3:

Solving for x gives:

Thus, x = 79.8dB(A)

The required fan noise reduction is then 91 - 79.8 = 11.2dB(A)

(b) USA criteriaUsing equation 4.39 with the integral replaced with a sum and with withLB = 90 and L = 5, we obtain:

In solving for x, we must remember the proviso that combined fan andsaw noise levels of less than 90dB(A) at any time do not contribute to thenoise exposure results in a value of x as close to 90dB(A) as possible.Assuming a precision of 0.1dB(A), the allowed fan noise plus saw idlenoise is 89.9dB(A). This is because if the fan noise plus saw idle noise

is greater than 90dB(A), the overall is greater than 90dB(A). ThusL )

Aeq

the maximum allowed fan noise is. Thus the required fan noisex ' 10log10 108.99 & 108.8 ' 85.4dB(A)

reduction is 91 - 85.4 = 5.6dB(A).

Problem 4.4

(a) LAeq,8h = 10Log10(1/8)[2 ×1095/10 + 6 ×1070/10] = 89.0 dB(A)

(b) EA,8 ' 32 × 10(89 & 100) /10 ' 2.54 Pa2 @h

(c) We may assume that the 70 dB(A) does not contribute significantly, so we


HDI ' 10log10mt

0

10Lp /20

dt

59.5 ' 10log10 10110/20 × T ' 55 % 10log10 T

LAeq,8h ' 10log1018

4.5 × 1010.5 % 1.5 × 109.5 ' 102.6dB(A)

L )

Aeq,8h ' 16.667

× log10 (1 /8) 100.3 × (105 & 90) /5 × 4.5 % 100.3 × (95 & 90) /5 × 1.5 % 90

' 101.4dB(A) (USA criteria)

Ta ' 6 × 2&(102.6 & 90) /3 ' 0.33 hours

need to find the allowed exposure to 95 dB(A). This is given byTa ' 8 × 2& (95 & 90) /3 ' 2.52 hours

(d) SPL due to machine only is: 10log10 [109.5 & 109.1 ] ' 92.8 dB(A)

Problem 4.5

In this case,

where T is the number of years to cross the hearing loss criterion.

Thus, T . 3 years, and he will be 23 years old before he joins the old folks(assuming that he is in the 20% more sensitive part of the population).

Problem 4.6

(a) Using equation 4.3 in the text:

(European criteria)

Using equation 4.41 in the text with the integral replaced with asummation sign:

(b) European criteria

Using equation 4.42 in the text,

Criteria 87

Ta ' 6 × 2&(101.43 & 90) /5 ' 1.23 hours

HDI ' 10log10ji

Ti × 10Lpi /20

'10log10 5 × 108.5/2 % 3 × 109/2 % 6 × 109.5/2 % 1 × 1010/2 % 10 × 108/2

' 58.6

L )

Aeq,8h ' 16.667

× log1018

2 × 100.3 × (91 & 90) /5 % 2 × 100.3 × (96 & 90) /5 % 90

' 88.9dB

LAeq ' 10log1018

2.4 × 1085/10 % 1.6 × 1088/10

% 2 × 1091/10 % 2 × 1096/10

' 91.8dB(A)

USA criteria

Problem 4.7

From figure 4.4(b) in the text, there is a 22% risk of developing a 22dBhandicap.

Problem 4.8

(a) USA criteriaUsing equation 4.41 with LB = 90 and L = 5, we obtain:

Daily noise dose = 2(L

)

Aeq & 90) /5' 0.86

No reduction in exposure time is necessary.

(b) European criteriaUsing equation 4.3 or 4.39 with LB = 90 and L = 3,


Ta '8

2(91.8 & 90) /3' 5.2hours

Daily noise dose = 2(LAeq & 90) /3

' 1.53Allowable exposure time:

Thus a reduction of 2.8 hours is required.

Problem 4.9

Number of impacts per day = 80 × 60 × 8 × 0.6 = 23,040B-duration = 100 msecB-duration × number of impacts = BN = 2.3 × 106

Peak SPL = 125dB

Allowable level for BN = 2.3 × 106 is obtained from fig 4.5 in the text.

European criteriaLa = 112.5dB, and noise dose = 2(125 - 112.5)/3 = 18

U.S.A. criteria La = 121dB, and noise dose = 2(125 - 120.5)/5 = 1.9

Allowable BN for 125dB peak - see fig 4.5 in text.

European criteria BN = 1.40 × 105, and noise dose = 23.04/1.4 = 16.4

U.S.A. criteria BN = 1.29 × 106, and noise dose = 2.304/1.29 = 1.8

The small differences in results obtained using the two methods (allowablelevel vs allowable BN) are due to difficulties in reading the figure any moreaccurately.

The operator is overexposed according to both criteria.

Required work day decrease

European criteriaAssuming that the press accounts entirely for the exposure of the employee,

Criteria 89

then from figure 4.6, the allowed BN product is 1.4 × 105. In terms of hours,this is equal to: 1.4 × 105 /(80 × 60 × 100) = 0.29 hours of press operation.Accounting for the background noise, the exposure will be controlled by thisfor a minimum of 7.71 hours (8 - 0.29). This corresponds to a noise dose of(7.71/8) × 2(85 - 90)/3 = 0.3. Thus the allowable time of press operation is 0.7 ×0.29 = 0.20 hours. [Iterating again does not affect the result significantly].

Thus required workday decrease = 4.8 - 0.2 = 4.6 hours of press operation.

USA criteriaBackground noise of 85dB(A) does not contributeFrom figure 4.6, allowable BN product = 1.2 × 106. In terms of operatinghours, this is equal to: 1.29 × 106 /(80 × 60 × 100) = 2.69 hours of pressoperation. The background noise when the press is not operating does notcontribute to the exposure according to USA criteria.

Thus required work day decrease = 4.8 - 2.7 = 2.1 hours of press operation.

Problem 4.10

Number of impacts per day = 40,000B-duration = 60 msecB-duration × number of impacts = BN = 2.4 × 106

Peak SPL = 135dB

Allowable level for BN = 2.4 × 106 is obtained from fig 4.6 in the text.

European criteriaLa = 112dB, and noise dose = 2(135 - 112)/3 = 200.

U.S.A. criteria La = 121dB, and noise dose = 2(135 - 121)/5 = 7.

Allowable BN for 135dB peak - see fig 4.6 in text.

European criteriaBN = 1.2 × 104

and noise dose = 240/1.2 = 200


U.S.A. criteriaBN = 3.4 × 105

and noise dose = 240/34 = 7

Allowable number of impacts

USA criteriaThe background noise of 87dB(A) contributes nothing to the daily noise dosebecause it is less than 90dB(A). Thus the allowed number of impacts = 3.4× 105 /60 = 5700. Thus required decrease = 40,000 - 5,700 = 34,300.

European criteriaAssuming that the press accounts entirely for the exposure of the employee,then from figure 4.6, the allowed BN product is 1.2 × 104. The allowablenumber of impacts is then 1.2 × 104 /60 = 200. However this represents sucha small part of the 8-hour day, that the 87dB(A) background can be consideredto dominate the exposure for almost 8 hours, resulting in a noise dose of 0.5due to this alone. Thus the allowable dose due to the impact noise is 0.5,which corresponds to 100 impacts.

Problem 4.11

Use figure 4.7 in the text.

Expected level - "very loud voice to shout"Required level - "shout".

Problem 4.12

See figure 4.7 in text. The answers are:No. No. No. Too Loudly.

Problem 4.13

The one third octave band levels would have to be first combined into octaveband levels by logarithmically adding together three third octave bands for

Criteria 91

Lp250 ' 10log10 1060/10 % 1065/10 % 1063/10 ' 67.9 dB

each octave band result. The three bands to add would be one with a centrefrequency the same as the octave band and one band above and one below thatone. For example, if the 200Hz, 250Hz and 315Hz one third octave bandlevels were 60dB, 65dB, and 63dB respectively, then the 250Hz octave bandlevel would be:

Problem 4.14

A-weighted overall sound levels are inadequate for noise level specificationand control because they give no indication of the frequency content of thenoise which is necessary for assessing annoyance and determining the type ofnoise control approach which may be feasible. It is preferable to have data asoctave band levels for control purposes, although often for specificationpurposes, NR, NC or RC numbers are adequate as they take into account thespectral content of the noise. However overall dB(A) numbers are adequatefor the purposes of assessing hearing damage risk and for comparing noise(either occupational or environmental) with permitted levels according to localregulations.

Problem 4.15

(a) A-weighted levels are calculated and tabulated below

Frequency(Hz)

63 125 250 500 1k 2k 4k 8k

Lp (dB re 20µPa)

100 101 97 91 90 88 86 81

A-weighting -26.2 -16.1 -8.6 -3.2 0.0 1.2 1.0 -1.1

A-weightedlevel

73.8 84.9 88.4 87.8 90 89.2 87 79.9


Lp ' 10log10 107.38 % 108.49 % 108.84

% 108.78 % 109 % 108.92 % 108.7 % 107.99

' 96.1 dB(A)

Ta ' 8 × 2& (96.1 & 90) /3 . 1.95 hours

L ' 38 % 0.3[28.5 % 35.3 % 28.5 % 33 % 35.3 % 38 % 33] ' 107.5 sones

The overall A-weighted level is:

(b) Using equation 4.42 in the text, the allowable number of hours is:

(c) NR level of noise - plot levels on NR curves as shown below, where itmay be seen that NR = 91.

(d) Loudness level

Frequency(Hz)

63 125 250 500 1k 2k 4k 8k

Lp

(dB re 20µPa)100 101 97 91 90 88 86 81

Sones 28.5 38 35.3 28.5 33 35.3 38 33

The overall level in sones is calculated using equation 2.33. Thus:

Criteria 93

P ' 40 % 10log10 S

log10 2' 107.5 phons

x ' 10log10 1096.1/10 & 10(96.1 & 1.5) /10 ' 90.8dB(A)

0

10

20

30

40

5045

35

15MAF

63 125 250 500 1k 2k 4k 8k

Octave band center frequency (Hz)

Oct

ave

band

sou

nd p

ress

ure

leve

l (dB

re

20

Pa)

25

From equation 2.32:

Note that the above equation is inaccurate for levels above 100 phons.

(e) Following example 1.4, the contribution of the machine to the overalllevel is x dB(A) where x is defined as:

Problem 4.16

(a) The levels are first plotted on NC and NCB curves

The result is NC = 33 and NCB = (38 + 30 + 20 + 16)/4 = 26


0

10

20

30

40

5045

35

15

10

0

6316 12531.5 250 500 1k 2k 4k 8kOctave band center frequency (Hz)

Oct

ave

band

sou

nd p

ress

ure

leve

l (dB

re

20

Pa)

25

NCB

The system is sufficiently quiet for churches holding less than 250 people(see table 4.8 in the third edition of the textbook and table 4.2 in the firstedition or look up AS2107-1987), however for larger churches, the levelshould be about 5dB lower.

(b) This NCB level is exceeded by more than 3dB in 125Hz, 250Hz and500Hz bands so it will sound rumbly. Note that RC criteria would resultin a neutral classification (not rumbly or hissy).

Best fit between 125Hz and 500Hz is NCB = 33. No levels in the octavebands between 1000Hz and 8000Hz are above NCB = 33 so sound is nothissy.

This conclusion can be checked by plotting on RC curves. As can be seenfrom the following RC plot, the noise will be neutral (not rumbly orhissy -see 3rd. Edn. text, page 159) as no levels in bands below 500 Hzexceed the RC level by more than 5 dB. Note RC = (38+30+20)/3 = 29.Although this conclusion is different to that drawn using NCB curves, itcan be seen that the RC classification is close to rumbly.

Criteria 95

LA ' 10log10 1053/10 & 1048/10 ' 51.3dB

45

35

25

30

40

50

(RC)

16 31.5 63 125 250 500 1k 2k 4k 8k10

20

30

40

50O

ctave

ban

d s

oun

d pre

ssure

leve

l (dB

re

20

Pa)


(c) Optimum spectrum levels of added masking noise would be equal to theRC-33 levels (as this corresponds to the highest existing spectrum levels)with the existing levels logarithmically subtracted from it. For example,in the 63Hz band, the RC-33 value is 53dB and the desired added levelis:

Desired spectrum levels of masking noise are listed in the table below.

Octave bandcentrefrequency (Hz)

63 125 250 500 1000 2000 4000 8000

Sound pressurelevel (dB)

48 48 43 38 30 20 16 12

RC-30 values 50 48 43 35 33 28 23

Desired addedlevels

51.3 0 0 0 30 27 22


LA ' 10log10 10(60 & 26.2) /10 % 10(55 & 16.1) /10 % 10(55 & 8.6) /10

% 10(50 & 3.2) /10 % 10(55 & 0) /10 % 10(55 % 1.2) /10

% 10(50 % 1.0/10 % 10(45 & 1.1) /10

' 59.9dB(A)

0

10

20

30

40

5045

35

15

10

0

6316 12531.5 250 500 1k 2k 4k 8kOctave band center frequency (Hz)

Oct

ave

band

sou

nd p

ress

ure

leve

l (dB

re

20

Pa)

25

NCB

Problem 4.17

(a) The NCB value for the noise is (37 + 33 + 33 + 32)/4 = 34

(b) The line of best fit for the NCB curve between 125 Hz and 500 Hz is 32or 33 NCB. Three high frequency bands exceed this curve so the noisesounds hissy. No bands below 500 Hz exceed 34 NCB so noise is notrumbly. See following figure.

Problem 4.18

(a) A-weighted level is:

Criteria 97

45

35

25

30

40

50

(RC)

16 31.5 63 125 250 500 1k 2k 4k 8k10

20

30

40

50

60

Oct

ave

ban

d so

und

pre

ssur

ele

vel (

dB r

e 20

P

a)


NR level from following figure (NR curves) = 58.

(b) The levels are plotted on RC curves below, where it can be seen that thespectrum would sound hissy.

(c) The noise level is 59.9dB(A). The dB(A) adjustments to the base levelof 40dB(A), the resulting allowable noise levels and the expected publicreactions are:


day: +20 -2 = 58dB(A) (marginal public reaction)evening: +20 -5 -2 = 53dB(A) (little public reaction)night: +20 -10 -2 = 48dB(A) (medium public reaction)

See table 4.11 in the text for public reaction estimates.

(d) The noise reductions between inside and outside and the resulting insidelevels are in the table below.


63 125 250 500 1000 2000 4000 8000

Exterior soundpressure levels

60 55 55 50 55 55 50 45

Expected noisereduction (dB)

5 5 8 10 14 16 20 21

Interior soundpressure levels

55 50 47 40 41 39 30 24

The spectrum in the last line of the table is plotted in the figure belowand represents an NR value of 42.

Criteria 99

From Table 4.12 in the text, daytime base level = NR 30 and nighttimebase level = NR 25.

Daytime adjustments = +5 +10, which give an allowable NR = 45nighttime adjustment = +10, which gives an allowable NR = 35

Thus we would expect complaints during the evening and night but notduring the day if the windows are closed.

When the windows are open, 5dB is added to levels in all octave bandsand the NR value of the interior noise becomes NR 47. This would resultin a few complaints during the day and an increase in the number ofnighttime complaints.

(e) Factory could be built provided it only operated during the day. If thenoise occurred only 25% of the time:

.LAeq ' 10log10 0.25 × 1059.9/10 ' 53.9dB(A)

From the results of (c) above, it can be seen that the factory could nowoperate in the evening as well, but not at night. However from the resultsof (d) above, and table 4.11 in text, the NR criteria would indicate that itshould still only operate during the day.

Problem 4.19

From table 4.10 in the text, the acceptable level for nighttime operation isLAeq = 40 - 10 + 15 = 45dB(A). From table 4.11 in the text, the expectedcommunity response would be widespread complaints.

φ 'Ar

ej( ωt & kr)

p ' jωρAr

e j(ωt & kr)

u 'A

r 2e j(ωt & kr) %

jkAr

e j(ωt & kr)

'Ar

e j(ωt & kr) 1r% jk

'p

jωρ1r% jk '

pρc

1 &j

kr

5

Solutions to problems relating tosound sources and outdoor soundpropagation

Problem 5.1

(a) Intensity, I = W/S = 10-2 /(4π × 0.52 ) Watts/m2 = 3.18 mWatts/m2

(b) and thus theprms ' Iρc ' 3.18 × 1.206 × 343 × 10&3 ' 1.15Pa

pressure amplitude = prms/2 = 1.62 Pa.

(c) For outwardly travelling spherical waves in free space, equation 1.40cmay be written as:

Using equations 1.6 and 1.7, the acoustic pressure and particle velocitymay be written respectively as:

and

Sound sources and outdoor sound propagation 101

*u* '*p*ρc

1 %c

2π f r

2

'1.62

1.206 × 3431 %

3432 × π × 400 × 0.5

2

' 4.06mm/s

Q '4πW

k 2ρc'

Wc

π f 2ρ

'0.01 × 343

π × 4002 × 1.206' 2.38 × 10&3 m 3 /s

Thus: *u* '*p*ρc

*1 &j

kr*

or,

(d) Lp ' 20log10 prms % 94 ' 95.2dB

(e) Lw ' 10log10 W % 120 ' 100dB

(f) Source strength, Q, may be calculated using equation 5.12 in the text.Thus:

Problem 5.2

(a) From equation 5.13(b) in the text, it can be seen that for a simple sourcethe r.m.s. pressure2 is inversely proportional to the distance2 from thesource. Thus the sound pressure level at 10m would be 110 -20log10(10/1) = 90dB.

(b) We can use the result of 5.1(c) above. The acoustic pressure amplitude

at 1m is and the amplitude atp ' 2 × 10&5 × 2 × 10110/20 ' 8.945 Pa10m is

. p ' 2 × 10&5 × 2 × 1090/20 ' 0.8945 Pa

At 1m, kr = 2πf/c = 2π × 100/343 = 1.832. At 10m, kr = 18.32. The


u 'pρc

1 %1

(kr)2

β ' &tan&1 1kr

' &tan&1 11.832

' &28.6E

β ' &tan&1 1kr

' &tan&1 118.32

' &3.1E

W 'Q 2 k 2ρc

4π'

6.282 × 10&8 × 1.8322 × 1.206 × 3434π

' 43.6µWatts

W 'Q 2 k 2ρc

4π'

6.282 × 10&8 × 14.6552 × 1.206 × 3434π

' 2.79 milliWatts

Lw ' 10log10 W % 120 dB

result of 5.1(c) may be written as:

Thus at 1m, and the phaseu '8.945

1.206 × 3431 %

1

1.8322' 24.6mm/s

relative to the acoustic pressure is:

At 10m, and the phaseu '0.8945

1.206 × 3431 %

1

18.322' 2.2mm/s

relative to the acoustic pressure is:

Problem 5.3

Source volume velocity = 4πr2urms = 4π × 0.012 × 0.5 = 6.28 × 10-4

At 100Hz, k = 2π × 100/343 = 1.832 m-1 and at 800Hz, k = 14.655 m-1.Using equation 5.12 in the text, the acoustic power radiated at 100Hz is:

and at 800Hz, the acoustic power radiated is:

The corresponding sound power levels are calculated using:

Thus at 100Hz, Lw = 76.4dB and at 800Hz, Lw = 94.5dB.


Q ' 4 π r 2 u ' 4π × 0.012 × 0.1 ' 1.257 × 10&4 m 3 /s

p 'Qkρc4πr

'1.257 × 10&4 × 0.916 × 1.206 × 343

4π × 10' 3.79 × 10&4 Pa

β ' & tan&1 1kr

φ 'Ar

e j(ωt & kr)

Problem 5.4

(a) The amplitude of the pressure fluctuations can be calculated using

equation 5.13(a) in the text. The amplitude of the volume velocity, ,Qis given by:

At 50Hz, k = 2πf/c = 2π × 50/343 = 0.916, and the amplitude of thepressure fluctuations is then:

(b) Using the equation from problem 5.2(c), the phase of the pressure minusthe phase of the particle velocity is given by:

At r = 0.5m, the above expression gives β = 65.4E and at r = 10m, β =6.2E.

This indicates that close to the source the acoustic pressure field isdominated by near field effects, whereas at 10m from the source, the nearfield effects will be small and the field may be approximated as apropagating plane wave.

Problem 5.5

The radiation impedance per unit area is equivalent to the specific acousticimpedance, Z which is simply p/u.

For outwardly travelling spherical waves in free space, equation 1.40c may bewritten as:


p ' jωρAr

e j(ωt & kr)

u 'A

r 2e j(ωt & kr) %

jkAr

e j(ωt & kr)

'Ar

e j(ωt & kr) 1r% jk

'p

jωρ1r% jk '

pρc

1 &j

kr

pu

' ρc 1 &j

ka

&1

' ρc1 %

jka

1 %1

(ka)2

' ρc(ka)2 % jka

(ka)2 % 1' ρc

ω2 a 2 % jωac

c 2 % ω2 a 2

1

r 2

MMr

r 2 MMr

%1

r 2sinθ

MMθ

sinθ MMθ

%1

r 2sin2θ

M2

Mψ2φ &

1

c 2

M2φ

Mt 2' 0

Using equations 1.6 and 1.7, the acoustic pressure and particle velocity maybe written respectively as:

and

Thus, setting r = a,

Problem 5.6

Follow the analysis in the text described by equations 5.1 to 5.12. In equation5.12, replace the mean square volume velocity Q2 with the product of half thevelocity amplitude and the surface area of the pulsating sphere; that is,

. The required result is then obtained.Q 2 '*U*2

2(4πa 2 )2

Problem 5.7

The wave equation is:


φ ' 2f )(ct & r) (h /r)cosθ

r 2 MφMr

' &2f ))(ct & r) (hr)cosθ & 2f )(ct & r)hcosθ

1

r 2

MMr

r 2 MφMr

' &2f )))(ct & r) (h /r)cosθ & 2f ))(ct & r) (h /r 2)cosθ

% 2f ))(ct & r) (h /r 2)cosθ

sinθMφMθ

' &2f )(ct & r) (h /r) sin2θ

1

r 2sinθ

MMθ

sinθMφMθ

' &2f )(ct & r) (h /r 3)2cosθ

1

r 2sin2θ

M2

Mψ2' 0

&1

c 2

M2φ

Mt 2' &2f )))(ct & r) (h /r)cosθ

&4f )(ct & r) (h /r 3)cosθ

p ' ρ MφMt

'A cosθ

kr1 &

j(kr)

e j(ωt & kr)

ur ' &Lφ 'A cosθkrρc

1 &2

(kr)2&

2 jkr

e j(ωt & kr)

(a) The solution given by equation 5.25 in the text is:

Substituting the solution into the various terms in the wave equationgives:

Adding all the above terms together gives:

which must equal zero to satisfy the wave equation. This is true providedthat r is very large compared to h.

(b) Equations 5.32 and 5.33 in the text may be written as:


φ '1ρ m

T

0

p dt ' &jAcosθ

ρck 2r1 &

jkr

e j(ωt & kr )

&MφMr

' &jAcosθ

ρck 2r 21 &

jkr

e j(ωt & kr )

%jAcosθ

ρck 2r

j

kr 2e j(ωt & kr )

%jAcosθ

ρck 2r1 &

jkr

(&jk)e j(ωt & kr )

r 2 MφMr

'Arcosθρck

1 &2jkr

&2

(kr)2e j(ωt & kr )

1

r 2

MMr

r 2 MφMr

' &Acosθ

r 2ρck1 &

2jkr

&2

(kr)2e j(ωt & kr )

&Arcosθ

r 2ρck

2j

kr 2%

4

k 2 r 3e j(ωt & kr )

&Arcosθ

r 2ρck1 &

2jkr

&2

(kr)2(&jk)e j(ωt & kr )

' &Acosθrρck

&1r%

2

k 2r 3& jk %

2j

kr 2e j(ωt & kr )

Using the first of the above equations and omitting the integrationconstant:

Checking the expression for u, by evaluating -Lφ, we obtain:

To verify that the expression obtained above for φ is a solution to thewave equation we substitute it into the wave equation and calculate theresult term by term as follows:


sinθMφMθ

'jAsin2θ

ρck 2r1 &

jkr

e j(ωt & kr )

1

r 2sinθ

MMθ

sinθMφMθ

'2jAcosθ

ρck 2r 31 &

jkr

e j(ωt & kr )

1

r 2sin2θ

M2

Mψ2' 0

&1

c 2

M2φ

Mt 2'ω2

c 2φ ' &

jk 2 Acosθ

ρck 2r1 &

jkr

e j(ωt & kr )

jAcosθρcr

&j

kr%

2j

(kr)3% 1 &

2

(kr)2%

2

(kr)2&

2j

(kr)3& 1 %

jkr

e j(ωt & kr )

φ 'f (ct & r)

r

r 2 MφMr

' & f (ct & r) & r f )(ct & r)

Adding all the above terms together gives:

which is equal to zero. Thus the solutions given by equations 5.32 and5.33 satisfy the spherical wave equation exactly.

Problem 5.8

Equation 1.35 in the text is:

To verify that the expression obtained above for φ is a solution to the waveequation we substitute it into the wave equation and calculate the result termby term.


1

r 2

MMr

r 2 MφMr

'f )(ct & r)

r 2%

f ))(ct & r)r

&f )(ct & r)

r 2

1

r 2sinθ

MMθ

sinθMφMθ

' 0

1

r 2sin2θ

M2

Mψ2' 0

&1

c 2

M2φ

Mt 2' &

f ))(ct & r)r

Q '4πW

k 2ρc'

4π × 0.5

4.582 × 1.206 × 343' 2.69 × 10&2 m 3 /s

WD ' ρck 4 h 2 Q 2

3π

'1.205 × 343 × 4.584 × (0.08 /2)2 × 2.6912 × 10&4

3π

' 0.0224watts

Adding all the above terms together gives 0, so equation 1.35 in the text is asolution.

Problem 5.9

The wavenumber, k, is equal to (2πf/c) = (2π × 250/343) = 4.58.

The source strength, Q, of the monopole may be calculated using equation5.12 in the text. That is:

The dipole acoustic power, WD, may be calculated using equation 5.29 in thetext to give:


p(r) 'jωρqe&jkr

4πr

p(r) ' p1 % p2 'jωρ4π

q1

r1

e& jkr1 %

q2

r2

e& jkr2

r1 . r % h cosθ and r2 . r & h cosθ

p(r, θ) 'jωρ4πr

e& j kr q1 e& jkhcosθ % q2 e jkhcosθ

' p1(r, θ) 1 %q2

q1

e2jkhcosθ

Sound power level = 10log10W + 120 = 103.5dB.

Problem 5.10

For a single source:

For 2 sources separated by a distance 2h, the total pressure, p, is the sum ofthe pressures p1 and p2 from each source. Thus:

As shown on page 179 in the text, for phase accuracy purposes, the followingapproximations are adequate:

where it has been assumed that h << r.Noting that for amplitude purposes, r1 . r2 . rThe above equation may be rewritten as:

When θ = θ0, p = 0, or

.1 %q2

q1

e2jkhcosθ0 ' 0

Thus, .q2

q1

' &e&2jkhcosθ0

Substituting this into the preceding equation for p gives:


p(r, θ) ' p1(r, θ) 1 & e&2jkh (cosθ0 & cosθ )

*p* ' 2*p1* 1 & cos(2khcosθ)

Q '4πW

k 2ρc'

4π × 0.01

9.162 × 1.206 × 343' 1.903 × 10&3 m 3 /s

ID ' ρck 4 h 2 Q 2

(2πr)2cos2θ

'1.205 × 343 × 9.164 × (0.005 /2)2 × 1.9032 ×10&6

(2π × 0.5)2× 0.7072

' 3.34µWatts /m 2

¢p 2 ¦ ' ρcI ' 1.205 × 343 × 3.33 × 10&6 ' 1.38 × 10&3 Pa 2

If θ0 = 90E, then p(r, θ) ' p1(r, θ) 1 & e2jkhcosθ

Taking the modulus of the preceding equation gives:

This function contains the directivity information and is plotted in the figureabove.

Problem 5.11

The wavenumber, k, is equal to (2πf/c) = (2π × 500/343) = 9.16.

The source strength, Q, of each monopole making up the dipole source maybe calculated using equation 5.12 in the text. That is:

(a) The dipole intensity at θ = 45E is given by equation 5.28 in the text andis:

(b)


Lp ' 10log10¢p 2 ¦p 2

ref

' 65.4dB

Frms '4πaA

3k 21 %

1

(ka)2

A

2'ρchQk 3

2π

A

2'

1.206 × 343 × (0.005 /2) × 1.903 × 10&3 × 9.163

2π' 0.241

Frms 'A

2

4π

3k 2'

0.241 × 4π

3 × 9.162' 12.0 mN

Q 2 'WM 4π

ρck 2

R

h = L =0.05m

0.1m

0.1m

1W

0.5W

0.5W

20m

(c) Required driving force can be calculated by taking the mean square valuecalculated using equation 5.39 in the text. Thus:

From equation 5.35 in the text:

Substituting in the previously calculated values for k and Q gives:

As ka is very small, equation 5.40 in the text may be used. Thus:

Problem 5.12

The arrangement is illustrated in thefigure.

As R is 1m off the floor, thedistance to it is

202 % 12 ' 20.02m

(a) The strength of each source may be calculated using equation 5.12 in thetext:


Q 2 '0.5 × 4π

1.206 × 343 × 2.292' 2.897 × 10&3

¢p 2 ¦ ' 5ρcWlong cos4θ

4πr 2× 2 ' Q 2 ρck 3hLcos2θ

πr

2

× 2

' 2.897 × 10&3 1.206 × 343 × 2.293 × 0.052 × cos2 (30)π × 20.02

2

× 2

' 1.271 × 10&4 Pa 2

Lp ' 10log101.271 × 10&4

4 × 10&10' 55.0dB

Lp ' 121.8 & 10log10(2π × 20.022) & 10log10(400 /413.66) ' 87.9dB

In this case, WM = 0.5W and k = 2π × 125/343 = 2.290. Thus:

The arrangement shown is a longitudinal quadrupole and the mean squaresound pressure at any location may be calculated using equations 5.54and 5.55 in the text. Note that the equation for the mean square pressureis multiplied by 2 in this case because the radiation is into half space.Thus:

The sound pressure level is then:

(b) 125Hz random noise will make the sources act independently and thepower radiated will be the arithmetic sum of the individual sources. ThusW = 1.5W and

.Lw ' 10log10 (1.5 /10&12 ) ' 121.8dB

Using equation 5.108 in the text, the sound pressure level may becalculated using S = 2πr2 and ρc = 413.6. Thus:


¢p 2D ¦

¢p 2M ¦

' 4(kh)2 cos2(90 & 1.145)

' 4(4.58 × 0.1)2 (0.02)2 ' 3.36 × 10&4

Reduction ' 10log10 3.36 × 10&4 &1' 34.7dB

0.1m

hole

speaker

5mO

0.2m

Problem 5.13

(a) T h e a r r a n g e m e n tapproximates a simple dipoleand is illustrated in the figure.The angle β = sin-1 0.1/5 =1.145E. The azimuthal anglegiven in the problem is irrelevant.

The wavenumber, k = (2π ×250)/343 = 4.58.

The sound pressure levels radiated by the hole alone (monopole) and hole+ speaker (dipole) may be calculated using equations 5.13a, 5.30b and5.29 in the text.

Using these equations, the ratio of the mean square pressures(dipole/monopole) may be written as:

Thus the reduction in sound pressure level due to the presence of thespeaker is:

(b) If a speaker is placed below the hole as well, a longitudinal quadrupoleis formed with L = h = 0.1. In this case, β = 0, and as can be seen fromequation 5.55 in the text (where θ = 90 - β), the theoretical mean squaresound pressure will be zero, implying a reduction of infinity dB.

Problem 5.14

In equation 5.62, (W/b) is effectively the power per unit length of source (asW is the power of each source separated by b) and in equation 5.70, W/D isthe same quantity. Thus the difference between the finite length and infinite


¢p 2 ¦ ' ρcW

2π2 r0 D[αu & αl]

Lp ' Lw & 10log10 (4πr0D ) % 10log10 (αu & αl ) % 10log10 (ρc /400)

' 130 & 10log10 (4π × 80 × 20) % 10log10 (0.248) % 10log10 (1.0342)

' 81.1dB

u

l20m

80m

A

pipe

length source is the quantity which is the ratio of the angle(αu & αl ) /π

subtended by the source at the observer in each case. Thus, by logicalargument, equation 5.65 can be rewritten as follows for a finite coherent linesource.

Problem 5.15

The arrangement is as shown in the figure.

αu ' &αl ' tan&1(10 /80) ' 0.124c

Finite length pipe, D = 20m, r0 = 80m and Lw = 130dB. Turbulent flow, soassume an incoherent source. Also assume incoherent addition of the directand ground reflected waves. Taking logs of equation 5.70 in the text gives forthe direct wave:

The ground reflected wave level is then 81.1 - 3 = 78.1dB. Thus the total


Lp ' 10log10 108.11 % 107.81 ' 82.9dB

¢p 2 ¦ ' [Wρc /4πr0D ] [αu & αl ] × 2

¢p 2 ¦ ' [2 × 1.206 × 343/(4 π × 200 × 50)] × 0.249 × 2

' 3.275 × 10&3 Pa 2

Lp ' 10log103.275 × 10&3

4 × 10&10' 69.1dB

u

l50m

200m

A

pipe

level at the receiver is:

Problem 5.16

The situation is as shown in the figure below. Equation 5.70 in the text maybe used to calculate the sound pressure level. The equation must be multipliedby the directivity factor (2 in this case).

Thus:

.αu ' αl ' tan&1 25200

' 0.124 radians

W = 2, r0 = 200, D = 50. Thus:


Ag ' &10log10 [1 % 10&2/10 ] ' &2.1dB

¢p 21 ¦ ' ρc

W

4πr 20

× DF

¢p 22 ¦ ' ρc

W4br0

× DF

¢p 21 ¦

¢p 22 ¦

'br2

π × r 21

'6 × 50π

' 95.5

From table 5.3 in the text, Aa = 19.3dB per 1000m, so for 200m, Aa = 19.3/5= 3.9dB. Sound intensity loss due to ground reflection is 2dB. Thus theground effect is given by equation 5.175b in the text as:

Thus, Ag + Aa = 1.8dB and the sound pressure level at the receiver is:

Lp = 69.1 - 1.8 = 67.3dB

Problem 5.17

The traffic may be treated as an infinite line source. At 1m r0 < b/π and themean square sound pressure is related to the source sound power, W, of onevehicle by:

where DF is the directivity factor for the source/ground combination.

At 50m, the sound pressure is related to the sound power, W, of one vehicleby:

Thus:

The level at 50m = level at 1m - 10log10 = 88 - 19.8 = 68.2dB(A).¢p 2

1 ¦¢p 2

2 ¦


I ' (p 2o /3ρcr 2) (2 % cosθ)

r

¢p 2 ¦ ' ρc W4br0

'413 × 2

4 × 7 × 250' 0.118 Pa 2

Lp ' 10log100.118

(2 × 10&5 )2' 84.7 dB

Lp ' 84.7 & 0.6 % 3.0 ' 87 dB (%6, &3 dB)

Problem 5.18

From Equation 5.62:

Concrete ground, so ground effect is Ag = -3 dB. Assume 20EC, airabsorption ranges from 2.6 to 2.8 dB per 1000 m. For 250 m air absorption -0.7 dB.From table 5.3, meteorological influence is +6, -3 dB. Assume no obstaclesblocking line of sight to the road from the residence.

So the sound pressure level at the residence is:

Problem 5.19

(a) The arrangement is shown in thefigure. We are given:

The sound power is obtained byintegrating the intensity over animaginary hemispherical surfacecentred at the centre of the speaker.

The sound power is then:


W ' mS

I dS 'p0

2

3ρc m2π

0

dψ mπ /2

0

(2 % cosθ)r 2 sinθ

r 2dθ

'p 2

0 2π

3ρc mπ /2

0

(2sinθ % cosθ sinθ ) dθ

'p 2

0 2π

3ρc.

π /2

0& 2cosθ & 0.25 cos2θ

'p 2

0 2π

3ρc2 % 1/2 ' 0.0127p 2

0 Watts

(b)& (c) As the speaker only radiates into a hemispherical space, thepresence of a baffle will have no influence on the powerradiated, regardless of whether the source is constantvolume or constant pressure.

Problem 5.20

(a) If the piston is assumed to be made up of an infinite number of pointmonopole sources, all pulsating in phase, then the sound pressure at anylocation can be calculated by summing the contributions from eachsource. This effectively means that an expression for the sound pressureat some distance, r, due to a monopole on the piston surface must beintegrated over the piston surface. If a baffle is present, the monopolesource must be replaced with a hemispherical source which radiates twicethe pressure.

(b) The function 2J1(x)/x vs x is plotted out in the figure below, where x = ka sinθ.


0 5 10 15 20

0.5

1.0

2J ( )1 xx

x

At 500Hz, ka = 2π × 500 × 0.1/343 = 0.91. Nodes in the radiationpattern occur when 0.91 sinθ = 3.8, 7, 10.1, etc. That is, there are nonodes.

At 2,500Hz, ka = 2π × 2,500 × 0.1/343 = 4.58. Nodes in the radiationpattern occur when 4.58 sinθ = 3.8, 7, 10.1, etc. That is, there is only onenode at θ = 60E.

At 10,000Hz, ka = 2π × 10,000 × 0.1/343 = 18.31. Nodes in theradiation pattern occur when 18.31 sinθ = 3.8, 7, 10.1, 13.3, 16.4, and19.6. That is, there are 5 nodes at θ = 12E, 23E, 33E, 47E and 64E.

Directivity patterns are shown for each of these cases in the figures onthe next page. The side lobes for the 10kHz case have been expanded forclarity. For sketching purposes, the figure shown at the beginning of part(b) (figure 5.7 in the text) may be used with the x-axis crossingsrepresenting the nodal locations of each lobe and the peaks in the curverepresenting the relative amplitude of each lobe. [Note that fig 5.6 willonly provide information for the first three lobes so it must be extendedfor the 10kHz case.]


RR ' ρcπa 2 × (2ka)2 /8 ' ρc(πa 2)24ω2 / (8πc 2) ' ρcS 2ω2 / (2πc 2)

σ 'RR

ρcS'

πa 2ω2

2πc 2' (ka)2 /2

W ' RRπa 2ρcU 2 /2

500Hz 2.5kHz 10kHz

Problem 5.21

(a) Using equations 5.95b and 5.96 in the text, the radiation resistance for apiston can be written as:

where it has been assumed that ka is sufficiently small that all but thefirst term of equation 5.96 in the text is negligible.

(b) The radiation efficiency at low frequencies is:

which described the solid line in figure 5.9 in the text for ka < 0.8.

(c) See fig 5.9 in the text.

Problem 5.22

(a) Piston radiating from an infinite baffle. k = 2πa/λ = 2 anda = 0.1m. From equation 5.98b in the text:


U ' ξω ' 2ξc /λ ' 2ξc /a ' 2 × 0.0002 × 343/0.1 ' 1.372m/s

W ' π × 0.12 × 1.206 × 343 × 1.3722 /2 ' 12.2W

I 'ρck 2

8π2r 2F 2(w)

I 'ρch 2U 2π2a 4

8π2r 2'

ρcU 2a 2

8r 2

'1.206 × 343 × 1.3722 × 0.01

2 × 22' 0.97W/m 2

Lp ' 10log10¢p 2 ¦p 2

ref

' 10log10ρcI

4 × 10&10' 120.0dB

and from fig 5.8, for ka = 2, RR = 1. The piston velocity amplitude, U, isgiven by:

Thus the radiated power is:

(b) From equation 5.84 in the text, the on-axis intensity is:

where w = kasinθ = 0. Thus, F(0) = Uπa2.

Thus:

(c) Radiation mass loading = πa2ρc[X(2ka)], where X(2ka) = 0.55 (see fig5.8 in the text). Thus the mass loading isπ × 0.01 × 1.206 × 343 × 0.55 ' 7.1 kg/s

(d) Sound pressure level at 2m:

Problem 5.23

(a) The arrangement is shown in the figure on the next page where it can beseen that h = L = 102.5mm.


WM ' (Q 2H ρck 2 /4π) × 2

Wlong ' [ (2k 3hLQL )2ρc /5π ] × 2

Wlong

WM

'4k 6h 2L 2Q 2

H ρc4π

20πQ 2H ρck 2

'45

k 4h 2L 2

'45

(2π)4 f 4 (0.1025)4

3434' 9.94 × 10&12 f 4

hole

speaker speaker

200mm

205mm

(b) Power radiated by original opening (assuming a constant volume velocitysource in an infinite baffle) is (see equation 5.12 in the text and multiplyby 2 to account for radiation into half space):

The power radiated by the longitudinal quadrupole may be calculatedusing equation 5.54 in the text. Again the equation in the text must bemultiplied by 2. Thus:

where QL = QH/2. Thus:

Frequency, Hz dB reduction(-10log10(Wlong/Wm)

63125250500

3826142


¢p 2long ¦

¢p 2M ¦

'[5ρcWlong cos4θ /4πr 2 ] × 2

[WM ρc /4πr 2 ] × 2' 5cos4θ

Wlong

WM

ΔLp ' &10log10 [5cos4θWlong /WM ]

Lp ' Lw & 10log10 [4πr 2] ' 120 & 10log10 [2π × 100] ' 92dB

(c) The ratio of the mean square pressures may be obtained using equations5.13b and 5.55 in the text. Thus:

The reduction in sound pressure level is then:

Values of sound pressure reduction (dB re 20µPa) are tabulated belowfor the required values of θ.

frequency(Hz)

θ = 0 θ = π/4 θ = π/2

63125

3119

3725

44

When the speakers are turned on, the sound field amplitude distributionhas 2 lobes with maxima at θ = 0, π and minima at θ = π/2, 3π/2.

Problem 5.24

(a) The directivity index due to the hard floor is 3dB.

(b) The expected sound pressure level due to an omni directional source ona hard floor is:

The actual sound pressure level is 110dB so the directivity due to thesource characteristics is -18dB.


¢p 2 ¦ ' 2ρcWπHL

tan&1 HL

2r H 2 % L 2 % 4r 2

'2 × 1.206 × 343 × 0.01 × S

S × πtan&1 0.25

2 × 25 0.25 % 0.25 % 4 × 252

' 2.63 × 10&4 Pa 2

Lp ' 10log102 × 2.633 × 10&4

4 × 10&10' 61.2dB

hole

2m

25mProblem 5.25

The radiated sound poweris W = IS = 0.01 × S =0.01S Watts. Thearrangement is shown inthe figure.

With no ground reflection, the on-axis sound pressure level may be calculatedusing equation 5.105 in the text (as we may assume that the opening behaveslike an incoherent plane source). Here, H = L = 0.5, S = HL and r = 25. Thus:

Assuming a similar travel distance for the ground reflected wave, the totalsound pressure level at the receiver is:

Interestingly, in this case, the receiver is sufficiently far from the source forthe source to appear as a point source and the same result would have beenobtained if equation 5.106 had been used instead of equation 5.105.


ρfR1

'1.206 × 2000

2.25 × 105' 0.011

βR1

ρf

1/2

' 1.721

0.011

1/2

' 16.6

Ag ' &10log10 1 % 10&AR /10

' &2.4dB

hole

3m

150m

1.5m

x y

R

Problem 5.26

(a) Sound power level, Lw = 10log10W + 120 = 123dB.

(b) The arrangement for calculating the ground effect is shown in the figure.

From similar triangles, x = 2y. Thus x = 100 and y = 50.

tanβ = 3/100, so β = 1.72E.

For grass covered ground, R1 = 2.25 × 105 (middle of range). Thus:

From figure 5.20 in the text, AR = 1.3dB. The ground effect is then:

Thus the effect of the ground is to increase the level at the receiver by2.4dB.

(c) Loss due to atmospheric absorption. From table 5.3 in the text,Aa = 15.5dB per 1000m (25% RH and 20EC). So for a distance of 150m,Aa = 15.5 × 0.15 = 2.3dB.


¢p 2 ¦ ' 1.206 × 343 × 2

2π × 1502' 5.852 × 10&3 Pa 2

Lp ' 10log105.852 × 10&3

4 × 10&10' 71.7dB

5

90

10000

10000 - xx

r1

r2

r

(d) The opening should be treated as an incoherent plane source and equation5.105 in the text used to calculate the sound pressure level. However aswe found in the previous problem, the receiver is far enough from thesource for it to appear as a point source (see figure 5.11) and we may useequation 5.106. Thus:

The sound pressure level is then:

Aa + Ag = -0.1, thus the sound pressure level at the receiver is equal to71.8dB.

(e) Adding a second opening will add 3dB to the sound pressure levels at thereceiver as it may be assumed that the sound fields from the two sourcesare incoherent. Thus the sound pressure level at the community locationof (b) above would be 74.8dB.

Problem 5.27

The arrangement is illustrated in the figure. Using similar triangles:


x5'

10000 & x90

90x ' 50000 & 5x

x ' 50000/95 ' 526.31579m

r1 ' 52 % (526.315)2 ' 526.33954

r2 ' 902 % (10000 & 526.31579)2 ' 9474.111701

r ' 852 % 100002 ' 10000.36124

r1 % r2 & r ' 0.0900m

f 'cλ

'1500

2 × 0.0900' 8.3kHz

Destructive interference will occur if λ/2 = 0.0900m. This occurs at afrequency, f, given by:

Problem 5.28

We would NOT expect to measure 84dB(A) at the operator's position due toreflected energy from the nearby wall. As the sound is predominantly in the500Hz to 2000Hz band, the operator is not in the hydrodynamic near field ofthe machine (although he/she could be in the geometric near field). Also themachine is far enough from the wall for its sound power to be unaffected bythe wall. Assuming incoherent addition of direct and reflected waves,assuming that geometric near field effects are negligible assuming that themachine does not act as a barrier to the reflected sound and assuming that theloss on reflection from the wall is negligible, the level at the operator'sposition may be calculated as the logarithmic sum of the direct and reflectedwaves. The reflected wave path length is 6m and the direct path length is 2m.Thus the sound pressure level of the reflected wave is


Lp ' 10log10 108.4 % 107.45 ' 84.5dB(A)

Rp '(Zs /ρc) & 1

(Zs /ρc) % 1

Rp '(Zs /ρc)cosθ & cosψ

(Zs /ρc)cosθ % cosψ

*Rp* '* (Zs /ρc)cosθ & 1*

* (Zs /ρc)cosθ % 1*

. Thus the total sound pressure level expected84 & 20log10(6 /2) ' 74.5dB

at the operator's position is:

Problem 5.29

(a) Specific acoustic impedance is a complex quantity characterised by anamplitude and a phase and is the complex ratio of acoustic pressure toacoustic particle velocity at any point in an acoustic medium, includingthe interface between two different media. Characteristic impedance isequal to ρc. For an acoustic medium where the viscous and thermallosses are small (such as air or water), it is a real quantity and is equal tothe specific acoustic impedance of a plane wave propagating in anacoustic medium of infinite size.

(b) The absorption coefficient (assuming plane incident waves) is defined interms of the reflection coefficient, Rp, as α = 1 - *Rp*

2. Also, the normalspecific acoustic impedance of the surface of an acoustic medium ofinfinite extent is the characteristic impedance of the medium, for aninfinitely thick medium. Thus equation 5.129 in the text may be usedwith θ = 0 to give:

If θ is not equal to 0, then

If it is assumed that the wavenumber in the material is much larger thanthat in air, cosψ = 1 and:

The maximum absorption coefficient will occur when the modulussquared of the reflection coefficient is a minimum; that is, when


*Rp*2 '

(2cosθ & 1)2 % 9cos2θ

(2cosθ % 1)2 % 9cos2θ'

13cos2θ & 4cosθ % 1

13cos2θ % 4cosθ % 1

α ' 1 & *Rp*2 ' 1 &

13 × 0.0769 & 4 × 0.2774 % 113 × 0.0769 % 4 × 0.2774 % 1

' 1 & 0.286 ' 0.71

pp0

' e&αx

Decay rate ' 20log10 eα ' 20 × 0.4343ln eα ' 8.686α (dB/m)

Lp ' Lw & 10log10 (2πr 2 ) & Ag & Aa & Am'110 & 65.5 % 3 & 2.7 × 0.75 &Am

' 45.5 dB (%8, &6 dB)

is a mimimum. Differentiating the above expression wrt cos2 using thechain rule, and setting the result equal to zero, we obtain the minimumvalue when cosθ = 0.2774 or θ = 74E. The corresponding maximumvalue of the absorption coefficient is given by:

Problem 5.30

Power radiated by window = 0.1 W = 110 dBConcrete ground, so Ag = -3 dBAssume 20 EC temperature, so Aa ranges from 2.6 to 2.8. Use Aa = 2.7.Range due to meteorological conditions is: (+8, -6 dB) from Table 5.10.Assume no barriers between the source and receiver.The receiver is far enough away for the window to be treated as a point sourcein a baffle. Thus:

Problem 5.31

The pressure amplitude at any location x is given by:

where is the amplitude at x = 0. The decibel decay rate per unit distancep0

is 20log10 of the reciprocal of the above expression when x = 1 and is thusgiven by:

Important factors are air temperature and humidity.


A 2

600

r - 600 r

Problem 5.32

The situation is illustrated in the figure below. We need to find the radius ofcurvature, r, of the wave and hence the distance, d. We may use the aircraftas the reference frame. Thus we assume a coordinate system movinghorizontally at the speed of the aircraft and then later on calculate the distancethat the aircraft travels during the time it takes for the sound to reach theground (with an assumed aircraft speed). The wave which hits the ground atgrazing incidence will be the one heard first. The total sonic gradient is 1/10= 0.1s-1. Using equation 5.190 in the text, the radius of curvature of the waveis thus 343 × 10 = 3430m. The distance d is then:


d ' r 2 & (r & 600)2 ' 34302 & 28302 ' 1940m

θ0 ' cos&1 3430 & 6003430

' 34.4E

t ' mθ0

0

r dθc0 & 0.1h

' mθ0

0

dθ(c0 /r) & 0.1 % 0.1cosθ

' 10 × mθ0

0

dθcosθ

' 10 loge [ secθ % tanθ ]34.4E0 ' 6.4seconds

We now need to take into account the speed of the aircraft and the distance itwill travel in the time the sound wave travels to the observer.

Let the angle subtended at the centre of the circular arc shown in the previousfigure be θ0. The speed of sound as a function of θ (with θ = 0 correspondingto ground level and θ = θ0 corresponding to the aircraft level) is given by cθ =c0 - 0.1h, where h = height above the ground and c0 = 343m/s is the speed ofsound at ground level. The value of θ0 is given by

Thus the time taken for the sound to travel from the aircraft to the ground is

If the aircraft travels at 400km/hour (not given in the question), then it wouldtravel 710m in 6.4 seconds.

Thus the aircraft emerges from ground shadow (1940m - 710m) = 1230mfrom the observer.

6

Solutions to problemsrelating to sound power,its use and measurement

Problem 6.1

(a) This is discussed in detail on p246–247 in the text.

(b) For a constant power source in the corner of the room, the radiatedpower would be concentrated over a one eighth sphere instead of asphere and the sound pressure level in the direct field would thus beincreased by 9dB. However, there would be no change in thereverberant field sound pressure level.

For a constant volume velocity source in the corner of the room, theradiated power would be increased by a factor of 8 due to there being3 reflecting surfaces and in addition the power would beconcentrated over a one eighth sphere instead of a sphere and thedirect field sound pressure level would thus be increased by 18dB.However, the reverberant field sound pressure level would beincreased by only 9dB, corresponding to the power increase.

For a constant pressure source in the corner of the room, the directfield radiated sound pressure level would be unchanged, but thereverberant field sound pressure level would be reduced by 9dB,corresponding to a reduction of 9dB in the radiated power.

(c) A good approximation would be a constant pressure source modelbecause the noise is originally generated by a fluctuating pressure,the amplitude of which is controlled by the aerodynamics and not theacoustics of the problem.


10log10W ' 10log10¢p 2 ¦ % 10log10 S & 10log10(ρc)

10log10W

Wref

' 10log10¢p 2 ¦p 2

ref

% 10log10 S % 10log10(400) & 10log10(ρc)

Lw ' Lp % 10log10 S

Lw ' Lp % 10log10 S ' 85 % 10log10(2π × 22)

' 85 % 14 ' 99dB

Problem 6.2

(a) Sound power level is a measure of the rate of total energy radiatedby an acoustic source while sound pressure level is a measure of thefluctuating sound pressure at a particular location. Sound powerlevel is a source property whereas sound pressure level depends onthe measurement location as well as the strength and size of thesource.

(b) Beginning with and taking logs of both sides gives:W ' ¢p 2 ¦S /ρc

Dividing both sides by 10-12 and remembering that the referencesound power level is 10-12W and the reference sound pressure levelis 2 × 10-5Pa, the preceding equation may be written as:

If the quantity ρc is approximated as 400, then the precedingequation becomes:

(c) Using the above equation, the sound power level would be:

Assumptions: ρc = 400 and the sound pressure level measurementswere made in the acoustic far field of the machine.

(d) The desirable quantity is usually sound power level as it indicates theamount of acoustic energy which will be added to an environmentand allows the increase in sound pressure level to be calculated at


r > 3λ /2π; r > 3R; r > 3π R2 /2λ

any location in the far field of the source as a result of theintroduction of the machine (provided the sound radiation isomnidirectional or directivity information is also given). Soundpower is also independent of the presence of nearby reflectingsurfaces, provided that these are more that a quarter of a wavelengthfrom the acoustic centre of the machine or noise source.

On the other hand, the sound pressure level at a specified location isaffected by the presence of reflecting surfaces and also does notnecessarily allow the sound pressure level at other locations to becalculated. However, if only the noise exposure of the operator ofa machine is of concern, then perhaps it is better to specify soundpressure level at the operator's location measured in the presence ofspecified reflecting surfaces than sound power level as the operatormay not be in the acoustic far field of the source.

Problem 6.3

(a) This is discussed in detail on pages 249-251 in the text.

(b) Assuming that "negligible" is a factor of 10, the frequency would begiven by (see figure 6.1 in the text).γ ' 10/κ and therefore2r / R ' 10λ / (π R ) ' 10 × 343/(π R f )f ' 10 × 343/(2πr) ' 3430/(2 × π × 1) . 550Hz

(c) r = R = 1, so γ = 2 and κ = π × 550/343 . 5. From figure 6.1, for thefar field to be dominant, γ = 15. Thus r = 7.5m is the distance atwhich the far field would be dominant.

Problem 6.4

(a) The anechoic room should be sufficiently large that sound pressuremeasurements can be made in the far field of the source. Thus thefollowing criteria should be satisfied (see eq. 6.5, p.250 in text).

The lowest frequency of interest is 44Hz (see table 1.2 on p43 in thetext) and the highest frequency is 11,300Hz.

Sound power use and measurement 135

2 × 6 (4.5 % 0.75 % 7.8 /4) by (3.72 % 0.2 % 7.8 /4) >by (3.72 % 0.6 % 7.8 /4) >

' 14.4m × 11.7m × 6.3m

1.3λ3 ' 1.334363

3

' 210m 3

W '¢p 2 ¦Sρc

'¢p 2 ¦2πr 2

ρc

frequency(Hz)

λ (m) 3λ/2π 3R 3πR2 /2λ

R = 1.5 R = 0.4 R = 0.6

4411,300

7.80.03

3.720.015

4.5, 1.2, 1.8 1.36349

0.124.8

0.2255.9

The last criterion in the above table for r represents the transitionfrom the geometric near field to the far field. Thus for highfrequencies it is not practical to take measurements in the far field;the geometric near field will suffice. As all measurements must betaken at least λ/4 from the room walls, the following minimuminterior room dimensions are needed.

As standard measurements use a hemispherical surface, the requiredroom dimensions are 14.4m × 14.4m × 7.2m high.

(b) For measurements in octave bands, the required room volume is:

Optimum dimensions are in the ratios 2:3:5. Thus 2x × 3x × 5x =210, or x = 1.91m. Thus the required dimensions are 3.8m × 5.7m× 9.6m.

Problem 6.5

The sound power is related to the average mean square pressure by:

or in terms of sound power level, Lw:


Lw ' Lp % 10log10(2πr 2) % 10log10400 /ρc

' 70 % 10log10(8π) % 10log10400 /413.6

' 83.9dB

W ' 10&12 × 10Lw/10

' 10&12 % 8.39 ' 0.245mW

3λ /2π; 3 R; and 3π R2 /2λ

¢p 2 ¦ ' (Qkρc)2

(4πr)2

The sound power, W, is thus:

Problem 6.6

(a) r.m.s. acoustic pressure:

prms ' 2 × 10&5 × 10Lp /20

' 2 × 103.3 & 5 ' 0.04Pa

(b) Source dimension, R = 0.1m; wavelength, λ = 1481/250 = 5.92m;r = 2m.The distance r must be larger than the following quantities for thelocation to be in the far field:

Substituting values for λ and l into the preceding expressions andevaluating gives 2.82, 0.3 and 0.008 respectively, so the location isnot in the far field but rather in the transition between thehydrodynamic near field and the far field.

(c) Intensity:

I 'p 2

rms

ρc'

0.042

998 × 1481' 1.08nano&watts /m 2

(d) Power, W ' I S ' 1.08 × 10&9 × 2 × π × 22 ' 0.027µW

(e) The mean square sound pressure is related to the source volumevelocity, Q, for a spherical source by equation 5.13a in the text as:


d '2Q

2πf × 4πa 2' 1.791Q / f

d '3.582r prms

ρ f 2

d '3.582 × 2 × 0.04

998 × 2502' 4.6 × 10&9 m

Lw ' Lp % 10log10 V & 10log10 T60 % 10log10 1 %Sλ8V

& 13.9dB

In this case, the source is a hemisphere and it is radiating intohemispherical space. Thus the same result is obtained as for aspherical source radiating into spherical space. The volume velocity,Q, is related to the surface displacement, d, by:

Combining the above two equations gives:

Substituting values for the variables gives:

Problem 6.7

Equation 6.13 in the text is:

From problem 6.4 the room size is 3.8m × 5.7m × 9.6m. The volume, V, is207.9m3 and the surface area, S, is 2(3.8 × 5.7 + 3.8 × 9.6 + 5.7 × 9.6) =225.7m2.

Using the above equation, the following table may be constructed.

f Lp 10log10T60 λ 10log10(1+Sλ/8V) Lw

631252505001000200040008000

851051009095989088

9.18.87.46.55.44.03.01.8

5.442.741.370.690.340.17

0.0860.043

2.41.40.70.40.20.10.10.0

87.6106.9102.693.299.1103.496.495.5

Overall 110.3


Lp(av) ' 10log101N j

i

10Lpi /10

' 86.8dB

Lw ' Lp % 10log10 S & Δ1 & Δ2

Problem 6.8

Average Lp = 10 log10 (1/5)[108.5 + 108.3 + 108.0 + 108.7 + 108.6] = 84.8dB

Area of measurement surface = 2(2 × 4 + 2 × 3) + 3 × 4 = 40m2.

Thus radiated power, (eq. 6.25 in text).Lw ' Lp % 10log10S & Δ1 & Δ2

Ratio of area of measurement surface to machine surface = 40/8 = 5, soΔ2 = 0.

Assuming that the machine is in a large enclosure, Δ1 = 0 as well. ThusLw = 84.8 + 10log1040 = 100.8dB.

Problem 6.9

The data may be used to construct the following table.

meas-ured

85 88 86 90 84 85 87 88 89 90 90 88 87 88 89 85

back-ground

80 82 80 81 80 79 81 79 80 81 83 83 82 80 80 79

machineonly

83.3

86.7 84.7 89.4 81.8 83.7 85.7 87.4 88.4 89.4 89 86.3 85.3 87 88.4 83.7

The sound power level may be calculated using equation 6.25 in the textwhich is:

where S = 40m2, and the factory volume V = 20 × 20 × 5 = 2000m2. Thus V/S= 50 and from table 6.4, p. 266 in the text, Δ1 = 2.5dB. Machine surface areais the area of a cube of dimensions 1m smaller than the test cube. Let the sideof the test cube = x. Then 5x2 = 40 and x = 2.828m. Thus the machine sizeis 0.828m × 0.828m × 1.828m high (as the machine is resting on the floor).The machine surface area is then Sm = 1.828 × 0.828 × 4 + 0.828 × 0.828 =


Lw ' 86.8 % 10log10 40 & 2.5 ' 100.3dB re 10&12 W

Lw ' 86.8 % 10log10 40 & 3 . 100dB re 10&12W

4R

'(1 /188) & (1 /428) [102/10]

102/10 & 1' 2.763 × 10&3

Lw ' Lp2 & 10 log10 [S &11 & S &1

2 ] % 10 log10 [10(Lp1 & Lp2 ) /10

& 1]

& 10log10ρc

400

Lw ' 84 & 10log10 188&1 & 428&1 % 10log10 102/10 & 1 & 10log10413.7400

' 84 %25.25 & 2.33 & 0.15 ' 106.8dB re 10&12W

6.7m2. Thus S/Sm = 40/6.7 = 5.9 and from table 6.3, p266 in the text, Δ2 = 0.

Assuming ρc = 400:

Problem 6.10

Second surface average = 86.8 - 2 = 84.8dB. Area ratio (surface1 to surface2)= 40/120 = 0.3333. From figure 6.3 or equation 6.27 in the text, Δ1 = 2.6dBand from problem 6.8, Δ2 = 0. Thus:

Problem 6.11

Machine surface area = 2(8 × 3 + 4 × 3) + 8 × 4 = 104m2. Area of test surface1m from machine = 2(10 × 4 + 6 × 4) + 10 × 6 = 188m2 = S1. Area of testsurface 3m from machine = 2(14 × 6 + 10 × 6) + 14 × 10 = 428m2 = S2. Usingequation 6.22 in the text:

Thus R = 1448 m2.

Sound power level, Lw is calculated using equation 6.24 in the text which is:

Thus:


Lp2 ' 10log101

11108.7 % 108.75 % 108.6 % 108.5 % 108.65

% 108.8 % 108.68 % 108.72 % 108.6 % 108.58% 108.53

' 86.6dB

Lp1 ' 10log10 1090/10 & 1080/10 ' 89.5dB

Lp2 ' 10log10 1086.6/10 & 1080/10 ' 85.5dB

Δ1 ' 89.5 & 85.5 & 10log10 [100.4 & 1] % 10log10 [1 & 96/280]' 0.4dB

However, this excludes the near field correction term of equation (6.25) in thetext, which for this case is -1 dB. So if we used equation 6.25 and also appliedthe correction due to ρc not equal to400, the result would be Lw = 105.8 dB.

Problem 6.12

(a) The average noise level measured on the larger surface is calculatedby logarithmically averaging the given values. Thus:

The background level may be subtracted from the overall averagedlevels as this will give the same result as subtracting it from theindividual levels and then averaging.

Thus the noise level on test surface 1 due only to the machine is:

and the noise on test surface 2 due only to the machine is:

(b) Reverberant field correction, Δ1, use equation 6.27 in the text. Firstcalculate the areas of the test surfaces.Surface 1, S1 = 2(6 × 2.5 × 2) + 6 × 6 = 96m2.Surface 2, S2 = 2(10 × 4.5 × 2) + 10 × 10 = 280m2. Thus:

(c) Correction for non-normal sound propagation is Δ2. Machinesurface area, Sm = 2(5 × 2 × 2) + 5 × 5 = 65m2.S1 /Sm = 96/65 = 1.48. Thus from table 6.3 in the text, Δ2 = 1.


Lw ' Lp % 10log10 S & Δ1 & Δ2

' 89.5 % 10log10(96) & 0.4 & 1 ' 107.9dB

Lw ' Lp % 10log10(2πr 2) % 10log10400ρc

' 75 % 17.5 & 0.1 ' 92.4dB(A)

Lp ' 10log10 1085/10 & 1082/10 ' 82dB(A)

Lw ' 87.4 & 10log10 3 ' 82.6dB(A)

(d) Sound power level, Lw is calculated using equation 6.25 in the textwhich is:

Problem 6.13

(a) sound power level:

(b) Sound power level of existing machinery is 92.4 + 82 - 87 =87.4dB(A).

(c) Maximum allowable total reverberant sound pressure level generatedby the three new machines is:

Thus the maximum allowable sound power level is 87.4dB(A).

(d) If all new machines emit the same sound power level and the totalallowed is 87.4dB(A), the upper bound on the level generated byeach machine is:

Problem 6.14

(a) Advantages of sound intensity for sound power measurement:! reduces errors arising from presence of reflecting surfaces;! reduces errors from near field effects resulting from measurements

taken close to the source;! reduces errors caused by background noise generated by sources

other than the one under test;! Allows good results to be obtained at low frequencies.

Disadvantages of sound intensity for sound power measurement:


σ 'W

Sρc ¢v 2 ¦

! Usually more time consuming;! instrumentation is more expensive;! When the radiated sound field is complex, sound intensity

measurements can provide too much data which is time consumingto analyse and can be confusing.

(b) Advantages of sound intensity for transmission loss measurement:! reduces errors arising from flanking path transmission;! only requires a single reverberant room rather than two;! Allows good results to be obtained at low frequencies.

Disadvantages of sound intensity for transmission loss measurement:! Usually more time consuming;! instrumentation is more expensive;

(c) Advantages of sound intensity for localisation and identification of noisesources:! more reliable than a directional microphone due to greater spatial

resolution and the ability to measure very close to a source;! thus contamination from other nearby sources is reduced;

Disadvantages of sound intensity for localisation and identification ofnoise sources:! instrumentation is more expensive;

Problem 6.15

(a) "radiation efficiency" is a measure of the amount of sound powerradiated by a vibrating surface compared to that carried by a plane wavehaving the same mean square acoustic velocity (as the vibrating surface)and equal area. It can be expressed as:

(b) Referring to the above equation, it can be seen that the sound powerradiated by a surface of area S, mean square velocity and radiation¢v 2 ¦efficiency σ is:


W ' σSρc ¢v 2 ¦

Lw ' 10log10 ¢v 2 ¦ % 10log10 S % 10log10σ % 146 dB re 10&12W

To identify possible paths of sound transmission between rooms, theradiation efficiencies of all walls ceiling and floor can be used togetherwith their measured mean square velocities and the above equation tocalculate the relative contributions of each surface to the overall soundpower transmitted into the room, thus allowing the flanking paths to beidentified and ranked.

(c) The radiation efficiency of a surface is close to one when the bendingwavelength of bending waves in the surface is less than or equal to thewavelength of the radiated acoustic waves; that is, at frequencies equalto or above (and in practice just below, for finite size surfaces) thecritical frequency of the surface. In this case there will always be someradiation angle at which the bending waves on the surface will match thetrace acoustic wavelength in the surrounding medium with a resultingstrong coupling between the surface vibration field and the radiatedacoustic field.

Problem 6.16

(a) Equation 6.31 in the text is:

The critical frequency, fc ' 0.551c 2 / (cL h) ' 0.551 × 3432/(5400 × 0.003) ' 4001 Hz

10log10S = 0.Ph/S = 4 × 0.003/1 = 0.012.


Lp ' Lw & 10log10(2πr 2) % 10log10ρc

400

Lp ' 10log10 104.43 % 105.39 % 104.81 ' 55.3dB re 20µPa

LpA ' 10log10 10(4.43 & 0.86) % 10(5.39 & 0.3) % 104.81 ' 52.8dB(A)

Thus the following table may be constructed.

Octave bandcentre

frequency(Hz)

f/fc 10log10σ 10log10¢v 2 ¦ Lw (dB re 10-12W)

2505001000

0.060.120.24

-19.8-18.2-16.0

-54.0-46.0-54.0

72.281.876.0

(b) From figure 5.11 in the text we can see that for , the panelr / HL ' 10will appear as a point source and the sound pressure level is given by:

The quantity 10log102πr2 = 28 and 10log10(ρc/400) = 0.1.

Thus the following table can be constructed.


Lw (dB re 10-12W) Lp (dB re 20µPa)

2505001000

72.281.876.0

44.353.948.1

The overall sound pressure level is:

(c) The A-weighted sound pressure level is:

(d) r.m.s. acceleration levels (approximate).


a = 2πfv (see following table)

f (Hz) v (mm/s) a (m/s2)

2505001000

252

3.115.712.6

M ' (20 % 25 % 30 % 32) /177 ' 0.6

7

Solutions to problems relatingto sound in enclosed spaces

Problem 7.1

Direct field

The direct field of a sound source is defined as that part of the sound fieldwhich has not suffered any reflection from any room surfaces or obstacles.

Reverberant field

The reverberant field of a source is defined as that part of the sound fieldradiated by a source which has experienced at least one reflection from aboundary of the room or enclosure containing the source.

Problem 7.2

250Hz octave band, bandwidth from table 1.2 on p43 in the text is353 - 176 = 177Hz. Thus the modal overlap, M, is (from equation 7.24 in thetext) given by:

Problem 7.3

(a) If nz = 0, the pressure distribution in the room is uniform and themonopole source will excite the mode. For nz = 1 and nz = 3, there willbe a node at Lz/2 and the monopole will not excite the mode. For nz = 2,there will be an antinode at Lz/2 and the mode will be excited by amonopole source.

(b) A dipole will be ineffective at exciting the nz = 0 and nz = 2 modes

Sound in enclosed spaces 147

f1,1,1 '343

21

102%

1

52%

1

22' 93.9Hz

because the phase of each part of the dipole is opposite to the other andthe modal response is not. On the other hand, one would expect goodexcitation of the nz = 3 mode because the phase on one side of a node is180E different to that on the other side and this can be matched by thedipole.

(c) At a rigid wall, the particle velocity is zero. The required derivation isdescribed on pages 278 and 279 in the text.

(d) The required shape is shown in the figure below where the nodes arerepresented by lines and the relative phases of acoustic pressure are

shown by plus and minus signs.

Using equation 7.17 in the text:

Problem 7.4

(a) Cut-on frequency is the frequency at which the wavenumber, κ becomesreal and the mode begins to propagate down the duct without decayingin amplitude (assuming a rigid, hard walled duct).

(b) From the equation given in the problem, for a single mode the acoustic


κ3,2 ' (ω /c)2 & [45π2 /L 2z ]

ω ' c [ 45π /Lz ]

κ3,2 ' [5π2 /L 2z ]& [45π2 /L 2

z ] ' ± j 40π /Lz

'c

1 & 45π2c 2 / (ω2 L 2z )

'c

1 & const /ω2

c32

f

c

fco

pressure amplitude is given by . Thus the dB decay perp ' p0 e&jκmnx

unit distance is:

, where is the sound pressureΔ ' 20log10

p0

p1

' 20 × 0.4343 × jκmn p1

amplitude at 1m. For m = 3, n = 2, and Ly = Lz/3:

Cut-on frequency is thus given by:

Thus at 1/3 of the cut-on frequency, κ3,2 is given by:

Thus Δ = 172/Lz (dB/m).

(c) Phase speed, c3,2 = ω/κ3,2 = ω

(ω2 /c 2) & 45π2 /L 2z


f 'c2

16.2

'343

2×

16.2

' 27.7Hz

ψ '¢p 2 ¦ρc 2

¢p 2max ¦ ' (2 × 10&5 )2 ×1080/10 ' 0.04Pa 2

¢p 2 ¦ ' 0.04 × cos2 πxL

The variation of c3,2 as a function of frequency is shown in the figure.Near the mode cut-on frequency it can be seen that the phase speedapproaches infinity and at high frequencies it approaches the speed ofsound in free space.

Problem 7.5

(a) Room dimensions are 4.6m × 6.2m × 3.5m. The lowest resonancefrequency is the axial mode corresponding to the longest roomdimension. Thus:

(b) Pressure distribution follows a half cosine wave in the 6.2m direction asshown in the figure and calculated from equation 7.19 in the text. Thepressure is uniform across a given cross section defined by the 4.6m ×3.5m dimensions. Energy density is given by equation 7.32 in the text as:

The sound pressure level in a room corner is 80dB and this correspondsto the maximum sound pressure. Thus:

The sound pressure at any other location is (from equation 7.19 in thetext):

Thus the energy in the room is given by:


E ' mV

ψ dV ' 4.6 × 3.5mL

0

¢p 2 ¦ρc 2

dL

'4.6 × 3.5 × 0.04

1.206 × 3432 mL

0

cos2 πxL

dx

' 4.54 × 10&6 x2%

Lsin(2πx /L)4π

L

0

dEE

' &ScαV

dt

mE

E0

dEE

' &mt

0

ScαV

dt

logeE & logeE0 ' &Scαt

V

E ' E0 e&Scαt /V

¢p 2 ¦ ' ¢p 20 ¦e&Scαt /V

Lp &Lp0 ' 4.343cαtS /V ' 4.343cαt /L

Substituting L = 6.2 gives E = 1.4 × 10-5 Joules.

(c) The analysis procedure used on p290 and 291 in the text may be usedhere. Alternatively, we may start with the given equation and write:

Integrating gives:

which gives:

or

As E % +p2,, we can write:

Thus:


60 ' 4.343cαT60 /L

α '60

4.343×

LcT60

'60

4.343×

6.2343 × T60

'0.250

T60

¢p 2 ¦4ρc

' I

Wa '2 × 0.04

4 × 1.206 × 343× 0.05 × 16.1 ' 38.9µW

Wa 'dEdt

'EScα

V'

1.407 × 10&5 × 343 × 0.056.2

' 38.9µW

f '343

2×

0.58615.5

' 18.3Hz

and

Therefore:

(d) T60 = 5 seconds, so = 0.25/5 = 0.05.αPower consumption, . and Sw = 4.6 × 3.5 = 16.1 m2. TheWa ' 2IαSw

quantity, I, is the sound intensity in the direction of one wall. As shownon p287 in the text, the effective intensity, I, in one direction is related tothe acoustic pressure in a diffuse field by:

Substituting 0.04Pa2 for the maximum mean square pressure and theabove equation into the equation for Wa gives for the absorbed power:

Alternatively, the expression given in part (c) may be used to give:

Problem 7.6

(a) Substituting L = 2.7, c = 343, a = 5.5 and the values of the characteristicfunction ψ given in the problem table into the equation for f given in theproblem gives for the lowest order resonance frequency (n = 0, m = 1 andnz = 0) gives:


f1,1,1 '343

21

2.72%

1.6975.5

2

' 82.7Hz

(b) The lowest order mode pressure distributionis shown in the figure at right, where thenodal plane which runs the full length ofthe cylinder is indicated. The sound field isin opposite phase from one side of thenodal line to the other. The sound pressurewill be at a maximum along two axial linesat the surface of the cylinder furthest fromthe nodal plane.

(c) The air particles are oscillating with avelocity in-quadrature with the localacoustic pressure as the mode is characterised by a standing wavegenerated by the interference of two acoustic waves travelling in oppositedirections across the nodal plane.

(d) This problem may be answered by inspection of the given table andequation. The modes are listed in the table below in order of ascendingfrequency, column 1 followed by column 2 etc.

1,02,00,13,0

4,01,15,02,1

0,26,03,11,2

7,04,12,20,3

5,13,21,36,1

4,22,37,10,4

5,23,31,46,2

4,32,47,25,3

3,46,34,47,3

5,46,47,4

(e) f = 343/(2 × 2.7) = 63.5Hz.

(f) Axial modes have 2 zeroes in the subscripts nz, n, and m. Tangentialmodes have one zero and oblique modes have none. These criteria canbe used to list any number of axial, tangential and oblique modes.

The resonance frequency of the first oblique mode is:

Modes with resonances below this with nz = 1 are found from the tablein the problem as those with ψ less than the value of ψ corresponding to


ψ < 82.7 × 2 × 5.5 /343 ' 2.651

area of circular regionarea of square region

'πr 2

4r 2'

π4

r

n = 1 and m = 1. For nz = 0, the value of ψ must be below that given by:

The modes with resonance frequencies below the (1,1,1) mode are listedbelow in the form (nz, m, n).

1,0,01,1,01,2,01,3,0

1,4,01,0,10,1,00,2,0

0,3,00,4,00,5,00,6,0

0,0,10,1,10,2,10,3,1

0,0,2

That is, there are 17 modes with resonance frequencies below that of thefirst oblique mode. Of these, 9 are axial and 8 are tangential modes.

Problem 7.7

(a) A 2-D space would be one where thedimensions in 2 directions were very muchlarger than in the third direction. Anexample would be a large factory with alow ceiling in the frequency range belowthe first floor/ceiling axial resonancefrequency.

Following the procedure for a 3-D spacedescribed on p285-287 in the text, a 2-Dsquare region enclosing a circle is considered as shown in the figure.

Time for sound to travel through the circular region (length ofencompassing square) = 2r/c. Energy in the square region of length 2r(and unit thickness) as a result of a wave travelling normally to any of the

sides is , where I is the incident wave intensity. The energy per unitI2rc

perimeter in the circular region as a result of an incident wave 2r wide is:


ΔE 'π4

I2rc

E ' m2πr

ΔE dx ' m2π

0

ΔEr dθ 'π2

Ir 2

c m2π

0

dθ 'I r 2π2

c

ψ 'ES

'I r 2π2

cπr 2'

πIc

I 'ψcπ

ψ 'Ic

'¢p 2 ¦ρc 2

ψ 'πIc

'¢p 2 ¦ρc 2

I '¢p 2 ¦πρc

1

1 A

The total energy in the circular region due to waves from all directionsis found by integrating the preceding expression over the perimeter of thecircle. Thus:

Energy density, ψ, is given by:

Thus:

Consider a plane wave travelling unit distance. Theenergy in unit area is the intensity multiplied by thetime it is present which is, E = I/c. The energy isalso equal to the energy density multiplied by unitarea. Thus ψ = E. Thus for a plane wave:

Equation 7.32 in the text also gives the relation between sound pressureand energy density. Thus for a diffuse 2-D field:

Rearranging gives:

(b) An example of a 1-D field would be the inside of a rigid tube closed atboth ends at frequencies below the first higher order mode cut-on


I '¢p 2 ¦2ρc

Wa ' IP α 'ψcπ

P α

W ' SMψMt

' W0 &ψcπ

Pα

X ' [πW0 /Pcα ] & ψ

dXdt

' &MψMt

' &1S

W0 &ψcπ

Pα

1X

'Pαcπ

W0 &ψcπ

Pα&1

L

1 A

frequency. Assuming unit cross-sectional area anda wave travelling from left to right over a distanceof L. The time for a wave to travel a distance, L, isL/c. The energy in the wave travelling from left toright is IL/c. The total energy (due to left and righttravelling waves) is thus 2IL/c. The total energy inthe volume is also ψL. Thus, I = ψc/2 and ψ =2I/c. Using this relation and equation 7.32 in the text gives for a 1-Dfield:

(c) Let S = area of 2-D room ofunit height as shown in thefigure. Rate of energyabsorbed, Wa, around theperimeter, of length P andabsorption coefficient is given by:α

Rate of change of energy in the reverberant field = rate of supply, W0 -rate absorbed, Wa. Thus:

Introducing the dummy variable:

into the preceding equation, we may write:

and


1X

dXdt

' &PαcπS

mX

X0

dXX

' mt

0

&PαcπS

logeX & logeX0 ' &PαctπS

X ' X0 e&Pαct /πS

ψ ' ψ0 e&Pαct /πS

¢p 2 ¦ ' ¢p 20 ¦e&Pαct /πS

Wa ' 2 I α ' 2ψc2α

L

Thus:

Integrating gives:

Thus:

and

For a decaying sound field, W0 = 0 at t = 0. Using the precedingequation, X = X0 when t = 0. Also the definition of the dummy variable,X, above can be used to show that when W0 = 0, X = X0 = -ψ0. The sameequation may be used to show that as W0 = 0, then at any time t, X = -ψ.From the preceding discussion and the equation above we may write:

We know that ψ % +p2,. Thus:

(d) Let L be the length of the 1-D tube of unitcross sectional area.Rate of energy absorbed, Wa, at the endsof the tube of length L and absorptioncoefficient is given by:α

Rate of change of energy in the reverberant field = rate of supply, W0 -


W ' LMψMt

' W0 & ψcα

X ' [W0 /cα ] & ψ

dXdt

' &MψMt

' &1L

W0 & ψcα

1X

' αc W0 & ψcα &1

1X

dXdt

' &αcL

mX

X0

dXX

' mt

0

&αcL

logeX & logeX0 ' &αctL

X ' X0 e& αct /L

rate absorbed, Wa. Thus:

Introducing the dummy variable:

into the preceding equation, we may write:

and

Thus:

Integrating gives:

Thus:

and

For a decaying sound field, W0 = 0 at t = 0. Using the precedingequation, X = X0 when t = 0. Also the definition of the dummy variable,X, above can be used to show that when W0 = 0, X = X0 = -ψ0. The sameequation may be used to show that as W0 = 0, then at any time t, X = -ψ.From the preceding discussion and the equation above we may write:


ψ ' ψ0 e& αct /L

¢p 2 ¦ ' ¢p 20 ¦e& αct /L

cx ' ccosθ and cy ' csinθ

Nr 'ccosθ

Lx

%csinθ

Ly

Nav '1π /2 m

π /2

0

Nr dθ '2cπ m

π /2

0

cosθLx

%sinθLy

'2cπ

sinθLx

&cosθLy

π /2

0

'2cπ

1Lx

%1Ly

0

Ly

Lx

y

x

We know that ψ % +p2,. Thus:

(e) Mean free path, 2-D space. Sound propagation in any direction in the 2-D plane is equally likely (see figure). Due to symmetry, we need toconsider only one of the 4 quadrants and find an average path lengthbetween reflections for sound propagating in these directions.

Consider a soundwave propagatingwith speed c, in anangular direction ofψ from the x-axis asshown in the figure.Resolving into x andy components, wehave:

The number of reflections per unit time for a wave travelling in the θdirection is given by:

Averaging over all directions in one quadrant (the average for the otherthree quadrants would be the same) we find that the average number ofreflections per unit time is:


Λ 'π2

Lx Ly

Lx % Ly

'πSP

Lw ' Lp % 10log10 V & 10log10 T60 % 10log10 (1 % Sλ/8V) & 13.9

' 95%10log10(105)&10log10(2.5)%10log10 1% 142 × 0.6868 × 105

&13.9

' 95 % 20.2 & 4.0 % 0.5 & 13.9 ' 97.8dB re 10&12W

η 'acoustic powerelectrical power

'10&12 × 1097.8/10

10' 6 × 10&4 ' 0.06%

c sin

c cos

But Nav = c/Λ, where Λ is the mean free path. Thus:

For a 1-D space (see figure),the mean free path, which isthe d is tance be tweenreflections is equal to L, thelength of the space.

Problem 7.8

(a) Room size = 7m × 5m × 3m. Volume = 105m3, surface area = 2(7 × 5 +7 × 3 + 5 × 3) = 142m2. Wavelength of sound, λ = 343/500 = 0.686m.The sound power in the reverberation room is related to the soundpressure level by equation 6.13 in the text. Thus:

Alternatively, we could assume that the correction term, (1 + Sλ/8V) isalready included in Lp. In this case, Lw = 97.8 - 0.5 = 97.3dB.

Power conversion efficiency is given by:

(b) Fire box dimensions are: 10 × 12 × 20m.Axial resonances - first mode in each direction given by:


f 'c

2L'

8642 × 10

,864

2 × 12,

8642 × 20

' 43.2Hz, 36Hz, 21.6Hz

p ' 2prms ' 2 2 × 10&5 × 10155/20 ' 1.59kPa

I 'p 2

i

ρc'

(1.125 × 103 )2

(1 % 1 & α )2ρc

(i) As the frequency of instability is 36Hz, it seems likely that it isassociated with the axial mode across the width in the 12m direction.

(ii) The acoustic pressure will be largest at the two side walls normal tothe 12m dimension, because on reflection the pressure amplitude isdoubled.

(iii) Amplitude of cyclic force acting on walls.Acoustic pressure is 155dB. Amplitude is then:

Side wall area = 10 × 20 = 200m2.Force on each wall = 1.59 × 200 = 318kN.

(iv) Power absorbed by 2 side walls.Pressure at wall is the sum of the incident and reflected pressures.

Thus, . However, .(pi % pr )rms ' 1.125 × 103 pr ' (1 & α )pi

So, [1 % (1 & α ) ] pi ' 1.125 × 103

The absorbed power is controlled by the incident sound intensitywhich is:

The absorbed power is , where Aw = area of one wall.Wa ' 2α IAw

For a 1-D sound field, . Thus: ¢p 2 ¦ ' ¢p 20 ¦e&cα t /L

Lp0 & Lp ' 4.343c α t /L

and . ButT60 '(60 /4.343) × L

cα


T60 '2.2Q

fn

'2.2 × 30

36' 1.833secs

α '(60 /4.343) × 12

864 × 1.833' 0.105

Wa '2 × 0.1047 × (1.1246 × 103 )2 × 200

(1 % 1 & 0.1047)2 × 1.16 × 864' 14kW

η '13953

800 × 103' 1.7%

Lp0 & Lp ' 4.343c α t /L

60 ' 4.343 × 343 × 0.05 × T60 /5

Lp0 & Lp ' 4.343P αc t / (πS)

60 ' 4.343 × 20 × 0.25 × 343 × T60 / (π × 25)

Thus:

The absorbed power is then:

(v) Power conversion efficiency is:

Problem 7.9

(a) Taking logs of the equation given in the problem for a 1-D field, wehave:

Thus:

which gives, T60 = 4.0 seconds.

(b) Again taking logs of the equation given in the problem gives:

P = 4 × 5 = 20m and S = 5 × 5 = 25m2. Thus:


(c) S = 2(5 × 5 + 5 × 5 + 5 × 5) = 150m2


60 ' 1.086 × 150 × 343 × 0.1833 × T60 /125

W ' IA '¢p 2 ¦4ρc

A '1.265

4 × 413× 30 ' 0.023 W ' 103.6 dB re 10&12 W

V = 5 × 5 × 5 = 125m3

α '50 × 0.05 % 100 × 0.25

150' 0.183

Using equation 7.50 in the text, we obtain:


Problem 7.10

In a reverberant field the sound intensity in any particular direction is equalto that in any other direction resulting in a net active intensity (averaged overall directions) of zero and a large reactive intensity. Thus in the situationunder consideration here, the active intensity will only be contributed to by thedirect field of the source and the reactive intensity field will dominate theactive field by a large amount, especially at large distances from the source.Due to the dominance by the reactive field together with limitations on thephase accuracy between the two microphones of any measurement system,accurate measurements of the active field will not be generally feasible in areverberant room.

Problem 7.11

(a) Energy density in a reverberant field is and the given SPL is 95¢p 2 ¦ /ρc 2

dB.Thus, and the energy density is¢p 2 ¦ ' (2 × 10&5 )2 × 109.5 ' 1.265 Pa 2

then: Energy density = 1.265/(413 × 343) = 8.9 × 10-6 J/m3

(b) Sound power incident on a wall is given by:


T60 '55.25 × 400

343 × 360 × 0.1' 1.8seconds

W '360 × 0.1 × 4 × 10&4

4 × 1.206 × 343 × (1 & 0.1)' 9.67µW

I '4 × 10&4

4 × 1.206 × 343' 2.42 × 10&7 W/m 2

r 'Sα

8π (1 & α )

1/2

'36

8π × 0.9

1/2

' 1.26m

Problem 7.12

(a) Room 10m × 10m × 4m, S = 2(10 × 10 + 10 × 4 × 2) = 360m2,V = 10 × 10 × 4 = 400m3. Using equation 7.51 in the text forreverberation time, we may write:

(b) Lp = 60dB corresponds to +p2, = 4 × 10-10 × 1060/10 = 4 × 10-4 Pa2. Usingequation 7.41 in the text we obtain:

Sound power level, Lw = 10log10(9.67 × 10-6) + 120 = 69.9dB re 10-12W.

(c) Using equation 7.33 in the text:

(d) From equations 7.40 to 7.42 in the text, the reverberant field level isequal to the direct field level when Dθ/(4πr2) = 4/R = .4 (1 & α )/(Sα )Assuming the acoustic centre of the source is within a quarter wavelengthfrom the hard floor, Dθ = 2 and the distance, r, at which the fields areequal is:

Problem 7.13

(a) Room 3.05m × 6.1m × 15.24m,S = 2(3.05 × 6.1 + 3.05 × 15.24 + 6.1 × 15.24) = 316.1m2,V = 3.05 × 6.10 × 15.24 = 283.54m3. Using equation 7.51 in the text forreverberation time, we may write:


α '55.25VScT60

'55.25 × 283.54316.1 × 343 × 2

' 0.072

Lw ' Lp % 10log104(1 & α)

Sα% 10log10

ρc400

' 74 & 10log104 × 0.928

316.1 × 0.072& 0.15 ' 81.7dB re 10&12W

316.1α ' 246.1 & 246.1α

T60 '55.25V

Scα'

55.25 × 283.54316.1 × 343 × 0.438

' 0.33seconds

c 'γRTM

'1.4 × 8.314 × 1473

0.035' 700 m/s

Using equation 7.42, the sound power output may be written as:

(b) To lower the reverberant field by 10dB, we need to increase R by a factorof 10. Old = 24.61m2, thus required new =Sα / (1 & α) Sα / (1 & α)246.1m2. Thus in the new situation, expanding the above equation gives:

which results in a new required value of = 0.438. The old value of α αis 0.0722, so the increase in absorption needed is:

= 316.1(0.438 - 0.0722) = 115.5m2.Δ Sα

(c) Using the value of = 0.438 obtained above and equation 7.51 in theαtext, we obtain:

Problem 7.14

(a) Sound power level = 10 log10W + 120 = 10 log103.1 + 120 = 124.9 dB re 10-12 W

(b)


Rb 'Sα

(1 & α)'

400 × 0.080.92

' 34.78m 2

αn '100 × 0.08 % 300 × 0.5

400' 0.395

Lp ' Lw % 10log104(1 & α )

S α%10log10

ρc400

' 124.9 % 10log104 × 0.95

100 × 0.05% 10log10

700 × 0.29400

' 120.8 dB

ρ 'γP

c 2'

1.4 × 101.4 × 103

7002' 0.29 kg/m 3

Surface area, S = πdL + πd2/2 = π × 4 × 6 + π × 16/2 = 32π = 100 m2

Problem 7.15

Lp ' 10log10 (109.5 % 109.7 % 109.9 ) ' 102.1 dB(A)

Lp ' Lw % 10log10Q

4πr 2%

4(1 & α )Sα

Thus, 102.1 & 110 ' 10log102

4πr 2%

3.64320 × 0.09

and, 10&7.9/10 &3.64

320 × 0.09'

1

6.28r 2

Thus, r '1

6.28 × 0.0358' 2.1 m

Problem 7.16

Room 10m × 10m × 5m, S = 2(10 × 10 + 10 × 5 × 2) = 400m2,V = 10 × 10 × 5 = 500m3. The room constant before treatment is

Adding sound absorbing treatment to the walls and ceiling results in a newmean absorption coefficient calculated as follows:


Rn 'Sα

(1 & α)'

400 × 0.3950.605

' 261.2m 2

Δ Lp ' 10log10

Dθ

4πr 2%

4R

old

& 10log10

Dθ

4πr 2%

4R

new

' 10log102

4π32%

434.78

& 10log102

4π32%

4261.2

' 6.0dB

α '48 × 0.15 % 132 × 0.05

180' 0.0767

¢p 2R ¦ ' 4 × 25 × 10&3 × 1.206 × 343 (1 & 0.0767)

180 × 0.0767' 2.766 Pa 2

LpR ' 10log10 (2.766) % 94 ' 98.4dB

Thus the new room constant after treatment is:

Using equation 7.42 in the text allows the difference in sound pressure level(assuming that the sound power is constant) to be calculated as follows(assuming the acoustic centre of the machine is within a quarter wavelengthof the hard floor):

Problem 7.17

(a) Room 8m × 6m × 3m, S = 2(8 × 6 + 8 × 3 + 6 × 3) = 180m2.The mean Sabine absorption coefficient may be calculated using equation7.78 in the text to give:

Using equation 7.41, we can write:

The reverberant field sound pressure level is then:

(b) Direct and reverberant fields equal (see equation 7.42 in the text) when


r 'Sα

16π (1 & α )

1/2

'180 × 0.0767

16π × (1 & 0.0767)

1/2

' 0.55m

10log10 ¢p 2 ¦ ' 10log10Wρc

πa 2& 8

Lp (reverb) ' Lw % 10log10ρc

400& 10log10 (πa 2 ) & 8

' 130 % 0.14 & 18.95 & 8 ' 103.2 dB

Lp (direct) ' Lw % 10log10ρc

400& 10log10 (4πr 2 )

' 130 % 0.14 & 24.97 ' 105.2 dB

. Assuming that the acoustic centre of the source is wellDθ /4πr 2 ' 4/R

above the floor, Dθ = 1 and the distance, r, at which the two fields areequal is:

Problem 7.18

(a) This is a flat room as on pages 314-319 in the text. Pressure reflectioncoefficient amplitude = 0.7, β = 0.72 and α ' 1 & 0.72 ' 0.51Room height, a = 5 m and distance r = 5 m. Thus r/a = 1. From thefigure, the reverberant field sound pressure is given by:

Thus the reverberant field sound pressure level is:

The direct field sound pressure level is:

(b) Sabine room, area, and volume,S ' 2[10 × 10 % 10 × 5 × 2] ' 400 m 2

V ' 10 × 10 × 5 ' 500 m 2

The total sound pressure level (direct plus reverberant) in the room is:


Lp ' Lw % 10log10Q

4πr 2%

4(1 & α)Sα

% 0.15

C ' 10log10 0.3 %SE(1 & α )

Si α' 10log10 0.3 %

300 × 0.49400 × 0.51

' 0.1 dB

Lp ' 87.4 & 24.9 ' 62.5 dB

Lp ' Lw % 10log10D

4πr 2%

4(1 & α )S α

' 130 % 10log101

4π × 25%

4 × 0.49400 × 0.51

' 111 dB

r 'DS α

4π × 4(1 & α )'

400 × 0.5116π × 0.49

' 2.9 m

T60 '55.25V

Sc α'

55.25 × 500400 × 343 × 0.51

' 0.39 seconds

Lp ' Lw % 0.14 & 10log10 2πr 2 ' 130 % 0.14 & 42.8 ' 87.4 dB

(c) Direct and reverberant fields equal when .D /4πr 2 ' 4(1 & α) /S αThus:

(d)

(e) Treat the room like an enclosure. Thus, the sound level at the receiverwithout the enclosure at a distance of 50 + 5 m is:

The enclosure noise reduction is given by NR = TL - C, where

Thus the noise reduction = 24.9 dB and the SPL at 50 m is:

Problem 7.19

Q = 2, r = 1, α = 0.08, Lw = 95dB, S = 400. Thus:


Lp ' 95 % 10log102

4 × π%

4(1&0.08)400 × 0.08

% 0.15 ' 89.5dB

Lw ' Lp % 20log10 r % 10log10(4π) & 3 % 10log10400ρc

' 80 % 9.54 % 11.00 & 3.01 & 0.15 ' 97.4dB

Lp ' Lw % 10log10(4 /R) % 0.15

10log10 R ' 97.38 & 85 % 0.15 % 6.02 ' 18.55

328 α(1 & α)

' 71.62

Problem 7.20

(a) Reference source on hard asphalt (DI = 3), Lp = 80dB at 3m. Soundpower level is:

(b) Assuming a negligible direct field, the room constant may be calculatedusing equation 7.42 in the text (with a 0.15dB correction for ρc differentfrom 400). Thus:

Substituting in values for the parameters gives:

Thus, R = 71.6.

(c) Room 14m × 6m × 4m, S = 2(14 × 6 + 14 × 4 + 6 × 4) = 328m2.Using equation 7.43 in the text gives:

Thus, = 0.18.α

(d) Existing Lp = 75dB and the allowable total = 80dB. Thus the allowablecontribution from the three new line printers is

. The allowable contribution from each10log10 (108 & 107.5 ) ' 78.3dB

line printer is then . The allowable sound10log10107.83

3' 73.5dB

power level is the obtained using equation 7.42 in the text (using only thereverberant field part) as follows:


Lw ' 73.6 & 10log104

71.6& 0.15 ' 86.0dB

R '328 × 0.261(1 & 0.261)

' 116m 2

Δ Lp ' 10log1011671.6

' 2.1dB

Lw ' 74.29 & 10log104

116& 0.15 ' 88.8dB

(e) Area of ceiling = 14 × 6 = 84m2, corresponding = 0.5.αArea of floor and walls = 328 - 84 = 244m2,corresponding = 0.179.α

Average . Thus: α '0.5 × 84 % 0.179 × 244

328' 0.261

Using equation 7.120 in the text, the reduction in sound pressure level isthus:

Thus the level in the room before addition of the new printers will nowbe 75 - 2.1 = 72.9dB. The allowable sound pressure level of the threenew line printers together is now . The10log10 (108 & 107.29 ) ' 79.1dB

allowable contribution from each line printer is then

. The allowable sound power level is the10 log10107.91

3' 74.3dB

obtained using equation 7.42 in the text (using only the reverberant fieldpart) as follows:

Problem 7.21

(a) Considering only the reverberant field of the machine we may useequation 7.41 in the text. The surface area of the factory is S = 2(10 × 10 × 3) = 600m2. An Lp of 83dB corresponds to a +p2, of

. Using equation 7.41, the required4 × 10&10 × 1083/10 ' 7.98 × 10&2 Pa 2

mean absorption coefficient is given by:


α(1 & α)

'4 × 0.01 × 1.206 × 343

7.981 × 10&2 × 600' 0.346

600 × 0.257 ' 100 × 0.01 % 500x

90 ' 100 % 10log102

4πr 2%

4208

Thus . However this would be the required absorptionα ' 0.257coefficient if the floor were lined as well. If we assume that the concretefloor has an absorption coefficient of 0.01, and we let the requiredabsorption coefficient of the walls and ceiling be x, then we can useequation 7.78 in the text to write:

which gives x = 0.306. Thus the required absorption coefficient for thewalls and ceiling is 0.31.

(b) The room constant, R, is from part (a) 0.346 × 600 = 208m2.The sound power level, Lw = 10log10(0.01) + 120 = 100dB.For a total Lp of 90dB we may write (assuming a directivity factor, Dθ =2 as the source is assumed close to a hard floor and other surfaces aremore absorptive):

Solving the above gives r = 1.40m as the radius around the machinewithin which the sound pressure level will exceed 90dB.

Problem 7.22

(a) From Table 4.10 in the text, the allowable community noise level toensure minimal risk of complaints is 40 + 15 - 10 = 45dB(A). If the onlynoise is in the 500Hz octave band then the -3.2dB A-weighting at thisfrequency, results in an allowable level of 48.2dB in that band. Theexisting level is 44dB so the allowed increase is 4.2dB. A reverberantfield sound pressure level corresponds to a sound power level (seeequation 7.42 in the text) of:


Lw ' 88 & 10log104(1 & α)

Sα& 0.15 dB

α '55.25 × 2500

343 × 2.1 × 1450' 0.132

Lw ' 88 & 10log104(1 & 0.132)1450 × 0.132

& 0.15 ' 105.3dB

Lw ' 10log10 10109.5/10 & 10105.3/10 ' 107.4dB

Sα ' 500 × 0.5 % 950 × 0.132 ' 375.4m 2

Room 25m × 20m × 5m,S = 2(25 × 20 + 25 × 5 + 20 × 5) = 1450m2.V = 25 × 20 × 5 = 2500m3

T60 = 2.1 secs. From equation 7.51 in the text:

Thus:

This is the sound power level of the existing equipment. Thus theallowable total sound power of existing + new equipment is 105.3 + 4.2= 109.5dB. Thus the allowed power level for the new equipment is:

As there are 5 new machines, the allowed power level for each isLw = 107.4 - 10log10(5) = 100.4dB

Assumptions

! Only absorption is due to floor, walls and ceiling! Air temperature of 20EC.! The relative contribution of direct and reverberant sound energy to

the community noise levels will be the same for the new machinesas for the old machines.

(b) If ceiling tile with = 0.5 were added then the new is:α Sα

corresponding to = 0.259α

Old R = and(1450 × 0.132) / (1 & 0.132) ' 220.5m 2

new R = .(1450 × 0.259) / (1 & 0.259) ' 506.8m 2

Thus the allowed increase in sound power level for the same reverberant


Δ Lw ' 10log10(506.8) & 10log10(220.5) ' 3.6dB

Wa ' ψcSα /4 ' 0.282 × 343 × 0.144 × 0.1 /4 ' 0.35W

Lp ' Lw % 10log104(1 & α)

Sα%10log10

ρc400

Lp ' 84 % 10log104 × 0.85

216 × 0.15% 0.15 ' 74.4dB

Dθ

4πr 2'

1

4π × 1.52' 0.035

field sound pressure level is:

Assuming that the reverberant field dominates the direct field, the newallowed power level of each machine is 100.4 + 3.6 = 104dB.

Problem 7.23

(a) Energy density, ψ = ¢p 2 ¦ /ρc 2

Thus ψ = (2 × 10&5 )2 × 10140/10 / (1.205 × 3432) ' 0.28J /m 3

(b) Enclosure surface area,S = 2(0.2 × 0.15 + 0.2 × 0.12 + 0.15 × 0.12) = 0.144m2.Power flow into walls (equation 7.39a) is given by

(c) The power generated is equal to the power absorbed by the walls. Thusthe power required to drive the source is 0.348/0.2 = 1.7W.

Problem 7.24

(a) Reverberant field Lp from taking logs of equation 7.41 in the text.

S = 2(10 × 6 + 10 × 3 + 6 × 3) = 216m2, = 0.15, so:α

(b) Compare As the acoustic centre of the source isDθ

4πr 2with

4(1 & α)Sα

well above the hard floor, Dθ = 1; so:


4(1 & α)Sα

'4 × 0.85

216 × 0.15' 0.105

Lp ' 84 % 10log10 0.0354 %4 × 0.85

216 × 0.15% 0.15 ' 75.6dB(A)

Lp ' 84 % 10log10 0.0354 %4 × 0.5

216 × 0.5% 0.15 ' 71.5dB(A)

r '216 × 0.15

4 × 4 × π × 0.85' 0.87m

Lp ' 90.2 % 10log104 × 0.85

216 × 0.15% 0.15 ' 80.6dB(A)

Lp ' 80.6 ' 84 & 10log10(4πr 2) % 0.15

Thus the reverberant field dominates.

(c) Using equation 7.42 in the text (and allowing for ρc not equal to 400) thesound pressure level corresponding to = 0.15 is:α

and the sound pressure level corresponding to = 0.5 is:α

This corresponds to a reduction of 4.1dB.

(d) From equation 7.42 in the text, when each machine is running separately,

the fields are equal when . Thus for each machine:Dθ

4πr 2'

4(1 & α)Sα

When the machines are running together, the reverberant fieldcontribution will be the sum of the two reverberant fields originatingfrom each machine. The total sound power output is

. The reverberant field sound pressure10 log10 108.9 % 108.4 ' 90.2dB

level is then:

Thus the required distance is that at which the direct field Lp is equal to80.6dB(A). For the original machine:


r '216 × 0.5

4 × 4 × π × 0.5' 2.07m

Lp ' 90.2 % 10log104 × 0.5

216 × 0.5% 0.15 ' 73.0dB(A)

Lp ' 73.0 ' 84 & 10log10(4πr 2) % 0.15

Lp ' 73.0 ' 89 & 10log10(4πr 2) % 0.15

Thus r = 0.42m (distance from original machine at which originalmachine direct field = reverberant field with both machines running).

For the new machine, Lp ' 80.6 ' 89 & 10log10(4πr 2) % 0.15

Thus r = 0.75m (distance from new machine at which new machine directfield = reverberant field with both machines running).

(e) Increase α to 0.5. From equation 7.42 in the text, when each machine isrunning separately, the fields are equal when

. Thus for each machine,Dθ

4πr 2'

4(1 & α)Sα

The reverberant field sound pressure level when both machines arerunning is,

Thus the required distance is that at which the direct field Lp is equal to73.0dB(A). For the original machine,

Thus r = 1.02m (distance from original machine at which originalmachine direct field = reverberant field with both machines running).

For the new machine,

Thus r = 1.81m (distance from new machine at which new machine directfield = reverberant field with both machines running).

(f) Assumptions:! Each machine is radiating omni-directional sound.! Both machines exhibit similar frequency spectra.! The frequency averaged absorption coefficient is obtained using a

sound source with a frequency spectrum similar to that of themachines under test.


T60 '55.25 × 900

600 × 343 × 0.1' 2.4secs

R '600 × 0.11 & 0.1

' 66.67m 2

Lp ' Lw % 10log10(4 /R) % 10log10 4 % 0.15' 94 % 10log10(4 /66.7) % 10log10(4) % 0.15' 94 & 12.22 % 6.02 % 0.15 ' 88.0dB

(87.95)

Lp ' 10log10 107.5 % 108.80 ' 88.2dB (88.16)

Lp ' 10log10 107.5 % 107.795 ' 79.7dB

ΔLp ' 88.16 & 85 ' 3.16dB ' 10log10 Rf /Ri

Problem 7.25

(a) Room 10m × 15m × 6m,S = 2(10 × 15 + 10 × 6 + 15 × 6) = 600m2.V = 10 × 15 × 6 = 900m3. The room reverberation time can be calculatedusing equation 7.51 in the text. Thus:

(b) Room constant from equation 7.43 in the text:

(c) For each machine, Lw = 94dB. Thus for 4 new machines,Lw = 94 + 10log104. Thus the reverberant field sound pressure level dueto the 4 new machines is:

The existing reverberant field level is 75dB prior to installation of thenew machines. Thus the total level after installation is:

(d) When quiet machines are installed, the sound power and the soundpressure level contribution due to the new machines is reduced by 10dBto 94 and 78.0 respectively. Thus the total reverberant field level afterinstallation of quiet machines is:

(e) The design goal is 85dB. Thus the required reduction due to ceiling tileif untreated machines are used is:


0.187 '0.1(600 & x) % 0.6x

600

D

4πr 2'

4(1 & α )S α

or r 'DSα

16π (1 & α )

α '(1450 & 500) × 0.1 % 500 × 0.5

1450' 0.238

Thus . The initial room constant is 66.67m2.Rf /Ri ' 103.16/10 ' 2.07

Thus the new requirement is R = 138m2. The required averageabsorption coefficient is then given by

. Thus . To achieve this, let x600 × α ' 138 & 138 × α α ' 0.187square metres of room surface be covered by tile. Then:

which gives x = 105m2. Area of ceiling = 10 × 15 = 150m2. So coveringthe ceiling with ceiling tile would be adequate. This would cost $50 + 3× 150 = $500 which is much less than the machine noise control and isthus the preferred option.

Problem 7.26

(a) Distance at which direct and reverberant fields are equal is given by:

The room surface area, S, is given by S ' 2(20 ×25 % 20 ×5 % 25 × 5) ' 1450m2

So r '2 × 1450 × 0.1

16π × 0.9' 2.5m

(b) In this case, D=4 instead of 2 and r = 2.53 × 21/2 = 3.6m

(c) If the ceiling were covered with ceiling tiles, the new sabine absorptioncoefficient would be:

So r '2 × 1450 × 0.238

16π × 0.762' 4.2m


Lw ' Lp & 10log10(4 /R) & 0.15

Lp1 & Lp2 ' 10log10D

4πr 2%

4(1 & α1 )

S α1

& 10log10D

4πr 2%

4(1 & α2 )

S α2

' 10log102

4π × 0.52%

4 × 0.91450 × 0.1

& 10log102

4π × 0.52%

4 × 0.7621450 × 0.238

' &1.795 % 1.901 ' 0.1dB

(d) The operator is only 0.5m from the machine so he/she is in the directfield. The reverberant field contribution at this distance is small so theceiling tiles will have only a very small effect. Could calculate this (butnot necessary).

Problem 7.27

(a) Room 5m × 5.5m × 3m, S = 2(5 × 5.5 + 5 × 3 + 5.5 × 3) = 118m2.V = 5 × 5.5 × 3 = 82.5m3. Perimeter, L = 4(5.5 + 5 + 3) = 54m.

Assuming that measurements of Lp are made far enough from themachine that the direct field is negligible compared to the reverberantfield, the sound power level is given by:

This allows the following table to be constructed.

Octave bandcentre

frequency (Hz)Lp α R Lw

6325010004000

75858470

0.010.020.020.03

1.1922.4082.4083.649

69.682.681.669.5

Overall 85.4

(b) The reverberation time may be calculated using equation 7.51 in the textand the modal density may be calculated using equation 7.21. Equations


2.569 × 10&5f 2 % 1.575 × 10&3f % 0.0197 ' 7.673

f ' &1.575 × 10&3

2 × 2.569 × 10&5

%(1.575 × 10&3 )2 % 4 × 2.569 × 10&5 × 7.673

2 × 2.569 × 10&5

' &31 % 547 ' 516Hz

d '5.52

2

%52

2

%32

2

' 4.01m

7.23 and 7.24 are then used to calculate the modal overlap. Thefollowing table may be constructed.

Octave bandcentre

frequency (Hz)α T60 Δf

dNdf M

6325010004000

0.010.020.020.03

11.265.635.633.75

0.1950.3910.3910.590

0.222.027.3

417.4

0.040.7910.6

246.3

The reverberation time may be assumed constant between 250 and1000Hz, so we need to calculate the frequency wheredN/df = 3/0.391 = 7.67. That is:

Thus:

Thus the modal overlap is greater than 3 for frequencies above 520Hz.

(c) The effect of acoustic tile may be calculated using equation 7.42. Thequantity Lw will remain the same in each case so we need to calculate thechange in the second term in the equation.

Distance from room corner to centre is given by:

Assuming that the acoustic centre of the source is within a quarter of awavelength from the hard corner, Dθ = 8 and so


10log10 107.5 % 108.5 % 108.4 % 107

& 10log10 10(7.5 & 0.39) % 10(8.5 & 0.38) % 10(8.4 & 0.46) & 10(7 & 0.41)

' 87.8 & 83.7 ' 4.1dB

. The following two tables (one with noDθ

4πr 2'

8

4π × 4.012' 0.0396

tile and the other with 25m2 of acoustic tile) may be constructed.

Octave bandcentre

frequency(Hz)

α(no tile) R

Dθ

4πr 2%

4R

10log10

Dθ

4πr 2%

4R

6325010004000

0.010.020.020.03

1.1922.4082.4083.649

3.41.71.7

1.14

5.32.3 2.30.6

In the following table (which includes the effects of acoustic tile, theoverall absorption coefficient is calculated using equation 7.78 in thetext.

Octavebandcentre

frequency(Hz)

α(wall)

α(tile)

α(overall) R

Dθ

4πr 2%

4R

10log10

Dθ

4πr 2%

4R

Improve-ment(dB)

63250

10004000

0.010.020.020.03

0.080.150.200.25

0.0250.0480.0580.077

3.05.957.279.84

1.370.720.590.45

1.4-1.5-2.3-3.5

3.93.74.64.1

The noise reductions in each octave band are given by the last column inthe above table.

(d) Assuming that the overall space average level will be reduced by thesame amount as the level at the centre of the room, the differencebetween new and old overall levels is:


Lw ' 95 & 8.3 % 22.5 % 0.2 & 13.9 ' 95.5dB re 10&12W

Lp ' 95.5 % 10log102

4π × 0.52%

4(1 & 0.022)193.2 × 0.022

% 0.15

' 97.6dB re 20µPa

T60 '55.25V

Scα'

55.25 × 179.7193.2 × 343 × α

'0.150α

Problem 7.28

(a) Using equation 6.13 in the text and substituting in the appropriate valuesgives:

(b) Using equation 7.42 in the text, (and allowing for ρc … 400) we can write:

Problem 7.29

(a) Room 6.84m × 5.565m × 4.72m,S = 2(6.84 × 5.565 + 6.84 × 4.72 + 5.565 × 4.72) = 193.2m2.V = 6.84 × 5.565 × 4.72 = 179.7m3. The room reverberation time can becalculated using equation 7.51 in the text. Thus:

The results for each third octave band are given in the following table.


λ ' 179.7 /4.6 1/3 ' 3.393m

λ ' 179.7 /1.3 1/3 ' 5.171m

One third octave bandcentre frequency (Hz)

α T60

63801001251602002503154005006308001000125016002000250031504000500063008000

0.0100.0100.0110.0110.0130.0150.0170.0170.0180.0180.0190.0200.0220.0250.0280.0310.0340.0370.0400.0440.0470.050

15.015.013.613.611.510.08.88.88.38.37.97.56.86.05.44.84.44.03.73.43.23.0

(b) Lowest 1/3 octave band given by V = 4.6λ3 (see p259 in text). Thus:

This corresponds to a frequency, f = 343/3.393 = 101Hz. Thus the roomis suitable for measurements down to and including the 100Hz 1/3 octaveband.

(c) Lowest octave band given by V = 1.3λ3 (see p259 in text). Thus:

This corresponds to a frequency, f = 343/5.171 = 66Hz. Thus the roommay just be suitable for measurements down to and including the 63Hz


Lw ' Lp & 10log10(4 /R) % 10log10(400 /ρc)

92.5 ' 87 & 10log10(4 /R) & 0.15

(4 /R) ' 10(87 & 92.5 & 0.15) /10 ' 0.272

octave band.

(d) For pure tone noise, the lowest acceptable frequency is given by (see p259 in text). We need to solve for the frequency byf ' T60 /V 1/2

trial and error as illustrated in the table below.

One third octave bandcentre frequency (Hz)

T60 correspondinglowest acceptable

frequency

6380

100125160200250315400500630

15.015.013.613.611.510.08.88.88.38.37.9

578578550550506472443443430430419

From the table, it can be seen that the lowest acceptable frequency fortonal noise is 430Hz.

Problem 7.30

(a) Lw = 92.5dB, Lpr = 87dB for reference source. The room constant iscalculated using only the reverberant part of equation 7.42 in the text andallowing for ρc = 413.6. Thus:

and so R = 14.7m2.


Lp ' 10log10 108.5 & 108.1 & 10log10(4)' 76.8dB(A) re 20µPa

Lw ' 76.77 & 10log10(4 /14.7) & 0.15

' 82.3dB(A) re 10&12W

Lw ' 76.77 & 10log10(4 /29.4) & 0.15 ' 85.3dB(A) re 10&12W

10log10

Dθ

4πr 2%

4(1 & α1)

S α1

& 10log10

Dθ

4πr 2%

4(1 & α2)

Sα2

(b) Existing Lp is 81dB(A). Allowed total Lp = 85dB(A). 4 new machines,so allowed Lp from each new machine is:

The corresponding allowed sound power level of each machine is then:

Assumptions:

! Direct field small compared to reverberant field at measurementlocations.

! ρc = 413.6! Spectral content of noise from new machines is similar to that

of existing noise. If not, then calculations should be done inoctave bands.

(c) Doubling room constant gives the allowed sound power level of

Same assumptions as for part (b).

Problem 7.31

Room 15m × 15m × 5m,S = 2(15 × 5 × 2 + 15 × 15) = 750m2.

Using equations 7.42 and 7.43, the noise reduction is given by:

The mean absorption coefficient before treatment is calculated usingequation 7.78 and is:


α1 '15 × 15 × 0.01 % 525 × 0.1

750' 0.073

α2 '15 × 15 × 0.01 % 525 × 0.7

750' 0.493

ΔLp ' 10log102

4 × π × 16%

4(1 & 0.073)750 × 0.073

& 10log102

4 × π × 16%

4(1 & 0.493)750 × 0.493

' &11.1 % 18.1 ' 7.0dB

and after treatment it is:

Assuming that the acoustic centre of the machine is within a quarter ofa wavelength of the floor, Dθ = 2; thus the noise reduction at the specifiedlocation is:

Problem 7.32

Room: Lw = 105dB, V = 100m3, S = 130m2, T60 = 1.5secs.Office: Lw = 85dB, V = 80m3, S = 100m2, T60 = 0.75secs.Partition area: 15m2.

For the room and the office, the quantity is calculated using equation 7.51Sαin the text and the reverberant sound pressure level existing in each space iscalculated by combining equations 7.42 and 7.43 with the direct field term ofequation 7.42 omitted. Thus the following table may be constructed.

Room Office

Sα 10.739 17.182

α 0.0826 0.1718

Lp ' Lw % 10log104(1 & α)

Sα% 0.15

100.5dB 78.0dB


Lp ' 10log10 107.9 & 107.8 ' 72.1dB

¢p 2 ¦ ' 4 × 10&10 × 10116/10 ' 159.2Pa 2

1 & αα

'159.2 × 50

4 × 1.206 × 343 × 1' 4.812

T60 '55.25 × 30

343 × 50 × 0.172' 0.56secs

α '10 × 0.8 % 40 × 0.172

50' 0.298

¢p 2 ¦ ' 4 × 1 × 1.206 × 343 × (1 & 0.298)50 × 0.298

' 77.96Pa 2

Allowable level in office = 78 + 1 = 79dB. Allowable level due to newmachine is:

The level in the room is 100.5dB, so the noise reduction required of the wallis 100.5 - 72.1 = 28.4dB which should be rounded up to 30dB forspecification purposes.

Problem 7.33

(a) Room volume V = 30m3, surface area, S = 50m2

Reverberant field mean square pressure is:

Using equation 7.41 in the text we obtain:

Thus = 0.172α

(b) Using equation 7.51 in the text:

Thus, T10 = 0.56/6 = 0.094secs.

(c) New is given by:α

Using equation 7.41, we obtain for the new mean square sound pressure:


Lp ' 94 % 10log10(77.96) ' 112.9dB

The corresponding sound pressure level is then:

which corresponds to a reduction of 116 - 112.9 = 3.1dB.

(d) Adding 3 more sources increases the existing number by a factor of 4.Providing all sources produce the same sound pressure level, the increasein total sound pressure level over that corresponding to one source wouldbe .ΔLp ' 10log10(4) ' 6dB

Problem 7.34

porous acoustic fibrous material (fibreglass, rockwool, ceramic fibre, steelwool):

Absorptionmechanisms: viscous friction losses due to difference in

velocities of air particles adjacent to fibres andthose in the centre of the gap between fibres.

Absorptioncharacteristics: good at mid to high frequencies but at low

frequencies need a large thickness of material tobe effective.

Applications: air handling duct mufflers, pipe lagging,reverberant space absorption.

Avantages: Inexpensive, high absorption coefficient in mid tohigh frequency range.

Disadvantages: fibre loss can be a health hazard so care has to betaken to properly contain the material; susceptibleto powdering in the presence of vibration; not oil,water or chemical resistant.

Porous plastic and rubber (polyurethane foam etc.):Absorptionmechanisms: viscous friction losses due to difference in

velocities of air particles adjacent to capillarywalls and those in the centre of the capillaries.

Absorption


characteristics: good at mid to high frequencies but at lowfrequencies need a large thickness of material tobe effective.

Applications: vehicle cabins, pipe lagging, reverberant spaceabsorption.

Avantages: High absorption coefficient in mid to highfrequency range; no health risk due to fibres.

Disadvantages: not oil, water or chemical resistant; expensive, fireand smoke hazard; will not tolerate hightemperatures.

Helmholtz resonators:Absorptionmechanisms: viscous friction losses around neck of resonator

associated with large air particle motion in thecentre of the neck and zero motion at the walls ofthe neck at resonance.

Absorptioncharacteristics: good in a narrow band of frequencies around the

design frequency.Applications: electrical transformer enclosures, vehicle mufflers,

mufflers for tonal noise generated by largeindustrial fans.

Avantages: can be made to be immune to moisture, oil andchemicals. Can withstand high temperatures; nohealth risk.

Disadvantages: Narrow frequency range of effective absorption.

Resonant panelsAbsorptionmechanisms: panel and backing cavity damping losses.Absorptioncharacteristics: good in a narrow band of frequencies around the

design frequency.Applications: electrical transformer enclosures, auditoria,

concert halls, reverberant spaces.Avantages: can be made to be immune to moisture, oil and

chemicals. Can withstand high temperatures; nohealth risk.

Disadvantages: Narrow frequency range of effective absorption.


α 'j Siαi

j Si

'

2 × 6.84 × 5.565 × 0.02 % 2 × 5.565 × 4.72 × 0.05%2 × 6.84 × 4.72 × 0.06

2 × 6.84 5.565 % 2 × 5.565 × 4.72 % 2 × 6.84 × 4.72

' 8.023 /193.2 ' 0.042

Problem 7.35

We require = 0.8 at 125Hz. Referring to figure 7.8 in the text (p.307) it isαclear that we need to use curve "D" which implies the use of sound absorbingmaterial behind the panel. From figure 7.9, it can be seen that the 125Hz linecrosses curve "D" when the cavity depth is 110mm and the panel mass is2kg/m2. Thus this is the required design. Note that the guide notes in thecaption of figure 7.8 should also be included as design specifications.

Problem 7.36

Following equation 7.78 in the text:

Problem 7.37

Room volume = π × 72 × 2.5 = 384.85m3. The optimum reverberation timesare calculated using equation 7.121 in the text with K = 5. The calculatedvalues are listed in the following table.

Octave bandcentre frequency (Hz)

T60

125250500

1000200040008000

1.451.060.960.960.960.960.96


(b) Using equation 7.51 in the text, the recommended is 65m2 at 500HzSαand above, 59m2 at 250Hz and 43m2 at 125Hz.

(c) Area ceiling = π × 49 = 154m2. At 125Hz, the Sabine absorptioncoefficient of tile is 0.2. Thus if all the ceiling were covered themaximum would be 154 × 0.2 = 30.8m2 (assuming that the floor andSαwalls contributed a negligible amount). Alternatively, if it is assumedthat the floor and walls are of concrete with = 0.01, then the totalα

= 30.8 + 0.01 × (154 + 2π × 7 × 2.5) = 33.4m2. The recommended α Sαat 125Hz is 43m2 (from part (b)), so the ceiling tile would NOT beadequate.

(d) A compromise would be to use sufficient tiles to achieve as closely aspossible the required absorption over the range 500 to 1000Hz and thendesign a panel absorber with a maximum absorption at 125Hz and noabsorption at 500Hz. The required absorption in the range 500 to4000Hz is 65m2. Thus the optimum amount of tile is 65/0.8 = 81m2,which will be OK at 500 and 2000Hz, a bit much at 1000Hz and a bitlittle at 4000Hz, but nevertheless a good compromise. We are then leftwith an area of ceiling of 154 - 81 = 73m2 for panel absorbers. Theamount of absorption needed at 125Hz is 43 - 3 - 81 × 0.2 = 24m2. Sothe panel absorber must have an absorption coefficient of 24/73 = 0.33at 125Hz and nothing at 500Hz.

Choosing curve H from figure 7.8 in the text and allowing for the factthat the 125Hz band includes the peak, the average for the 125Hz bandαis approximately 0.35. The absorption coefficient of the panel at 500 andabove is likely to be 0.08 and at 250Hz it will be 0.12. To optimise therequired absorption coefficients, we can vary the relative areas. The areaof floor and walls is 264m2. Thus the amount of absorption due to thefloor and walls is 2.6m2 in the 125, 250 and 500Hz bands and 5.3m2 inthe 1000, 2000 and 4000Hz octave bands. Including this and with anarea of 70m2 of tile and 65m2 of panel, the amount of absorption in theoctave bands from 125Hz to 4000Hz is 41, 53, 64, 70, 67, and 63m2

which is a little low at 125Hz and 250Hz (optimum is 43 and 59respectively) and a little high in the other bands (optimum at 500Hz andhigher frequencies is 62m2). Choosing the area of panel = 75m2 and thearea of tile = 65m2, gives the following amounts of absorption: 42, 51,61, 67, 63, 60m2 which is close enough. Note that there are many otheradequate solutions to this problem.


Lw ' Lp % 10log10(2πr 2) & 0.15 ' Lp % 21.85

dNdf

'4 × π × 1252 × 30

3433%

π × 125 × 62

2 × 3432%

408 × 343

' 0.264

Problem 7.38

(a) sound power of the source is given by:



63 125 250 500 1k 2k

Lp (dB re 20µPa) 90 85 78 73 70 65

Lw (dB re 10-12W) 112 107 100 95 92 87

(b) Room dimensions 5m × 3m × 2m. Using equation 7.17 in the text, wehave:

f1,0,0 '343

2×

15

' 34.3Hz

f0,1,0 '343

2×

13

' 57.2Hz

f0,0,1 '343

2×

12

' 85.8Hz

f2,0,0 '343

2×

25

' 68.6Hz

f1,1,0 '343

2× (1 /5)2 % (1 /3)2 ' 66.7Hz

Thus the 3 lowest order modes are in the 31.5Hz and 63Hz third octavebands.

Room volume, V = 5 × 3 × 2 = 30m2

Area, S = 2(5 × 3 + 5 × 2 + 3 × 2) = 62m2.Perimeter, L = 4(5 + 2 + 3) = 40The modal density is given by equation 7.21 in the text. Thus:


N '4 × π × 1413 × 30

3 × 3433%

π × 1412 × 62

4 × 3432%

40 × 1418 × 343

' 19modes

N '4 × π × 1133 × 30

3 × 3433%

π × 1132 × 62

4 × 3432%

40 × 1138 × 343

' 11.4modes

From table 1.2 on p43 in the text, the 125Hz third octave bandwidth is141 - 113 = 28Hz, so the number of modes in the band is28 × 0.264 = 7 to 8 modes.Alternatively equation 7.20 could be used to calculate the number ofmodes occurring below 141Hz and 113Hz and then take the difference.The number of modes below 141Hz is:

The number of modes below 113Hz is:

Thus the number in the 125Hz octave band is between 7 and 8 modes.

(c) Equations 7.51, 7.43 and 7.42 (with a correction for ρc = 413.6) in thetext, and the knowledge that S = 62m2, may be used to construct thefollowing table.

Octave bandcentre

frequency (Hz)63 125 250 500 1k 2k Overall

Lw

(dB re 10-12W) 112 107 100 95 92 87

T60 5.5 5 4 3 2 1.5

S α 0.877 0.964 1.206 1.608 2.411 3.215

α 0.0141 0.0156

0.0194

0.0259

0.0389

0.0519

R 'Sα

1 & α

0.890 0.979 1.230 1.650 2.509 3.391

Lp(reverb) 118.7 113.3 105.3 99.0 94.2 87.9

A-weighting -26.2 -16.1 -8.6 -3.2 0.0 1.2

Lp (dB(A)) 92.5 97.2 96.7 95.8 94.2 89.1 102.8


Assumptions:

! A-weighting assumed uniform across each octave band when infact it varies continuously with frequency.

! Direct field contribution assumed negligible.! Reflections from and absorption of surfaces other than room

boundaries is excluded.

(d) If 2 more generators were added, the sound pressure level would increaseby .10 log10(2 % 1) ' 4.8dB(A)

(e) Ceiling tiles added with area = 15m2. Remaining room surface area = 62- 15 = 47m2 = Sfloor, walls. The following table may be constructed, where:

.αoverall ' αnew 'S αfloor walls % S αceiling

Sfloor walls % Sceiling

Octave bandcentre

frequency (Hz)

63 125 250 500 1k 2k Over-all

Lw

(dB re 10-12W) 112 107 100 95 92 87

Sαfloor walls0.66 0.73 0.91 1.22 1.83 2.44

αceiling0.15 0.25 0.55 0.85 1.0 1.0

Sαceiling2.25 3.75 8.25 12.75 15 15

αoverall ' αnew0.0469 0.0722 0.1477 0.2253 0.2715 0.2813

Rnew 'Sαnew

1 & αnew

3.05 4.83 10.75 18.03 23.10 24.27

Lp(reverb) 113.3 106.3 95.9 88.6 84.5 79.3 114.2

A-weighting -26.2 -16.1 -8.6 -3.2 0.0 1.2

Lp (dB(A)) 87.1 90.2 87.3 85.4 84.5 80.5 94.5

Similar assumptions as made in part (c).


αst ' 1 & e&55.25V /ScT60

Problem 7.39

Room 20m × 15 × 4m,S = 2(20 × 15 + 20 × 4 + 15 × 4) = 880m2.V = 20 × 15 × 4 = 1200m3.

The desired reverberation times are calculated using equation 7.121 on p. 329in the text, with 10% increase at 250Hz, 50% increase at 125Hz and 100%increase at 63Hz. The existing mean statistical absorption coefficient may becalculated using equation 7.56 in the text rearranged to give:


Octavebandcentre

frequency(Hz)

ExistingT60

Existingmean αst

DesiredT60

Desiredmeanαst

Requiredincrease in

αst

631252505001000200040008000

3.02.62.32.12.02.02.01.8

0.07060.08100.09110.09930.10400.10400.10400.1149

1.861.401.020.930.930.930.930.93

0.1110.1450.1940.2110.2110.2110.2110.211

0.0400.0640.1030.1120.1070.1070.1070.096

There are many possible solutions to achieve the desired mean absorptioncoefficients. One alternative is to look for the frequency at which theadditional absorption required is the largest and choose a material which hasa maximum absorption coefficient at this frequency. In this case, themaximum increase in absorption is needed at 500Hz, so a material thicknessof 25mm should be chosen. The amount of material required is thencalculated on the basis of achieving the optimum reverberation time in theoctave bands most important for speech; namely, 500Hz to 2000Hz. In thiscase, it would seem that the required increase in mean absorption coefficientis 0.107 which would satisfy the requirements at and above 1000Hz, with acompromise of a little less than desired at 500Hz. let x be the fraction of roomsurface area to be covered with absorbing material. Using equation 7.58, we


(1 & 0.211) ' (1 & 0.85)x × (1 & 0.104)(1 & x)

log10 (0.79) ' x log10(0.15) % (1 & x)log10(0.896)

W '¢p 2

i ¦S4ρc

W 'τ ¢p 2

i ¦S4ρc

¢p 2i ¦ ' 4× 10&10 × 1088/10 ' 0.252Pa 2

W '1.995 × 10&3 × 0.252 × 1.5

4 × 1.206 × 343' 0.457µW

¢p 2 ¦ ' 1.206 × 343 × 4.565 × 10&7

2 × π × 602' 8.349 × 10&9 Pa 2

have:

Taking logs of both sides, we obtain:

which gives x = 0.0711. So the required area of 25mm thick acoustic materialis 63m2.

Problem 7.40

Assume that the house is approximately in a direction along the normal axisfrom the window. The power incident on the window is the intensity in thedirection of the window multiplied by the area of the window. Thus

The power radiated through the window is then

where = . T he reverberant sound pressureτ ' 10&TL /10 10&2.7 ' 1.995 × 10&3

level is 88dB. Thus

The power radiated through the window is then

For an incoherent plane source, the on-axis sound pressure at the receiver is

given by equation 5.106. The quantity . Thus fromr / HL ' 60/1.5 ' 40figure 5.11, it is clear that we can treat it as a hemispherically radiating pointsource producing a sound pressure described by equation 5.106 in the text.Thus:


Lp ' 10log10(8.349 × 10&9) % 94 ' 13.2dB re 20µPa

q ' 1000 ×π × 0.0022

4 × 0.07' 6.7mm

q ' 1000 ×π × 0.0022

2 × 0.07 × 2 3' 5.1mm

This corresponds to a sound pressure level of :

As the ground is hard asphalt, we may add 3dB to the level to account for theeffect of ground reflection. Thus the expected level at the house is 16dB.

Problem 7.41

(a) First calculate distance between holes. Could assume parallel orstaggered holes as shown in the two figures to follow.

Let q be the distance between holes as shown in the figures. Choosing asegment of plate as shown in the figure we can calculate the ratio of holes tototal area of the segment and set this equal to 0.07. This gives a hole spacingfor the parallel holes of:

and for the staggered holes,

For the purposes of this problem we will use q = 6.7mm. Using equationE.7 in the 2nd. edn. of the text on p528, we have:


fmax2 × π × 0.1

343tan fmax ×

2 × π × 0.1343

'7 × 0.1 /100

0.003 % 0.85 × 0.002 × (1 & 0.22 × 0.002 /0.0067)

0.00183 × fmax tan fmax × 0.00183 ' 1.5256

a(X) '3.1696(1.63259 & 3.1696) × 0.06592 & 0.16632 × 1.632592

3.16962 × 0.06592 % 0.16632 × 1.632592

' &0.80849

Rewriting gives:

Can solve by trial and error, choosing values of fmax until the LHS =1.5256 as illustrated in the table below.

fmax(Hz) LHS

100500600550540543

542.9

0.03391.18922.14671.59021.50001.52631.5254

Thus the frequency of maximum absorption is 543Hz.

If we used Equation 7.77 in the 3rd edn. we would get 691 Hz, but theerror is greater than 15% because the condition, fL/c < 0.1 is not satisfied.

(b) Specific normal impedance is given by equation C.43. To evaluate thisequation we need to use equation C.41 and to evaluate that we needequations C.3 and C.4. Referring to equation C.15, X = 1.206 ×543/10000 = 0.0655. Thus, T1 = 1.63259, T2 = 0.06590, T3 = 3.1696, T4

= 0.1663. Referring to equation C.9:

and


b(X) '1.632592 × 0.0659 × 0.1663

3.16962 × 0.06592 % 0.16632 × 1.632592' 0.24893

a(X1) '3.0785(1.5415 & 3.07850) × 0.056362 & 0.165262 × 1.54152

3.07852 × 0.056362 % 0.165262 × 1.54152

' &0.84133

b(X1) '1.54152 × 0.05636 × 0.16526

3.07852 × 0.056362 % 0.165262 × 1.54152' 0.23297

κ ' (1 & 0.4 × (&0.49807 % j0.23297))&1 ' (1.19923 % j0.093188)&1

ρm ' (1 & 0.8085 % j0.2489)&1 ' (0.1915 % j0.24893)&1

Zm

ρc' ρmκ

'1

(0.1915 % j0.24893) × (1.19923 & j0.093188)

'1

0.25285 % j0.28068

' 1.7717 & j1.9667 ' 2.6471e& j0.8375

' 1.6270e&j0.4188 ' 1.4864 & j0.6616

X1 = 0.856 × 1.206 × 543/10000 = 0.05605. Thus,T1 = 1.54150, T2 = 0.05636, T3 = 3.07850, T4 = 0.16526. Referring toequation C.9:

and

Thus τ = -0.84133×0.592 + j0.23297 and σ = -0.80849 + j0.24893. Thequantities κ and ρm may be calculated using equations C.5 and C.6 asfollows:

and

Using equations C.3 , we obtain:


km '2πfmax

c

ρm

κ'

2π × 543343

1.19923 & j0931880.1915 % j0.24893

' 9.9468(1.19923 & j0.093188) × (0.1915 & j0.24893)

0.098638

' 9.9468 2.09319 & j3.2074 ' 9.9468 3.8300e& j0.99259

' 19.4663e& j0.49629 ' 17.1178 & j9.2692

j tan(kmR) '

e0.9269 cos(1.7118) % jsin(1.7118& e&0.9269 cos(1.7118 & jsin(1.7118)

e0.9269 cos(1.7118) % jsin(1.7118)% e&0.9269 cos(1.7118) & jsin(1.7118)

'&0.2995 % j2.8934&0.4107 % j2.1097

' 1.3480 & j0.1205

tan(kmR) ' &0.1205 & j1.3480

ZN

ρc'

(0.6616 % j1.4864)0.1205 % j1.3480

'2.0834 & j0.7127

1.8316

' 1.1375 & j0.3891

and using equation C.4b we obtain:

Before continuing, it will be useful to evaluate the quantity, tan(kmR).Using the previous result for km, and setting R = 0.1, we can write:

Thus:

Using the previous results and equation C.41 (assuming a rigid backingfor the porous material), we can write:

To calculate the overall impedance, we use equation C.43, but first weneed to evaluate R and tan(kR).

From equation 9.25 in the text, the effective length of the holes in theperforated sheet is:


R ' 0.003 %16 × 0.001

3 × π1 & 0.43 × 0.001/0.0067 ' 0.004588

tan(kR) ' tan(2 × π × 543 × 0.004588/343) ' 0.04567

t '2 × 1.8 × 10&5

1.206 × 2 × π × 543' 9.3538 × 10&5

Thus:

We also need to calculate the acoustic resistance of the holes usingequation 9.29 in the text. To evaluate this equation we need the followingquantities:

, k ' (2 × π × 543) /343 ' 9.9467

; A ' π × 0.0022 /4 ' 3.1416 × 10&6 m 2 D ' π × 0.002 ' 0.006283

ε = 1.0 as radiation from a baffle.

The quantity, h, is the largest of the half plate thickness or t. Thus:h = w/2 = 0.0015m.

The above quantities may be inserted into equation 9.29 to give:


ZP

ρc'

(100/7)(0.04567j % 4.1256 × 10&3)

1 %100 × 1.206 × 343

2 × π × 543 × 21.762 × 7× 0.04567 & 4.1256j × 10&3

'0.05894 % 0.6524j

1.00363 & 3.2836j × 10&4

' 0.05894 % 0.6548j

ZN

ρc%

ZP

ρc' 1.1375 & 0.3891j % 0.05894 % 0.6548j

' 1.1964 % 0.2657j ' 1.2255 e0.21854j

Ra A

ρc'

9.9467 × 9.3538 × 10&5 × 6.283 × 10&3 × 0.003

2 × 3.1416 × 10&6

× 1 % (1.4 & 1) 53 × 1.4

% 0.288 × 9.9467 × 9.3538 × 10&5 × log104 × 3.1416 × 10&6

π × 0.00152

%3.1416 × 10&6 × 9.94672

2 × π

' 4.00923 × 10&3 % 6.69555 × 10&5 % 4.9468 × 10&5

' 4.1256 × 10&3

We can now use equation C.43 to evaluate the overall impedance. The secondterm on the right is the impedance due to the perforated sheet and is:

Thus the total impedance is:

cosβ = 0.9762; cos2β = 0.9060; sinβ = 0.2168 and ξ = 1.2255.

Using the above data, the statistical absorption coefficient may becalculated using equation C.37 in the text as follows:


αst '8 × 0.9762

1.22551 &

0.97621.2255

× loge (1 % 2 × 1.2255 × 0.9762 % 1.22552 )

%0.9060

1.2255 × (0.2168)× tan&1 1.2255 × (0.2168)

1 % 1.2255 × 0.9762

' 6.373 × (1 & 0.7966 × 1.5881 % 3.410 × tan&1 [0.12097])

' 6.373 × (1 & 1.2651 % 0.4105) ' 0.93

NRC 'α250 % α500 % α1000 % α2000

4'

0.6 % 0.8 % 1.0 % 1.04

' 0.85

Problem 7.42

The NRC is given by:

So the material is adequate for the purpose.

Problem 7.43

Truck emits 110 dB. This is equal to: .W ' 10&12 × 10110/10 ' 0.1 watts r/a = 60/6 = 10.

From Fig 7.14, reverberant field pressure squared is:

10log10 ¢p 2R ¦ & 10log10

Wρc

πa 2' &4 dB

Thus, 10log10 ¢p 2R ¦ ' 10log10

0.1 × 413.6π × 36

& 4 ' &18.4 dB

so reverberant field pressure is, LpR ' &18.4 % 94 ' 75.6 dB

Direct field pressure: 10log10 ¢p 2D ¦ ' 10log10

0.1 × 413.6

4π × 602' &30.4 dB


So direct field pressure is, LpD ' &30.4 % 94 ' 63.6 dB

Total pressure, Lp ' 10log10 1075.4/10 % 1063.6/10 ' 75.7 dB

Assumptions:• Effective acoustic source location is in the centre of the cross section• Specularly reflecting surfaces• Ambient temperature of 20 EC

TL ' &10log10τ

8

Solutions to problems relating tosound transmission loss, acousticenclosures and barriers

Problem 8.1

(a) The Transmission Loss of a partition is an inverse decibel measure(bigger TL means a smaller amount of transmitted energy) of the amountof incident energy which is transmitted to the space on the side oppositethat on which the energy is incident. It is defined in terms of thetransmission coefficient, τ, which is the fraction of transmitted to incidentenergy, as follows:

It may be measured using two reverberant rooms with the panel to bemeasured acting as a partition between the two rooms with the spacearound the panel of high transmission loss construction so that all of thesignificant acoustic energy transmitted between the two rooms passesthrough the panel under test. The test is conducted by exciting the oneof the reverberant rooms with 1/3 octave band noise and then measuringthe difference in the space averaged sound pressure level in the tworooms. The appropriate mathematical analysis is embodied in equations8.13 to 8.16 in the text (page 342).

(b) Measurements often do not agree with theoretical calculations becausethe latter do not take into account the size of the panel exactly. Also theexperimental determination of space average sound pressure level is oftencharacterised by errors, especially at low frequencies when the soundfields in the two rooms are not sufficiently diffuse. Sometimes, energyis transmitted from one room to the next by way of paths not through thepanel (called "flanking"), resulting in Transmission Loss measurementswhich are too small.

Sound transmission loss, acoustic enclosures and barriers 205

TL ' 10log10 1 %πfmρc

2

& 5.5 (dB)

4 ' 10log10 1 %π × f × 7800 × 0.01

988 × 1481

2

& 5.5

f '100.95 & 1

2.8046 × 10&8' 16,800Hz

45o

45o

75 25

25

30

75

75All dimensions in mm

z0 zn

Problem 8.2

Mass Law Transmission Loss is obtained by combining equations 8.34 (withθ = 0E) and equation 8.35b, which gives:

Substituting in values for the variables gives:

Rearranging gives:

The Transmission Loss in air at this frequency is much greater because theimpedance of the panel compared to the characteristic impedance in thepropagating medium is much larger for air than water.

Problem 8.3

(a) First find location of neutral axis by taking moments about an axisthrough the centre of the angled section and shown as z0 in the figure. Inthe following equations bi is the length of the ith section. If the neutralaxis is denoted as zn where z is the vertical coordinate on the figure, then:


z0 & zn '

ji

bizi0

ji

bi

'75 × 15 % 2 × 25 × 2.5 & 25 × 10 % 75 × 15 & 75 × 15

75 % 3 × 25 % 75 % 2 2 × 30 %75

'1000384.9

' 2.6mm

B 'Eh

(1 & ν2)Rj

n

bn z 2n %

h 2 % b 2n

24%

h 2 & b 2n

24cos2θn

B1 '207 × 109 × 0.0012

0.91 × 0.310.15 0.01242 %

0.00122

12

% 0.05 0.00012 %0.0252

12% 0.025 0.01262 %

0.00122

12

% 0.075 0.01762 %0.00122

12

% 0.06 2 0.00262 %0.00122 % 2 × 0.032

24

' 8.805 × 108 (2.3082 × 10&5 % 2.6046 × 10&6 % 3.9720 × 10&6

% 2.3241 × 10&5 % 6.9427 × 10&6 )

' 5.27 × 104 kg m 2 s &2

Thus the neutral axis is 12.4mm from the centre of the top of the section.The section thickness, h = 1.2mm and the horizontal length, R, beforerepeating itself is 250 + 60mm = 0.31m.

The bending stiffness in the direction along the ribs may be calculatedwith E = 207GPa and = 0.3 using equation 8.10 in the text, which isνincorrect in the first printing of the text and should be:

Thus:


B2 '207 × 109 × 0.00123

12 × 0.91×

0.3850.31

' 40.7kgm 2 s &2

cB1 '5.27 × 104 × 4π2 × 106

11.62

1/4

' 650m/s

cB2 '40.7 × 4π2 × 106

11.62

1/4

' 108m/s

fc1 '3432

2π11.62

5.27 × 104

1/2

' 278Hz

fc2 '3432

2π11.6240.7

1/2

' 10,000Hz

Bab ' 0.5 5.27 × 104 × 0.3 % 40.7 × 0.3 %207 × 109 × 0.00123

3 × 2.6

' 7934

The stiffness in the direction across the ribs may be calculated usingequation 8.5 as follows:

(b) The bending wavespeed is calculated using equation 8.1 in the text. Thesurface mass of the panel is m = 7800 × 0.0012 × 0.385/0.31 = 11.62kgm-2 and the frequency is 1000Hz. Thus the lower and upper bendingwavespeeds corresponding to waves propagating parallel andperpendicular to the ribs respectively are:

(c) The lower and upper critical frequencies for the panel may be calculatedusing equation 8.3 in the text. Thus:

(d) Assuming that the enclosure wall edge condition is simply supported (agood approximation in practice for most enclosures), the first resonancefrequency of the panel may be calculated using equation 8.22 in the textwith i = n = 1. Thus:

and:


f1,1 'π

2 11.62

5.27 × 104

24%

40.7

24%

7934

24

1/2

' 28.4Hz

TLA ' 20log10(278 × 11.62) & 54 ' 16.2dB

TLB ' 20log10(278) % 10log10(11.62) & 10log10(278)

& 20log10[ loge(4)] & 13.2

' 19.1dB

TLC ' 20log10(5000) % 10log10(11.62) & 10log10(278)

& 20log10[ loge(20000/278)] & 13.2

' 34.4dB

TLD ' 10log10(11.62) % 15log10(10000) & 5log10(278) & 17

' 41.4dB

TLoct ' &10Log10 6 (1/3) [10&TL1/10

% 10&TL2/10

% 10&TL3/10

] >

(e) The sound transmission loss of the panel may be calculated using figure8.8b in the text. Point A is at 139Hz and the corresponding TL is givenby:

At point B (278Hz), the TL is:

At point C (5,000Hz), the TL is:

At point D (20,000Hz), the TL is:

These points are plotted on the following graph and interpolation is usedto find the octave band TL values. Strictly speaking, the curve shouldonly be used to find 1/3 octave values and the octave band levels mustthen be calculated from the following equation:


fc10.5fc1 0.5fc2

B

C

A

However, for most practical purposes, the results obtained that way arelittle different to the results obtained by reading the octave band levelsdirectly from the figure. However, in the case of isotropic panels, careshould be taken to avoid errors near the dip in the curve correspondingto the critical frequency. Following the figure, the octave band resultsare summarised in a table.


TransmissionLoss (dB)

63125250500

1000200040008000

915192125293337

Problem 8.4

(a) Only one half of the sine wave section needs to be considered.y1 = 20 sin(15π/40) = 18.48mm.


B1 '207 × 109 × 0.0016

0.91 × 0.04 × 10923.8 × 2 (18.48 /2)2 %

1.62 % 23.82

24

%1.62 & 23.82

24cos(101.8E) % 10 18.482 %

1.62

12

' 8.04 × 104 kg m 2 s &2

B2 '207 × 109 × 1.63 × 10&9

12 × 0.9157.640

' 111.8kg m 2 s &2

fc1 '3432

2π18

8.04 × 104

1/2

' 280Hz

fc2 '3432

2π18

111.8

1/2

' 7,500Hz

1

y1

15 10 15

b1 = 23.8mm, θ1 = 50.9Eb2 = 10mm, θ2 = 0b3 = 23.8mm, θ3 = 50.9El = 40, E = 207GPa

= 0.3ν

Using the corrected formof equation 8.10 in thetext, we can write thefollowing for the bendingstiffness for waves travelling parallel to the corrugations:

The bending stiffness for waves travelling perpendicular to thecorrugations can be calculated using equation 8.11 in the text as:

The surface mass is m = 7800 × 0.0016 × (23.8 + 23.8 + 10)/40 =18.0kg/m2.

The lower and upper critical frequencies are calculated using equation8.3 as:


f1,1 'π

2 18

8.04 × 104

34%

111.8

34

% 0.58.04 × 104 × 0.3

34%

111.8 × 0.3

34%

207 × 109 × 0.00163

34 × 3 × 2.6

1/2

' 12.5Hz

TLA ' 20log10(280 × 18.0) & 54 ' 20.0dB

TLB ' 20log10(280) % 10log10(18.0) & 10log10(280)

& 20log10[ loge(4)] & 13.2

' 21.0dB

TLC ' 20log10(3750) % 10log10(18.0) & 10log10(280)

& 20log10[ loge(15000/280)] & 13.2

' 34.4dB

TLD ' 10log10(18.0) % 15log10(7500) & 5log10(280) & 17 ' 41.4dB

Assuming that the enclosure wall edge condition is simply supported, thefirst resonance frequency may be calculated using equation 8.22 in thetext. Thus:

(b) The sound transmission loss of the panel may be calculated using figure8.8b in the text. Point A is at 140Hz and the corresponding TL is givenby:

At point B (280Hz), the TL is:

At point C (3,750Hz), the TL is:

At point D (15,000Hz), the TL is:

These points are plotted on the graph below and interpolation is used tofind the octave band TL values. Strictly speaking, the curve should only


TLoct ' &10log10 6 (1/3) [10&TL1/10

% 10&TL2/10

% 10&TL3/10

] >

fc10.5fc1 0.5fc2

B

C

A

be used to find 1/3 octave values and the octave band levels must then becalculated from the following equation:

However, for most practical purposes, the results obtained that way arelittle different to the results obtained by reading the octave band levelsdirectly from the figure. However, in the case of isotropic panels, careshould be taken to avoid errors near the dip in the curve correspondingto the critical frequency. Following the figure, the octave band resultsare summarised in a table.


TransmissionLoss (dB)

63125250500

1000200040008000

1319212326303538


fc '3432

2π36

111.8

1/2

' 10,600Hz

TLA ' 20log10(10,600 × 36) & 54 ' 57.6dB

TL ' 20log10(10,600 × 36) % 10log10(0.1) & 45 ' 56.6dB

0.5fc

A

(c) With viscoelastic damping the panel may be treated as isotropic with thesurface mass, m, now equal to 2 × 18 = 36kg/m2. The critical frequencyis:

On the isotropic panel curve, point A is at 5,300Hz (10,600/2) and the TLis:

At point B, the frequency is 10,600Hz and the TL is:

The results are plotted on the graph below from which can be read TLvalues as a function of 1/3 octave band centre frequency.

(d) Second panel, . The 1fc1 ' 0.55 × 3432 / (2000 × 0.013) ' 2500Hz

subscript is used because this critical frequency is smaller than that forthe other panel.

Surface mass, m = 0.013 × 1000 = 13kg/m2.


f0 ' 8036 % 13

36 × 13 × 0.1' 82Hz

TLA ' 20log10(36 % 13) % 20log10(81.86) & 48 ' 24dB

TLB1 ' 24 % 20log10(2500/82) & 6 ' 47.7dB

TLB2 ' 20log10(13 × 0.6) % 40log10(10600) & 99 ' 80dB

TLC ' 80 % 6 % 10log10(0.01) ' 66dB

Cavity resonance frequency is:

and the corresponding TL is:

At point B the frequency is 5300 Hz and the TL is:

and as there are rubber spacers, one panel may be considered to be pointsupported. Note that the panel with the higher critical frequency (thedamped corrugated panel in this case) must be the one which is pointsupported to obtain the high TL predicted. The value of TLB2 for line-point support is:

Assuming that there is sound absorbing material in the cavity, TLB = 80dB.

At point C, the frequency is 10,600Hz and the TL is:

At point D, the frequency is f1 = 55/0.1 = 550Hz.

The TL data are plotted in the following figure from which the 1/3 octavevalues can be read directly.

Problem 8.5

A double wall partition may perform more poorly than a single partition of thesame weight at resonance frequencies corresponding to acoustic modes in thecavity and also at the critical frequencies of the individual panels (if they arelightly damped).


f0 ' 80.41000 × 0.012 % 7800 × 0.0016

0.1 × 1000 × 0.012 × 7800 × 0.0016' 103.2Hz

TLA ' 20log10 (12 % 12.48) % 20log10(103.2) & 48 ' 20.0dB

fc1 '0.55 × 3432

2100 × 0.012' 2570Hz and 0.5fc1 ' 1280Hz

fc2 '0.55 × 3432

5400 × 0.0016' 7490Hz

f0 fR 0.5fc2

A

BD

C

Problem 8.6

Following the procedure on page 360 in the text, we have at point A:

and the corresponding Transmission Loss is:

The critical frequencies are:


TLB1 ' 20 % 20log10(2568/102.2) & 6 ' 42dB

TLB2 ' 20log10(12) % 10log10(0.6) % 30log10(7490)

% 20log10 1 %12.48 2570

12 7490& 78

' 61.7dB

TLC ' 61.7 % 6 % 10log10(0.01) ' 47.7dB

f0 fR 0.5fc2 fc2

A

BD

C

The Transmission Loss at point B is the larger of:

and:

At point C:

The frequency fR = 55/0.1 = 550Hz.

The Transmission Loss may thus be plotted as shown in the following figureand 1/3 octave (but not octave) values may be read directly from the figure.


0.001 'τw(18 & 0.25 & 2) % 0.25 × 10&28/10 % 2 × 10&25/10

18

τ '0.25 × 10&28/10 % 2 × 10&25/10

18' 3.73 × 10&4

τ '7.16 × 10&4 × 15.75 % 0.025 × 1 × 1 % 0.25 × 10&2.8 % 1.975 × 10&2.5

18

' 0.00238

TL ' &10log10 (2.38 × 10&3) ' 26.2dB

Problem 8.7

(a) TLoverall = 30dB, so τ = 10-30/10 = 0.001. If we let the required transmissioncoefficient of the wall be τw, then we can write:

From which τw = 7.161 × 10-4. Thus the required wall Transmission Lossis, TLw = -10log10(7.161 × 10-4) = 31.5dB.

(b) Maximum TL possible corresponds to τw = 0. Thus:

which corresponds to TL = 10log10(3.73 × 10-4) = 34.3dB

(c) 25mm crack under the door. Effective crack height is 50mm due toreflection in the floor, and τcrack = 1.0. Thus the overall transmissioncoefficient is:

which corresponds to a Transmission Loss of:

Problem 8.8

When designing an enclosure, always (if possible) include sound absorptivematerial on the ceiling and walls, thus resulting in an "average" enclosure withthe coefficients, C, as listed in the table below (along with the required noisereduction and corresponding wall TL given by TL = NR + C).


fc(steel) '0.55 × 3432

5400 × 0.003. 4000Hz ' fc2

fc(plaster) '0.55 × 3432

1600 × 0.025. 1600Hz ' fc1

m1 ' 760 × 0.025 ' 19kg/m 2 and m2 ' 7800 × 0.003 ' 23.4kg/m 2

f0 ' 8019 % 23.4

0.1 × 19 × 23.4' 78Hz

TLA ' 20log10(19 % 23.4) % 20log10(78) & 48 ' 22dB at 78Hz

TL63 ' TLA % 60log10(63 / f0) ' 27

20log10(m1 % m2) % 20log10 f0 % 60log1063 & 60log10 f0 ' 27 % 48

Octavebandcentre

frequency(Hz)

63 125 250 500 1000 2000 4000 8000

RequiredNR

14 18 25 35 50 40 40 40

C 13 11 9 7 5 4 3 3

requiredwall TL

27 29 34 42 55 44 43 43

100mm studs implies gap, d = 0.1m. Assume that the door and window havethe same TL as the walls.

To begin, try 3mm steel and 25mm plasterboard.

This is insufficient as we need 27dB at 63Hz. It seems that we need to lowerf0 below 63Hz, to put this point on the 18dB/octave slope.

That is:


20log10(m1 % m2) & 40log10f0 ' &33 (1)

40 log10f0 ' 40log1080 % 20log10(m1 % m2) & 20log10(m1 × m2) % 20

' 96 % 20log10(m1 % m2) & 20log10(m1 × m2)

20 log10(m1 × m2) ' 63dB or m1m2 ' 1410

fc1 '0.55 × 3432

1600 × 0.05. 810Hz

fc2 '0.55 × 3432

5400 × 0.005. 2400Hz

m1 ' 760 × 0.05 ' 38kg/m 2 and m2 ' 7800 × 0.005 ' 39.0kg/m 2

f0 ' 8038 % 39

0.1 × 38 × 39' 58Hz

TLA ' 20log10(38 % 39) % 20log10(58) & 48 ' 25dB at 58Hz

TLB1 ' 24.9 % 20log10(810 /57.7) & 6 ' 41.8dB

or

Using the equation for f0 on page 357 in the text and taking logs gives:

Substituting the above expression into equation (1) above gives:

Try 50mm plasterboard (same as gypsum board), m = 760 × 0.05 = 38kg/m2.Required steel weight = 1410/38 = 37kg/m2, which is 4.7mm thick. That is,use 50mm thick plasterboard and 5mm thick steel plate.

Assume a stud spacing of 0.6m and line-line support. Thus:


TLB2 ' 20log10(38) % 10log10(0.6) % 30log10(2400)

% 20log10 1 %39 × 8101/2

38 × 24001/2& 78 ' 57dB

TLC ' 57 % 6 % 10log10(0.005) ' 40dB

f0fc2 fc2

fR0.5fc20.5fc2

A

B

B

C

C

As the cavity is filled with sound absorbing material, TLB = 57dB. Assumea loss factor for the steel of 0.005 (see table on p.609 in text). Then:

The frequency at point D is fR = 55/0.1 = 550Hz.

The TL for this construction is plotted on the following figure (where thedashed line is the required Transmission Loss and the solid line is thepredicted Transmission Loss, calculated using Figure 8.9 in the text), where

it can be seen that the design is deficient between 1000Hz and 2500Hz. It isclear that the first critical frequency must be increased or the quantity TLB

(and hence TLC) must be increased. An easy solution to the problem is to usepoint supports for the steel panel but this may not be practical. Thus try75mm thick plaster board (m = 760 × 0.075 = 57kg/m2. The requiredthickness of steel panel is such that steel weight = 1410/57 = 25kg/m2, which


fc1 '0.55 × 3432

1600 × 0.088. 460Hz

fc2 '0.55 × 3432

5400 × 0.003. 4000Hz

m1 ' 760 × 0.088 ' 66.9kg/m 2 and m2 ' 7800 × 0.003 ' 23.4kg/m 2

f0 ' 8066.9 % 23.4

0.1 × 66.9 × 23.4' 61Hz

TLA ' 20log10(66.9 % 23.4) % 20log10(61) & 48 ' 27dB at 61Hz

TLB1 ' 26.8 % 20log10(460 /61) & 6 ' 38.3dB

TLB2 ' 20log10(66.9) % 10log10(0.6) % 30log10(4000)

% 20log10 1 %23.4 × 4601/2

66.9 × 40001/2& 78 ' 65dB

TLC ' 65 % 6 % 10log10(0.005) ' 48dB

is 3.2mm thick. This is an odd thickness, so try 88mm thick plasterboard (3× 25 + 13) m = 760 × 0.088 = 66.9kg/m2. Required steel weight = 1410/66.9= 21kg/m2, which is 2.7mm thick. That is, use 88mm thick plasterboard and3mm thick steel plate.

Assume a stud spacing of 0.6m and line-line support. Thus:

As the cavity is filled with sound absorbing material, TLB = 65dB. Assumea loss factor for the steel of 0.005. Then:

The frequency at point D is fR = 55/0.1 = 550Hz.

The TL for this construction is plotted on the previous figure as the grey line,where it can be seen that the proposed construction easily meets the noisereduction requirements and is even a little too good.


Problem 8.9

There is a 5mm air gap under door. For the purpose of calculating thetransmission coefficient, the reflection in the floor effectively doubles thewidth of the gap. However, once the transmission coefficient has beendetermined using figure 8.11 in the text, the area of gap in subsequentcalculations is determined without doubling the width.

Area of walls and ceiling = 2(4 × 2.5 + 3 × 2.5) + 4 × 3 = 47m2. Area underdoor = 0.005 × 1 = 0.005m2. Effective gap under door = 0.01m. τgap iscalculated using figure 8.11 in the text. We can construct the following tableusing equations 8.12, 8.65 and 8.66 in the text.


63 125 250 500 1000 2000 4000 8000

TLwall 27 45 51 56.5 60.5 52 49 63

Swallτwall 0.0938 0.00149 3.73×10-4

1.05×10-4

4.19×10-5

2.97×10-4

5.92×10-4

2.36×10-5

Sgapτgap 0.005 0.005 0.003 0.0016 0.001 5.5×10-4

2.5×10-4

1.3×10-4

τ 2.10×10-3

1.38×10-4

7.18×10-5

3.63×10-5

2.22×10-5

1.80×10-5

1.79×10-5

3.26×10-6

TL 27 39 41 44 47 47 47 55

From the table it can be seen that the effect of the gap under the door is tosignificantly reduce the effective wall TL and on comparing the results in theabove table with the table of problem 8.8, it can be seen that the requiredenclosure noise reduction will no longer be achieved in the 1000Hz octaveband.

Problem 8.10

The required air flow may be calculated using equation 8.84 in the text, inwhich ρ = 1.206kg/m2, Cp = 1010m2s-2C-1, ΔT = 3EC and H = 0.05 × 104W.Thus:


V '0.05 × 104

3 × 1010 × 1.206' 0.14m 3 /s

The required Insertion Loss specifications for the silencer would be the sameas the TL of the walls (not the noise reduction required of the enclosure as thisexcludes reverberant build-up in the enclosure) and this is found in problem8.8.

Problem 8.11

The following steps would need to be taken.

1. Check local noise regulations for allowable levels in the residential area.Measure existing levels in dB(A) on the perimeter of the supermarketproperty at the closest location to the proposed compressor location overan extended period (about a week with a statistical noise analyser, or ifthis is impractical use a sound level meter).

2. Choose as the design criterion for the compressor noise in the communitythe smallest of the regulation level and the lowest measured existing levelplus 5dB(A). If existing levels were determined using spot checks witha sound level meter, then the criterion may be more conservative, such asthe existing level plus 2dB(A).

3. Use table 4.8 to convert the allowable dB(A) level in the community toan NR level and then use the corresponding NR curve (figure 4.8) tospecify the allowable community noise levels in octave bands.

4. Calculate the noise reduction from the compressor site to the nearestcommunity location due to atmospheric absorption, turbulence, groundeffects and meteorological influences, or place a loudspeaker at theproposed compressor location and measure the noise reduction as afunction of distance from it.

5. Use the sound power data for the compressor and the excess attenuationdata together with equation 5.158 in the text to calculate the noise levelsdue to the compressor at the nearest community location in octave bands.

6. Thus calculate the required enclosure noise reduction and following that,


Lp ' Lw & 10log10(2πr 2) ' 105 & 10log10(2π × 802) ' 59dB

TL ' 10log10 1 %πfmρc

2

& 4

' 10log10 1 %π × 500 × 7800 × h

1.206 × 343

2

& 4 dB

the corresponding required wall transmission loss.

7. Check compressor cooling requirements and if necessary design linedinlet and outlet ducts (with forced air ventilation) with the same InsertionLoss as the Transmission Loss of the enclosure walls.

Problem 8.12

(a) Compressor is 80m from perimeter. The sound level at the receiver withno enclosure (assuming hard ground, zero reflection loss) may becalculated using equation 5.158 in the text with DIM = AE = 0, so that:

The required noise level = 38dB, so reduction required = 21dB at 500Hz.

Smallest panel of enclosure = 0.6m × 0.6m (stud spacing), so 500Hz iswell above the first panel resonance frequency (you can calculate thisusing equation 8.21 to be sure).

Critical frequency is:. Sofc ' 0.55 × 3432 / (5400 × h) ' 12/h Hz, where h is in metres

500Hz is in the mass law range for all choices of panel.

From equation 8.79 in the text, the required panel TL = NR + C, and foran enclosure lined on the inside, C = 7 (see table 8.4, p376 in the text).Thus the required enclosure TL = 21 + 7 = 28dB at 500Hz.

(b) The wall TL may be calculated in the mass law range by using equation8.36 in the text (with the assumption NOT made that fm/ρc > 1), with aconstant of 4 instead of 5.5 to account for octave band calculations.Thus:


f0 ' 80 2/(0.1 × 9.88) ' 114Hz

fc1 ' fc2 '0.55 × 3432

1600 × 0.013' 3100Hz

TLA ' 20log10(9.88 × 2) % 20log10(113.8) & 48 ' 19.0dB

TLB1 ' 19 % 20log10(3100/114) & 6 ' 42dB

When h = 0.001m, TL = 25.4dB which is too low. As we are in mass lawrange a doubling of panel thickness will increase the TL by 6dB whichis a bit high. Trying h = 0.0016m, gives TL = 29.5dB which is OK, sochoose a wall thickness of 1.6mm.

(c) Should consider the effect of a door and the design of appropriate doorseals, the need for cooling air, the introduction of the inlet air, the pipepenetration (to be isolated from the enclosure wall) for the compressedair and the need for vibration isolation of the enclosure from thecompressor.

Problem 8.13

Equation 8.79 in the text gives TL = NR + C. From Table 8.3, C = 5dB for anenclosure with surfaces lined with sound absorbing material. Thus,TL = 15 + 5 = 20dB

Problem 8.14

(a) The performance of a machinery noise enclosure should not be expressedas a single number dB(A) rating because the dB(A) performance will bedependent on the spectrum shape of the noise generated by the enclosedsource.

(b) We may use equation 8.79 in the text to relate TL and noise reduction andthus define the minimum required TL. The TL due to the panel may becalculated using the procedure on p360 in the text.

d = 0.1m, m1 = m2 = 760 × 0.013 = 9.88kg/m2.


TLB2 ' 20log10(9.88) % 10log10(0.6) % 30log10(3100)

% 20log10(2) & 78 ' 50dB

TLC ' 50.4 % 6 % 15log10(0.02) ' 31dB

fR ' 55/0.1 ' 550Hz

TLoct ' &10log10 6 (1/3) [10&TL1/10

% 10&TL2/10

% 10&TL3/10

] >

f0 fR 0.5fc2 fc2

A

BD

C

For line-line support:

Assuming that there is sound absorbing material in the wall cavity, TLB

= 50dB.

The corresponding TL curve is plotted in the figure below and the octaveband data are listed in the following table, where octave band results areobtained by using 1/3 octave values read from the graph and calculatedusing:

where TL1, TL2, and TL3, are the (1/3) octave band data read from thefigure.


V 'H

ρCp ΔT'

0.02 × 500001.206 × 1010 × 5

' 0.164m 3 /s

Lw ' Lp1 % TL % 10log10SE & C ' Lp1 % NR % 10log10SE

OctaveBand

CentreFrequency

63 125 250 500 1000 2000 4000 8000

TL (from plot)

14 21 35 41 47 43 36 52

C 13 11 9 7 5 4 3 3

NR 1 10 26 34 42 39 33 49

RequiredNR

10 15 20 25 30 35 40 13

It can be seen that the enclosure is deficient in the 63Hz, 125Hz, and4000Hz octave bands, so it is not adequate.

(c) ! Use double, staggered stud wall.! Use point-line or point-point support by placing rubber grommets

between the panel and stud at attachment points.! Use thicker panels.! Fix additional 13mm thick panels to existing panels with patches of

silicone sealant.! Use panels of different thickness.

(d) From equation 8.84 in the text:

(e) The sound power level is given by equation 8.70 in the text. Thus:

where SE = 2(5 × 2.5 + 4 × 2.5) + 4 × 5 = 65m2 and 10log10SE = 18dB.The power level calculations may be summarised in the following table.


Lpd ' Lw & 10log10S ' Lw & 16dB

¢p 2 ¦ ' ¢p 2d ¦ % ¢p 2

R ¦

OctaveBand

CentreFrequency

63 125 250 500 1000 2000 4000 8000

Lp1 80 83 78 73 70 60 60 60

NR 10 15 20 25 30 35 40 20

Lw 108 116 116 116 118 113 118 98

(f) Test surface 1m from machine of dimensions 2m × 1m × 1m, someasurement surface is 4m × 3m × 2m (machine assumed to be restingon the ground), having an area of S = 2(4 × 2 + 3 × 2) + 4 × 3 = 40m2 (4sides and 1 top). Machine surface area Sm = 2(2 × 1 + 1 × 1) + 2 × 1 =8m2. The sound pressure is related to the sound power by equation 6.25in the text with Δ1 = 0 because the measurements are made outdoors. Theratio, S/Sm = 5, so the near field correction Δ2 = 0 and the sound pressurelevel 1m from the machine surface is given by:

The results are tabulated in the following table:

OctaveBand

CentreFrequency

63 125 250 500 1000 2000 4000 8000

Lw 108 116 116 116 118 113 118 98

Lpd 92 100 100 100 102 97 102 82

(g) With the enclosure in place, the mean square sound pressure is equal tothe sum of the direct and reverberant field contributions. Sound pressurelevels are converted to mean square sound pressures using equation 1.78in the text and sound powers are converted to sound power levels andvice versa using equation 1.80 in the text. Thus the mean square soundpressure in the enclosure is:


¢p 2R ¦ ' 4Wρc

R

4R

'4

3.3310C /10

0.3 × SE

&1SE

The reverberant mean square pressure is:

where R is the room constant and W is the source sound power. Theenclosure constant C is given in terms of R by equation 8.82 in the textwhich can be rearranged (with the aid of equation 7.43) to give:

The results are summarised in the following table, where forconvenience, ρc = 400.

OctaveBand

CentreFrequency

63 125 250 500 1000 2000 4000 8000

C 13 11 9 7 5 4 3 3

Lw 108 116 116 116 118 113 118 98

W 6.3×10-2

0.40 0.40 0.40 0.63 0.20 0.63 6.3×10-3

4/R 1.21 0.76 0.47 0.29 0.18 0.14 0.10 0.10

Lpd 92 100 100 100 102 97 102 82

¢p 2d ¦ 0.63 4.0 4.0 4.0 6.3 2.5 6.3 0.063

¢p 2R ¦ 30.5 122 75.2 46.4 45.4 11.2 25.2 0.25

¢p 2d ¦ % ¢p 2

R ¦ 31 126 79 50 52 14 32 0.31

LpT 109 115 113 111 111 105 109 89

dB(A)correct.

-26 -16 -9 -3 0 1 1 -1

dB(A)sound

pressurelevels

92 100 100 100 102 97 103 88


Lp2 ' 60 % 10log10(65) % 10log102

4 × π × 2002& AE

' 24 & AE (dB)

(h) In the 2000Hz band, Lp1 = 60dB and the sound pressure level at distancer from the enclosure is given by equation 8.72 in the text (with Dθ = 2due to the hard ground surface) as:

AE is the excess attenuation given by equation 5.165 in the text.

Ag is included in Dθ above as the asphalt is hard resulting in essentiallyhemispherical spreading.

Aa = 10.8/5 = 2dB (see table 5.3 on page 225 in the text).

Am = +5, -4dB (see table 5.10)

Thus the range of variability is 18 to 27dB.

Problem 8.15

Noise level inside enclosure = 101dB at 1000Hz.Noise level on surface, 1m from enclosure = 91dB.Noise level at 50m distance = 70dB.

(a) Sound power level radiated by enclosure may be calculated usingequation 6.25 in the text.

Test surface 1 m from enclosure of dimensions 3m × 3m × 3m, someasurement surface is 5m × 5m × 4m (assuming that the enclosure isresting on the ground), having an area ofS = 2(5 × 4 × 2) + 5 × 5 = 105m2 (4 sides and 1 top). Machine surfacearea, Sm = 2(3 × 3 × 2) + 3 × 3 = 45m2. The sound pressure is related tothe sound power by equation 6.25 in the text with Δ1 = 0 because themeasurements are made outdoors. The ratio, S/Sm = 105/45 = 2.3, so thenear field correction Δ2 = 1 and the sound power level of the enclosureis given by:


Lw ' 91 % 10log10(105) & 1 ' 110dB

Lp ' 110 & 20log10(50) & 10log10(2π) ' 68dB

(b) We may use equations 5.158 and 5.161 with r = 50m and DIM = AE = 0so that:

(c) As the measured noise level is 70dB, the excess attenuation due toatmospheric absorption and meteorological influences is 68 - 70 = -2dB.The excess attenuation due to the ground effect is -3dB, so the totalexcess attenuation is -5dB.

(d) The machine is probably not well vibration isolated from the enclosure,causing the enclosure wall to vibrate and radiate noise. Vibration couldbe transmitted by way of the floor or by direct connection of parts of themachine (or attached pipework) to parts of the enclosure. Also, pipeworkor other equipment not included in the enclosure but attached to the noisymachine could radiate noise which was not apparent prior to installationof the enclosure.

Problem 8.16

Adequate internal absorption is necessary to prevent the build-up ofreverberant energy which will compromise the predicted acousticperformance.

Problem 8.17

(a) The enclosure should be vibration isolated to prevent the walls frombeing excited to vibrate and thus radiate sound which in turn willcompromise the enclosure performance.

(b) One possible disadvantage associated with vibration isolation from thefloor is increased difficulty in producing an adequate acoustic sealaround the base of the enclosure.

(c) Other factors which could degrade the enclosure performance are! poor seals around doors and windows


! inadequate TL performance of doors and windows! inadequate internal absorption! pipe penetrations in the enclosure walls not vibration isolated from

the walls or poorly sealed acoustically.! pipework or other equipment not included in the enclosure but

attached to the noisy machine could radiate noise which was notapparent prior to installation of the enclosure.

Problem 8.18

! acoustic absorbing material left out of wall cavity or too rigid andtouching both walls

! poor seals around windows and doors! glass in double glazing too thin! doors of insufficient acoustic performance! floor vibration transmitted to enclosure walls because of inadequate

vibration isolation! wall stud spacing incorrect! incorrect wall thickness or wall materials! poor seal at base of enclosure! poor bricklaying (if brick walls) leading to gaps in the mortar! change from original noise sources! tonal noise from the machine corresponding to a structural resonance or

an acoustic resonance (wall cavity or enclosure space).! pipework or other equipment not included in the enclosure but attached

to the noisy machine could radiate noise which was not apparent prior toinstallation of the enclosure.

A test to determine whether the problem was airborne or structure-bornewould be to turn the machines off and use loudspeakers in the enclosure togenerate the same noise levels inside the enclosure. If the exterior noise levelsare then the same with the loudspeakers operating as they were with themachine, then the problem is airborne flanking. If the noise external to theenclosure with the loudspeakers operating is lower, then the problem is likelyto be structure-borne vibration or radiation from equipment attached to thenoisy machine but not included in the enclosure.


LpA ' 10log10 10(6.3 & 2.62) % 10(6.7 & 1.61) % 10(6.2 & 0.86)

% 10(5.5 & 0.32) % 105.2 % 10(5 % 0.12) % 10(4.5 % 0.1) % 10(4.2 & 0.11)

' 59.2dB(A)

Problem 8.19

(a) The Noise Rating (NR) is obtained by plotting the un-weighted octaveband data on a set of NR curves (see fig 4.7, p118 in text) as illustratedin the following figure. The NR value is 52.5.

A-weighted sound level is given by:

(b) The acceptable noise level is obtained using table 4.10, p167 in the textand is Lp = 40 + 20 - 10 = 50dB(A). The actual level is over 9dB(A)above the allowable level, a situation which is not acceptable andaccording to table 4.11, it will generate widespread complaints from thecommunity. If the noise only occurs during the hours of 7am and 6pm,then the allowable level is Lp = 40 + 20 = 60dB(A) and the existing levelis thus acceptable.


0.5d1

'1.5

40 & d1

; d1 ' 10m

β ' tan&1 0.510

' 2.862E

ρfR1

'1.206 × 500

2 × 105' 0.003

0.5m

S

R

1.5m

40 - d1d1

(c) The octave band levels essentially follow the shape of the NR curve(except at 63Hz, where the level is much lower), so the noise will soundneutral.

(d) With no barrier, there are two propagation paths; the direct path and theground reflected path.

With a barrier, there are 8 paths as listed below:! over the top with no ground reflections! over the top with a ground reflection on the source side! over the top with a ground reflection on the receiver side! over the top with ground reflections on both sides

! around each end with no ground reflections! around each end with a ground reflection

(e) Attenuation of ground reflected wave (see figure)

Using similar triangles:

From table 5.2 on p209, R1 = 2 × 10-5 ; ρ = 1.206. Thus at 500Hz:


βR1

ρf

1/2

' 52E

d1

0.5'

20 & d1

3; d1 ' 2.857m

β ' tan&1 0.52.857

' 9.93E ; βR1

ρf

1/2

' 181E

0.5mS

20m 20m

1.5m3m

d1

R

S0.5m

R

1.5m

3m

20m20md2

From figure 5.20, the reflection loss on ground reflection is thus 3.9dB.

(f) Attenuation due to barrier. First calculate the reflection loss for eachpath which involves a ground reflection.

! Over the top, source side reflection


and from figure 5.20, p231 in the text, AR = 7.9dB

! over the top, receiver side reflection



d2

1.5'

20 & d2

3; d2 ' 6.667

β ' tan&1 1.56.667

' 12.7E ; βR1

ρf

1/2

' 231E

d3

0.5'

40 & d3

1.5; d3 ' 10m

0.5m

3m1.5m

R

20m20m

d3

S

S/

AA

A0.5m

20m 20m

d1

1.5m

3m

B

BBS

R

and from figure 5.20, p231 in the text, AR = 7.0dB

! around each end of the barrier


which is the same as if there were no barrier, so AR = 3.9dB

Now calculate Fresnel numbers for all paths over and around the barrier.

! Over the top (see figure)


A ) ' [0.52 % 2.8572 ]1/2 ' 2.9004m; A )) ' [32 % 17.1432 ]1/2 ' 17.404m

B ) ' [1.52 % 6.6672 ]1/2 ' 6.834m; B )) ' [32 % 13.3332 ]1/2 ' 13.663m

d ) ' [22 % 402 ]1/2 ' 40.05m

A ' [202 % 2.52 ]1/2 ' 20.156m; B ' [202 % 1.52 ]1/2 ' 20.056m

d ' [402 % 12 ]1/2 ' 40.01m

N1 ' (20.156 % 20.056 & 40.01) × 2.92 ' 0.58

N2 ' (2.900 % 17.404 % 20.056 & 40.05) × 2.92 ' 0.91

N3 ' (20.156 % 6.834 % 13.663 & 40.05) × 2.92 ' 1.76

N4 ' (2.900 % 17.404 % 6.834 % 13.663 & 40.01) × 2.92 ' 2.31

Δb1 ' 11.5 ; Δb2 ' 12.8 ; Δb3 ' 15.8 ; Δb4 ' 17.0dB

Top view

d

B

R

A

S

The distance d, between source and receiver depends on the path which isbeing considered. For example, for waves reflected from the ground on thesource side of the barrier only, the value of d (denoted d') below, is that fromthe image source to the receiver, etc.

From figure 8.14, p389 in the text:

It can be shown that in this case, the corrections of equation 8.88 are: Ab1 = 11.5 + 0.0 = 11.5; Ab2 = 12.8 + 0.1 = 12.9;Ab3 = 15.8 + 0.1 = 15.9; Ab4 = 17.0 + 0.2 = 17.2.

! Around the ends (see figure)


A ' B ' [202 % 52 % (1.5 & 0.5)2 /4 ]1/2 ' 20.622m; d ' 40.01m.

N5,6 ' (20.622 × 2 & 40.01) × 2.92 ' 3.6 ; Δb5,6 ' 18.7dB

A ) ' A )) ' [102 % 0.52 % 2.52 ]1/2 ' 10.32m

B ) ' [202 % 12 % 52 ] ' 20.64m; d ) ' [402 % 22 ]1/2 ' 40.05m

N7,8 ' (10.32 × 2 % 20.64 & 40.05) × 2.92' 3.6 ; Δb7,8 ' 18.7dB

NR ' 10log10 10&0/10 % 10&3.9/10

& 10log10 10&1.15 % 10&(1.29 % 0.79) % 10&(1.59 % 0.70)

% 10&(1.72 % 1.49) % 2 × 10&1.9 % 2 × 10&(1.90 % 0.39)

' 1.5 % 9.2 ' 10.7dB

AA0.5m 0.5m

20m 20m

1.5m BS

S

R

10m

3m

Direct waves, no reflection (one each side):

From figure 8.14 in the text:

Thus Ab5,6 ' 18.7 % 0.3 ' 19.0dB

Reflected waves (one each side)

From figure 8.14 in the text:

Thus Ab7,8 ' 18.7 % 0.3 ' 19.0dB

NR due to barrier is then calculated using equation 1.97 in the text as:


A

1.5m

B

d

48m

1.5m

2m

??

Problem 8.20

NR-50 octave band values may be read from figure 4.8, p155 in the text andthe following table may be generated.

Octavebandcentre

frequency(Hz)

63 125 250 500 1000

2000

4000

8000

Existingnoise

68 77 65 67 63 58 45 40

NR-50 76 67 60 54 50 48 46 44

Requiredreduction 0 10 5 13 13 10 0 0

Noise reduction is a function of Fresnel number which is directly proportionalto frequency. The important frequency is then 125Hz. If this is satisfied, thereduction at 500Hz will be OK. It is left to the reader to verify this.

Total of 8 paths (4 over top and 4 around sides) to consider to get an overallreduction of 10dB at 125Hz. Consider first the paths around the barrier edgesas the attenuation around these paths is independent of barrier height.


Around edge (elevation)

S h

g

j

ef

c1.5m

2m 23m 25m

1.5m

??

Plan

2m 23m 25m

b

5m

S R

As the source and receiver are at the same height, the location of the groundreflection will be mid-way (in a horizontal direction only) between the sourceand receiver.

Path with no ground reflectionPath h = (22 + 52)1/2 = 5.385mPath j = (482 + 52)1/2 = 48.260mA % B & d ' 5.385 % 48.260 & 50 ' 3.64mAt 125Hz, = 343/125 = 2.744m and N = (2/ ) × 3.64 = 2.65. From figureλ λ8.14 on p389 in the text, Δb = 17.0dB. The correction term given by equation8.88 in the text (assuming an omnidirectional source) is

, so .20 log10(53.64 /50) ' 0.6dB Ab ' 17.0 % 0.6 ' 17.6dB

Path with ground reflectionFrom the figure, and using similar triangles, b/5 = 25/48; b = 2.6mPath e = (252 % 1.52 % 2.6042 )1/2 ' 25.180mBy similar triangles, f/1.5 = 23/25; f = 1.38m


A , A1 3B , B1 2

h-1.5

d

2m 48m

B , B3 4 1.5m

1.5m 1.5m

h

h

A , A2 4

1.5m

2m 48m

Path g = (232 % 1.382 % 2.3962 )1/2 ' 23.166m

Path c = (0.122 % 22 % 52 )1/2 ' 5.387m

A % B & d ' 5.387 % 23.166 % 25.180 & [502 % 32 ]1/2 ' 3.64mThus, N = (2/ ) × 3.64 = 2.65. from figure 8.14 on p389 in the text,λΔb = 17.0dB. The correction term given by equation 8.88 in the text(assuming an omnidirectional source) is , so20 log10(53.74 /50.09) ' 0.6dB

. Losses due to ground reflection are zero.Ab ' 17.0 % 0.6 ' 17.6dB

The required barrier height can be found by trial and error. The results aresummarised in the following table, where the subscript "1" refers to wavestravelling over the barrier with no reflections, the subscript "2" implies areflection on the source side, the subscript "3" implies a reflection on thereceiver side and the subscript "4" implies a reflection on both sides. Thedistances "A" refer to the source side of the barrier (source to barrier top alongthe particular path specified by the associated subscript) and the distances "B"refer to distances on the receiver side. We will begin the trial with a barrierheight of 3.0m. The correction term given by equation 8.88 in the text(assuming an omnidirectional source) is added to the barrier attenuation Δb togive Ab.


NR ' 10log10 100 % 100 & 10log10 4 × 10&1.76

% 10&Ab1 /10

% 10&Ab2 /10

% 10&Ab3 /10

% 10&Ab4 /10

The barrier overall noise reduction is given by equation 1.97 in the text whichcan be expanded to give:

Barrier height 3.0m 3.5m 4.0m

A1 (m)A2 (m)A3 (m)A4 (m)B1 (m)B2 (m)B3 (m)B4 (m)

N1

N2

N3

N4

Δb1

Δb2

Δb3

Δb4

Ab1 (dB)Ab2 (dB)Ab3 (dB)Ab4 (dB)

2.504.922.504.92

48.0248.0248.2148.210.382.070.452.2810.116.110.517

10.216.610.617.5

2.8285.3852.8285.385

48.04248.04248.26048.2600.632.430.732.6611.917

12.017.112.017.512.217.6

3.2025.8523.2025.85248.06548.06548.31448.3140.9232.791.043.03612.617.313.118

12.817.913.318.7

NR (dB) 8.3 9.4 10.0

Thus the wall height should be about 4.0m.

Problem 8.21

The layout is illustrated in the following figure.


A ) ' 202 % 22 ' 20.0998m

A )) ' 302 % 32 ' 30.1496m

B ) ' 6.6672 % 32 ' 7.3109m

B )) ' 3.3332 % 1.52 ' 3.6550m

A ' 502 % 12 ' 50.0100m

B ' 102 % 1.52 ' 10.1119m

N1 '2

0.68650.0100 % 10.1119 & (602 % 0.52 )1/2 ' 0.35

N2 '2

0.68620.0998 % 30.1496 % 10.1119 & (602 % 3.52 )1/2 ' 0.76

2m

d1

50m 10m d2

1.5m

3m

A

A' A'' B'B''

B

From similar triangles, ; so d1 = 20.0md1

50 & d1

'23

; so d2 = 3.333md2

10 & d2

'1.53

λ = 343/500 = 0.686m

There are 4 paths which will contribute to the sound level at the receiver sowe need to calculate the Fresnel Number corresponding to each.

Path 1, A Y B

Path 2, A' Y A'' Y B


N3 '2

0.68650.0100 % 7.3106 % 3.6553 & (602 % 3.52 )1/2 ' 2.5

N4 '2

0.68620.0998 % 30.1496 % 7.3106 % 3.6553 & (602 % 0.52 )1/2 ' 3.5

NR ' 10log10 10&0/10 % 10&3/10

& 10log10 10&9.9/10 % 10&15/10 % 10&20/10 % 10&24.5/10

' 1.76 % 8.31 ' 10.1dB

Path 3, A Y B' Y B''

Path 4, A' Y A'' Y B` Y B''

As the wall completely surrounds the factory, no sound is diffracted aroundits edges. The correction term given by equation 8.88 in the text is greatest forpath 4 and is equal to 0.09dB which is negligible, so the correction will beignored.

From figure 8.14, the attenuations corresponding to N1 to N4 areΔN1 = 9.9, ΔN2 = 12, ΔN3 = 17, ΔN4 = 18.5. Adding 3dB to all groundreflections results in the following noise reductions corresponding to the 4paths: Path 1, 9.9dB; Path 2, 15dB; Path 3, 20dB; Path 4, 24.5dB. Fromequation 1.97 in the text, the noise reduction due to the enclosure is:

Problem 8.22

Referring to the following figures, we have for the reflection angles:

over top, source side,

tanβ1 '4

25 & x'

0.5x

; x ' 2.78m , β1 ' 10.2E

over top, receiver side,

tanβ2 '4

25 & y'

1.5y

; y ' 6.82m , β2 ' 12.4E


Over top (elevation)

Plan

R

S1 2

x

25m 25my

25m25m

7.5m

7.5m

0.5m

4m

1.5m

Around edge (elevation)

Plan7.5m

25m RS

S

S

Rh1

1m

325m25m

7.5m

25m

1.5m

4m

z

0.5m


around edge, tanβ3 '1.5

50 & z'

0.5z

; z ' 12.5m , β3 ' 2.3E

Flow resistivity of ground = R1 = 3 × 107 MKS rayls/m and ρ = 1.206kg/m3.With no barrier, β4 = β3 = 2.3E. The reflection loss, AR is calculated usingfigure 5.20 in the text and the results are tabulated in the table below.

Octave bandcentre

frequency (Hz)

ρfR1

R1

ρf

1/2

AR1 AR2 AR3 and AR4

631252505001000200040008000

2.5×10-6

5.0×10-6

1.0×10-5

2.0×10-5

4.0×10-5

8.0×10-5

1.6×10-4

3.2×10-4

6304473162231581127956

0.20.30.50.71.01.52.33.0

0.20.30.50.71.01.52.33.0

1.52.02.53.55.06.07.06.5

Path 1 - over top of barrier with no ground reflectionsd = 50.010m, A = 25.244, B = 25.125 and A + B - d = 0.359m

Path 2 - over top of barrier with ground reflection on the source sided = 50.040m, B = 25.125A = (2.782 + 0.52)1/2 + (42 + 22.222)1/2 = 25.402m andA + B - d = 0.487m

Path 3 - over top of barrier with ground reflection on the receiver sided = 50.010m, A = 25.244B = (6.822 + 1.52)1/2 + (42 + 18.182)1/2 = 25.598m andA + B - d = 0.802m

Path 4 - over top of barrier with ground reflection on both sidesd = 50.010m, A = 25.402, B = 25.598 and A + B - d = 0.990m

Paths 5&6 - around edges with no reflectionB = A = (0.52 + 252 + 7.52)1/2 = 26.106m and A + B - d = 2.201m


Paths 7&8 - around edges with reflection in groundIntersection height of diffracted wave with barrier edge= 1.5 × 12.5/37.5 = 0.5m.A = 2(0.52 + 12.52 + (7.5/2)2)1/2 = 26.120mB = 2(12 + 252 + 7.52)1/2 = 26.120m andA + B - d = 2.20m

Fresnel Number, . Values of N for each path areN '2f

343(A % B & d)

tabulated below.

Octave bandcentre

frequency (Hz)N1 N2 N3 N4 N5&6 N7&8

631252505001000200040008000

0.130.260.521.052.094.198.3716.7

0.180.360.711.422.845.6811.422.7

0.300.591.172.344.689.3518.737.4

0.360.721.442.895.7711.523.146.2

0.801.603.216.4212.825.751.3103

0.801.613.216.412.82651103

The noise reductions (NR or Δb) corresponding to the above Fresnel Numbersare calculated using figure 8.14, p389 in the text and are tabulated below. The correction term given by equation 8.88 in the text (assuming anomnidirectional source) will be less than 0.1dB overall and will be ignoredhere. The numbers in brackets indicate the sum of the ground reflection lossesand barrier diffraction loss. All quantities are in dB.


Octave bandcentre

frequency(Hz)

NR1 NR2 NR3 NR4 NR5&6 NR7&8

63125250500

1000200040008000

8.29.5

10.813.016.119.222.324

8.9 (9.1)10.1 (10.4)12.0 (12.5)14.8 (15.5)17.9 (18.9)20.9 (22.4)23.9 (26.1)

24 (27)

9.9 (10.1)11.8 (12.1)13.8 (14.3)17.0 (17.7)19.8 (20.8)23.0 (24.5)24 (26.3)24 (27)

10.0 (10.4)12.0 (12.6)14.5 (15.5)17.7 (19.1)20.7 (22.7)23.8 (26.8)24 (28.6)24 (30)

12.315.018.121.224242424

12.3 (13.8)15.0 (17.0)18.1 (20.6)21.2 (24.7)

24 (29)24 (30)24 (31)

24 (30.5)

The barrier noise reduction is given by equation 1.97 in the text and the resultsof the calculations are summarised in the table below. The subscript "A"refers to the condition with no barrier and the subscript "B" refers to thecondition with barrier. All quantities are in dB.

Octavebandcentre

frequency(Hz)

10 log10j 10&NRAi /10 &10log10

j 10&NRBi /10

NR SPL atreceiver

631252505001000200040008000

2.32.11.91.61.21.00.80.9

1.83.76.08.9

12.114.816.717.3

4.15.87.910.513.315.817.518.2

65.969.264.149.544.740.232.533.8

The overall A-weighted level is calculated using the octave band levels asdescribed on pages 101 and 102 in the text. The A-weighted level with thebarrier is calculated using the numbers in the last column of the precedingtable and is 58.2dB(A). The A-weighted level without the barrier is calculatedusing the octave band levels given in the problem and is 67dB(A). Thus thenoise reduction due to the barrier is 9dB(A).


ΔC ' K log10(2πfb /343)

Problem 8.23

If the barrier were a building, the additional noise reduction may be calculatedusing equation 8.98, p394 of the text which may be rewritten as:

The values of K are calculated using figure 8.17 in the text and trigonometryis used to calculate the angles, . The results are summarised in theθ and φtwo following tables. Values for ΔC are in dB and the subscript on C refersto the path number.

Path1

Path2

Path3

Path4

Paths5&6

Paths7&8

θ (degrees) 98 100 98 100 117 117

(degrees) 96 96 102 102 117 117φ

K 1.2 1.5 2.3 2.7 5.6 5.6

Octave bandcentre

frequency(Hz)

ΔC1 ΔC2 ΔC3 ΔC4 ΔC5,6,7&8

631252505001000200040008000

0.81.21.51.92.22.63.03.3

1.01.41.92.32.83.23.74.2

1.52.22.93.64.35.05.76.4

1.82.63.44.25.05.86.77.4

3.75.47.18.810.412.113.815.5

The noise reductions are calculated as before with the attenuation due to thebarrier thickness added to each path. The results (in dB) are summarised inthe table below.


Octave bandcentre

frequency(Hz)

NR1 NR2 NR3 NR4 NR5&6 NR7&8

631252505001000200040008000

9.010.712.314.918.321.825.327.3

10.111.814.417.821.725.629.831.2

11.614.317.221.525.229.533.033.4

12.215.218.923.327.732.635.437.4

16.019.425.230.034.436.137.839.5

17.522.427.833.539.442.144.846.0

The barrier noise reduction is given by equation 1.97 in the text and the resultsof the calculations are summarised in the table below. The subscript "A"refers to the condition with no barrier and the subscript "B" refers to thecondition with barrier.

Octave bandcentre

frequency(Hz)

10 log10j 10&NRAi /10 &10log10

j 10&NRBi /10

NR SPL atreceiver

631252505001000200040008000

2.32.11.91.61.21.00.80.9

3.66.08.6

12.015.719.322.924.5

5.98.110.513.616.920.323.725.4

64.166.961.546.441.135.726.326.6

The overall A-weighted level is calculated using the octave band levels asdescribed on pages 101 and 102 in the text. The A-weighted level with thethick building as the barrier is calculated using the numbers in the last columnof the preceding table and is 55.6(A) which is 2.6dB(A) less than for the thinbarrier. Thus the effect of thickening the barrier is to increase the noisereduction by 2.6dB(A) to a total of approximately 11dB(A).


IL ' 10log10

Dθ

4πr 2%

4S0α0

& 10log10

DθF

4πr 2%

4K1K2

S(1 & K1 K2)

' term1 & term2

S

A

d

BR

3m

1.5m

25m25m

0.5m

Problem 8.24

If the barrier is moved closer to the refrigeration unit, the noise reduction willincrease because the Fresnel numbers will increase due to the increased pathlength from source to receiver over the top of the barrier and around theedges.

Problem 8.25

An elevation view of the situation is shown in the figure. As the barrier isindoors, equation 8.109 on p402 in the text is appropriate. Thus:

Source and receiver are each 25m from barrier. For an omnidirectional source

on a hard floor Dθ = 2; and .Dθ

4πr 2' 6.366 × 10&5

r ' d ' [502 % 12 ]1/2 ' 50.01m

A ' (252 % 2.52)1/2 ' 25.125m B ' (252 % 1.52)1/2 ' 25.045m

A + B - d = 0.160m; S = 2 × 50 = 100m2 = area of gap between barrier andceiling.

S0 = 2(100 × 5 + 50 × 5 + 100 × 50) = 11,500m2; = 0.08.α0

Thus, 4 /S0α0 ' 4.348 × 10&3


α1 ' α2 '5750 × 0.08 % 150 × 0.15

5900' 0.082

S1α1 ' S2α2 ' 482.5m 2 and K1 ' K2 '100

100 % 482.5' 0.172

term1 ' 10log10 6.366 × 10&5 % 4.348 × 10&3 ' &23.6dB

term2 ' 10log10 6.366 × 10&5 F %4 × 0.1722

100(1 & 0.1722)

' 10log10 6.366 × 10&5 F % 1.29 × 10&3

S1 = S2 = S0/2 + 3 × 50 = 5,900m2.

Thus:

Referring to the first equation:

There is only one path over the top of the barrier, so F = (3 + 10N)-1, whereN is the Fresnel number for the path, calculated using equation 8.85 and figure8.13 on page 388 in the text. Thus term 2 can be written as:

The calculations which are a function of frequency are summarised in thefollowing table, where the barrier IL is given by IL = term 1 - term 2.

Octavebandcentre

frequency(Hz)

N = 2λ

(A % B & d)F

DθF

4πr 2 term 2 IL

631252505001000200040008000

0.0590.1170.2330.4660.9331.8653.7327.46

0.280.2400.1880.1310.0810.0460.0250.013

1.78×10-5

1.53×10-5

1.12×10-5

8.34×10-6

5.16×10-6

2.93×10-6

1.59×10-6

8.20×10-7

-29-29

-29.1-29.1-29.1-29.1-29.1-29.1

5.45.45.55.5

5.55.5

5.55.5


Lpd ' Lw & 10log10(4π × 16) ' Lw & 23dB

&10log10(1 & αf ) % 10log10(1 % 1.2)2 % 42

42

' &10log10(1 & αf ) % 1.1

&10log10 (1 & αc ) % 10log10(2 % 1.8)2 % 42

42

' &10log10(1 & αc ) % 2.8

S R

C

Problem 8.26

(a) The most dominantfive paths fromsource to receiver areshown in the figure.We are to ignore anypaths contributingless than 0.2dB to thetotal. That is, we will ignore paths which contribute 10dB or more lowerthan the direct path. Of course, the correct way to do this problem is touse equation 7.110 in the text for a flat room with a diffusely reflectingfloor (as there is furniture present) and specularly reflecting ceiling.However, here we are happy with an approximate solution. The directpath contribution is given by equations 5.158 and 5.160 in the text withDIM and AE = 0. Thus:

The second path (reflection from the floor) will be attenuated by:

The third path (reflection from the ceiling) will be attenuated by:

The fourth path (reflection from the floor then ceiling) will be attenuatedby:


&10log10(1 & αf ) (1 & αc ) % 10log10(1 % 3 % 1.8)2 % 42

42

' &10log10(1 & αf ) (1 & αc ) % 4.9

&10log10(1 & αf ) (1 & αc ) % 10log10(1.2 % 3 % 2)2 % 42

42

' &10log10(1 & αf ) (1 & αc ) % 5.3

The fifth path (reflection from the ceiling then floor) will be attenuatedby:

The contributions of each path to the final sound pressure level isobtained by subtracting the attenuations from the direct wave soundpressure level (arithmetically). The total sound pressure level is thenobtained by logarithmically combining the contributions from each path(assuming incoherent combination).

The results are tabulated in the table below. Note that paths involvingmore reflections will result in a contribution of less than 0.2dB to thetotal and are thus not included. Perhaps you could prove this by trial anderror.

frequency(Hz)

500 1000 2000

Lw 70 77 75

Lpd 47 54 52

Lp2 42.2 47.8 45.5

Lp3 38.2 38.2 29.2

Lp4 32.4 31 21.7

Lp5 31.9 30.5 21.2

Lpt 48.8 55.1 52.9


N '2 × f

c1.22 % 22 % 12 % 22 & 4 ' 3.31 × 10&3f

Ab ' Δb % 20log102.282 % 2.282

4' Δb % 1.2

Lpb ' Lpd & Ab

(b) When the barrier is present, the only contributions to the sound level onthe other side will be the wave reflected from the ceiling (Lp3) and thewave diffracted over the top, as all other paths are blocked by the barrier(do a sketch of the arrangement to prove this to yourself). For the wavediffracted over the top, we can write the following for the Fresnelnumber:

The attenuation, Δb is read from figure 8.14 in the text (point source) andcorrected using equation 8.88 to give:

The contribution, Lpb of the wave diffracted over the top of the barrier tothe sound field at the receiver is calculated using:

The results are summarised in the following table.

frequency(Hz)

500 1000 2000

Lpd 47 54 52

Lp3 39.9 33.9 24.4

N 1.65 3.3 6.6

Ab 16 19 22

Lpb 31 35 30

Lpt 39.0 39.9 33.3

reductiondue tobarrier

10 15 20


IL ' 10log10

Dθ

4πr 2%

4S0α0

& 10log10

DθF

4πr 2%

4K1K2

S(1 & K1 K2)

' term1 & term2

A B

S R 3 m1.5m

2m 3m

5m

10m

5m

2m20m

a

b

RS

50m

5m 5m

1m

d

Assumptions:

! The barrier is sufficiently wide that the contribution at the receiverposition due to refraction around the edges is negligible.

! All waves at the receiver combine together incoherently.

! The directivity of the source is uniform in all directions.

(c) The noise reduction could be increased by moving the barrier sufficientlyclose to either the noise source or receiver that the wave reflected fromthe ceiling will also be blocked. In addition the barrier Fresnel numberwill be increased, further reducing the energy diffracted over the top ofthe barrier.

Problem 8.27

(a) The situation is illustrated in the figure. As the barrier is indoors,equation 8.109 on p402 in the text is appropriate. Thus:

First we calculate the Fresnel Numbers for diffraction over the top andaround the edges using equation 8.85 and figure 8.13 on page 388 in thetext.


S ' 2 × 5 × 3 % 20 × 2 ' 70m 2

S1 ' S2 ' 30 % 1350 ' 1380m 2

α1 ' α2 ' (216 /2 % 30 αb) / (1350 % 30)

Diffraction over the topXR = XS = 5m; ZR = 1.5m; ZS = 1m; Y = 2m; h = 3mYR = 2 × 5/(5 % 5) ' 1

d ' [ (5 % 5)2 % (1 % 1)2 % (1.5 & 1)2 ]1/2 ' 10.210m

A ' [52 % 12 % (3 & 1)2 ]1/2 ' 5.477m

B ' [52 % 12 % (3 & 1.5)2 ]1/2 ' 5.315mFresnel Number, N = (2f/343)(A + B - d) = 3.393 × 10-3f

Diffraction around edge "a"XR = XS = 5m; ZR - ZS = 2m; Y = 0.5m; h - ZS = 5m;h - ZR = 3mYR = 0.5 × 5/(5 % 5) ' 0.25

d ' [ (5 % 5)2 % 4 × 0.252 % 4]1/2 ' 10.210m

A ' [52 % 0.252 % 52 ]1/2 ' 7.075m

B ' [52 % 0.252 % 32 ]1/2 ' 5.836mFresnel Number, Na = (2f/343)(A + B - d) = 1.575 × 10-2f

Diffraction around edge "b"XR = XS = 5m; ZR - ZS = 2m; Y = 0.5m; h - ZS = 5m; h - ZR = 7mYR = 0.5 × 5/(5 % 5) ' 0.25

d ' [ (5 % 5)2 % 4 × 0.252 % 4]1/2 ' 10.210m

A ' [52 % 0.252 % 52 ]1/2 ' 7.075m

B ' [52 % 0.252 % 72 ]1/2 ' 8.606m

Fresnel Number, Nb = (2f/343)(A + B - d) = 3.190 × 10-2f

S0α0 ' 2(5 × 20 % 5 × 50 % 20 × 50) × 0.08 ' 216.0m 2

r = d = 10.210m; DI = 5, thus Dθ = 3.16 andDθ/4πr2 = 2.51 × 10-3


term1 ' 10log10

Dθ

4πr 2%

4S0α0

' &16.8dB

and F ' ja,b,c

13 % 10Ni

DθF

4πr 2%

4K1K2

S(1 & K1K2)

The quantities K1 and K2 are calculated using equation 8.111 in the text.They are frequency dependent quantities, but all other variables neededfor their calculation have been evaluated above. We now proceed toevaluate the second term of equation 8.109 as a function of octave bandcentre frequency. Results for the second term as well as the overallInsertion Loss of the barrier are tabulated in the following table.

Octave bandcentre

frequency (Hz) 63 125 250 500 1000 2000 4000

N (Top) 0.21 0.42 0.85 1.70 3.39 6.79 13.57

Na 0.99 1.97 3.94 7.88 15.75 31.5 63.0

Nb 2.01 3.99 7.98 15.95 31.9 63.8 127.6

F 0.315 0.205 0.123 0.068 0.036 0.019 0.010

αb

(table 7.1)

0.08 0.25 0.83 1.0 1.0 1.0 1.0

α1 , α2 0.08 0.08 0.10 0.10 0.10 0.10 0.10

K1, K2 0.388 0.377 0.345 0.337 0.337 0.337 0.337

1.09×10-2

9.98×10-3

8.02×10-3

7.46×10-3

7.39×10-3

7.34×10-3

7.32×10-3

InsertionLoss (dB)

2.8 3.2 4.2 4.5 4.5 4.5 4.6


fr 'cL

πd'

5476π × 0.4022

' 4,334Hz

fc '0.55c 2

cL h'

0.55 × 3432

5476 × 0.00222' 5,323Hz

pipe 0.20m0.40m

Jacket

cL ' E / [ρm (1 & ν2 ) ] ' 71.6/[2700(1 & 0.342 ) ] ' 5476 m/s

(b) Moving the barrier closer to the source will increase the value of N whichwill reduce the value of F and thus the direct field contribution. Thevalue of K1 will be reduced and the term involving K1K2 will be reducedslightly. Thus we can expect an increase in the barrier insertion loss.This effect is expected to be only 1 or 2dB but calculations as were donein part (a) are necessary to verify this.

(c) Extending the barrier length will increase the barrier Insertion Lossbecause it will remove the contribution from waves diffracted around thebarrier ends and will also reduce the reverberant field contributionbecause of a smaller gap area and also because of an increase in theeffective room absorption. This effect should be slightly larger than theeffect in part (b) above but would still be restricted to a few dB. The costeffectiveness of this action can only be assessed by completing thecalculations as was done in part (a) of the problem and fully evaluatingthe benefit of the increased noise reduction vs the inconvenience ofrestricted passage.

Problem 8.28

This is a pipe lagging problem so wefollow the procedure on pages 404and 405 in the text with the erratacorrected in equations 8.116, 8.117and 8.119. The thickness of thejacket is h = 6/2700 = 2.22mm andthe diameter of the jacket, d = 0.2 +2 × 0.1 + 0.0022 = 0.4022m. Thequantity, 1000(m/d)1/2 = 3862. The longitudinal wavespeed is:

The ring frequency is:

The critical frequency is obtained using the equation on p337 as:


IL '40

1 % 0.12 /0.2log10

f 6 × 0.1132

' 25log10 ( f × 0.00587)

For the Insertion Loss calculations, we use equation 8.112 on p404 for octavebands of 4000Hz and below and equation 8.115 for the 8000Hz octave band.The results of the calculations are summarised in the table below at octaveband centre frequencies. Of course, if three 1/3 octave bands are averaged,then the result would be slightly different. In addition, equation 8.120 hasalso been used to generate an alternative set of Insertion Loss predictions.With equation 8.120, the Insertion Loss is given by:

which is valid for frequencies defined by .f $ 120/ 6 × 0.1 ' 155Hz

Octavebandcentre

frequency(Hz)

f/fr or

f/fc

Cr or Cc Xr or Xc

InsertionLoss

eqns. 8.112 - 8.119

InsertionLoss

eqn.8.120

631252505001000200040008000

0.01450.02880.05770.1150.2310.4610.9231.503

0.10900.21630.43260.86531.73053.46106.922015.849

789.51307173121202479271719914289

-18.87.617.925.529.419.422.615.9

--

4.211.7

19.226.7

34.341.8

The Insertion Loss results are somewhat different between the two methodsof calculation and as we shall see in the next problem, neither predictionscheme is particularly good. However, it is supposed that the true results liesomewhere between the two.

Problem 8.29

(a) The jacket cross section is shown schematically in the following figure.


x × 10&3 × 2700 % (1 & x) × 10&3 × 11300 ' 6

meff ' ρ1 h1 % ρ2 h2 ' 6.00kg/m 2

y '(hPb % hAl /2)EAl % EPb hPb /2

EAl % EPb

'(0.384 % 0.616 /2) × 71.6 % 16.5 × 0.384 /2

71.6 % 16.5' 0.598mm

Beff 'EAl hAl

12

h 2Al % (hPb % hAl /2 & y)2

1 & ν2Al

%EPb hpb

12

h 2Pb % (y & hPb /2)2

1 & ν2Pb

'71.6 × 0.616

120.6162 % 12 × (0.384 % 0.308 & 0.598)2

1 & 0.342

%16.5 × 0.384

120.3842 % 12 × (0.598 & 0.192)2

1 & 0.442

' 2.017 % 1.392 ' 3.41kgm 2 s &2

hPb

1mm hAl

Lead

Aluminum

y

neutralaxis

Let x mm be the thickness of the aluminium part of the jacket. Then:

Thus x = 0.616mm = thickness of Aluminium.Thickness of lead = 1 - 0.616 = 0.384mmEffective surface mass is given by:

The neutral axis location is given by:

The bending stiffness is given by:


cL '12h

Beff

meff

'12

0.001×

3.4096

' 2611m/s

fc 'c 2

2πmB

'3432

2π6

3.409' 24.84kHz

fr 'cL

πd'

2611π × 0.251

' 3311Hz

Assume that the effective Poisson's ratio is that of Aluminium and equalto 0.34. Thus the longitudinal wave speed is:

The critical frequency is given by equation 8.3 in the text as:

and the ring frequency is given on p404 in the text as:

The jacket insertion loss may be calculated using either equation 8.120or equations 8.112 to 8.119 in the text. We will use both methods hereas a comparison. The quantities used in the equations are m = 6kg/m2,R = 0.05m, d = 0.15 + 2 × 0.05 = 0.25m and D = 0.15m. The results ofthe calculations at the 1/3 octave band centre frequencies are summarisedin the following table. Note that the calculations using equation 8.120

are only valid for frequencies defined by .f $ 120/ 6 × 0.05 ' 220Hz


Octave bandcentre

frequency(Hz)

Xr or Xm Cr or Cc

Insertion Loss (dB)

Eqs. 8.112 - 8.119 Eq. 8.120

6380100125160200250315400500630800100012501600200025003150400050006300800010000

12791500169018692056222023802543270828593010315732803378343534033193225827143356413450385893

0.11370.14440.18050.22560.28880.36100.45120.5685 0.72190.90241.13701.44391.80482.25602.88773.60964.51205.68521.8682.3352.9423.7364.670

-2.44.18.5

12.115.718.621.323.926.328.430.231.531.730.121.124.931.424.130.029.520.730.437.0

-----

0.352.64.97.09.311.613.715.918.320.422.624.827.129.331.533.836.0

The data in the table are plotted in the following figure where the solidblack line represents the theory embodied in equations 8.112-8.119 andthe dashed black line represents the theory embodied in equation 8.120.In addition, some experimental data (solid grey line) from the text book(figure 8.18, second edition) are shown for comparison. It can be seenthat neither theory provides a very good prediction of the measured data.Unfortunately there are no better theories available at present.


(b) It is clear from the equations used for either calculation method thatincreasing the mass of the liner will increase the low frequency InsertionLoss. For the first prediction scheme, reducing cL will also result inincreased values for the low frequency Insertion Loss.

(c) One advantage of porous acoustic foam over rockwool is that the porousfoam will support the weight of the jacket indefinitely whereas rockwoolwill gradually compress and in a high vibration environment, it will turnto powder.

A disadvantage of foam is that it is not fireproof and if ignited it emitstoxic gas. Another disadvantage is that the foam is much more expensivethan rockwool.

RVL t

B

B

N holes

9

Solutions to problems relatingto muffling devices

Problem 9.1

Referring to the figure, we can imagine that each hole in the perforated sheetrepresents a neck of a Helmholtz resonator with the volume associated witheach neck being equal to the total volume behind the perforated sheet dividedby the number of holes. Let L be the depth of the backing cavity and letsconsider a section of sheet of dimensions B × B with a number of holes, N.

The total backing volume is then LB2

and the volume associated with one holeis LB2 /N. The percent open area isgiven by:

.P '100N

B 2

πd 2

4

Thus the effective resonator volume is:

,V '100L

Pπd 2

4'

100LAP

where A is the neck cross-sectional area.


f0 'c

2πA

LV'

c2π

P100L R

2R0 ' 0.85d(1 & 0.22d /a)

pi ' Ae j (ωt & kx ) ; pr ' Be j (ωt % kx % θ )

pT ' Ae j (ωt & kx ) % Be j (ωt % kx % θ )

uT '1ρc

Ae j (ωt & kx ) & Be j (ωt % kx % θ )

Z 'pT

uT

' ρc Ae & j kx % Be j (kx % θ )

Ae & jkx & Be j (kx % θ )' ρc A % Be j (2kx % θ )

A & Be j (2kx % θ )

' ρc A /B % e j (2kx % θ )

A /B & e j (2kx % θ )

AB

'108/20 % 1

108/20 & 1'

3.51191.5119

' 2.323

From equation 9.38 in the text:

where R = 2R0 + t is the effective length of the neck and 2R0 is the total effectiveend correction for the hole. Comparing the above equation with equation 7.77in the text gives the effective end correction of the perforate as:

The Helmholtz model is appropriate because the system is effectively a smallmass of air vibrating against a stiffness represented by the backing volume.

Problem 9.2

(a) Diameter, 0.4m, so higher order mode cut on frequency is. Thus at 200 Hz only plane waves propagate.f ' 0.586c /0.4 ' 500 Hz

Minimum pressure occurs when θ ' &2kx % π ; x ' 1.8 mAt 200 Hz, ;k ' 2πf /c ' 2π ×200/343 ' 3.664 = -10.048c. θ ' &2 × 3.664 × 1.8 % πAdding 4π, gives θ = 2.518c = 144.2E

At x = 2, 2kx = 14.656 and 2kx + θ = 4.608c = 264 degrees. Thus:

Muffling devices 267

R0 ' 0.61×0.05% 8 ×0.053π

1& (1.25×0.25) ×(1&0.639)2 ' 0.0078 m

Z ' 413 2.323 % cos(4.608) % jsin(4.608)2.323 & cos(4.608) & jsin(4.608)

' 413 2.219 % j(&0.995)2.427 & j(&0.995)

Z ' 413 2.219 & j0.9952.427 % j0.995

' 413 (2.219 & j0.995) × (2.427 & j0.995)(2.427 % j0.995) × (2.427 & j0.995)

' 413 4.395 & j4.6236.88

M ' ReZ/ρc '263.8413

' 0.64

Multiplying numerator and denominator by complex conjugate of thedenominator gives:

Thus, Z = 264 - j278

(b) The hole impedance is in parallel with the rigidReZA ' Re Z /Aorificeplate containing it. If the rigid plate impedance is effectively infinity, thecombined specific acoustic impedance is that of the hole and this is whatis measured by the standing wave. The acoustic impedance of the hole isthe specific acoustic impedance divided by the area of the hole only. Ifcross flow dominates resistance, M ' ReZAAorifice /ρcThus,

(c) End correction = no flow correction × (1 - M)2. Hole radius = 0.05 mThus total end correction is:

Problem 9.3

A quarter wave tuning stub works by changing the wall impedance of a ducton which it is mounted so that downstream propagating waves are reflectedback upstream. The viscous losses at the entrance to the stub also account forsome of the energy loss, because of the large particle motions near the edgesof the entrance. The stub can also act by changing the radiation impedance(and thus the amount of power radiated) of the source producing the noise.


0.2gfedcbaZ1

Zb Zd Zf

Za

Z1 Z2Zc

Z3Ze

Zg

Problem 9.4

(a) The impedance looking into the stub must be zero as we are ignoring theresistive component.

The equivalent electrical circuit is shown in the figure below:

Impedance looking in at location 3 = Z3 'Ze(Zf % Zg)

Ze % Zf % Zg

Impedance looking in at location 2 = Z2 'Zc(Zd % Z3)

Zc % Zd % Z3

Impedance looking in at location 1 = Z1 'Za(Zb % Z2)

Za % Zb % Z2

If we neglect resistive impedance, then Z1 = 0 and


. If we now substitute the expression for Z3Zb ' &Z2 ' &Zc(Zd % Z3)

Zc % Zd % Z3

into the preceding equation, we obtain:

(1)Zb Zc % Zd %Ze(Zf % Zg)

Ze % Zf % Zg

% Zc Zd %Ze(Zf % Zg)

Ze % Zf % Zg

' 0

We now need to substitute in physical dimensions in place of the

impedances. and the effective length of the holeZb ' Zd ' Zf ' jρωRA

is given by:

.R ' 2R0 '8d3π

1 & 1.25d

0.2

The density, ρ = 1.206, A = πd2/4 and ω = 200π.

The volumes between the orifices are given by:

andVe ' Vc 'πD 2

4L3

' 0.01047L

.Va ' Vg ' 0.5Ve ' 0.005236L

Required length, L = 0.5 × λ/4 = c/8f = 343/800 = 0.429m.Using equation 9.35, we obtain:

.Zc ' Ze ' &j1.206 × 3432

0.01047 × 0.4288 × 2 × π × 100' & j5.03 × 104

Za = Zg = 2Zc = -j1.01 × 105.

Equation (1) can now be solved by computer for d. The result isd = 44mm.

(b) The device with baffles would have a much larger resistive impedancethan the one without baffles because of all the cross-sectional changes atwhich there will be viscous losses. The larger resistive impedance willlower the quality factor and thus the peak attenuation, although thebandwidth of significant attenuation will increase.


Rs '414

0.001520.288 × 1.832 × 2.180 × 10&4 log10

4 × 0.00152

π × (2.180 × 10&4)2

' 146MKS Rayls

IL ' 20log10 1 %Zd

Rs

Problem 9.5

(a) To calculate the resistive impedance, equation 9.29 on p417 in the textmay be used. First we will evaluate the variables used in the equation.ρc = 343 × 1.206 = 414, A = π × 0.0442/4 = 0.00152 m2,k = 2π × 100/343 = 1.832,

t = h = ,2 × 1.8 × 10&5 / (1.206 × 2π × 100) ' 2.180 × 10&4

w = 0, and M = 0. Substituting these values into equation 9.29,ε ' 0we obtain:

(b) At the design frequency (100Hz), the Insertion Loss is found bysubstituting Rs for Zs in equation 9.46 in the text to give:

If the side branch is mounted an odd number of quarter wavelengths fromthe end of the duct, the reactive part of the impedance Zd willtheoretically be infinite (based on equation 9.14) and the side branchInsertion Loss will also be infinite as indicated by the above equation.However, in practice, the quarter wave tube is of finite cross section andZd cannot therefore be too large. This results in the Insertion Loss beingfinite and usually limited to 25 to 30dB.

Problem 9.6

(a) Closed end side branches have less viscous losses than open ended tubesbecause of one less cross-sectional change and corresponding edge.Thus the quality factor and therefore the peak attenuation for open endedtubes will be smaller, although the bandwidth of significant attenuationwill increase.

(b) Equation 9.46 in the text may be used here. The downstream ductimpedance for an infinite duct is simply ρc/A. Thus the ratio Zd/Zs inequation 9.46 is given by:


Zd

Zs

'1

j(ω & 104 /ω & jω × 10&4 )

ω2 ' 104 and ω ' 100rad /s

IL ' 20log10*1 %Zd

Zs

* ' 20log10*1 %104

ω* ' 40dB

end correction ' 0.61 × 0.03 %8 × 0.03

3 π(1 & 1.25 × 0.3) ' 0.0342m

ω0 ' c A / RV ' 3432.83 × 10&3

0.0742 × 6.28 × 10&3

1/2

' 844.6rad /sec

K

MC

When the Insertion Loss is at its peak, the side branch reactiveimpedance, Zs is zero. Thus:

The Insertion Loss is then:

Problem 9.7

The equation is simply a differential equation of the vibration of a mass actingagainst a spring. The quantity, M, represents thevibrating mass of air in the resonator neck, thequantity, C, represents the viscous damping lossesat the entrance and exit of the neck caused bymotion of the air particles relative to the edges ofthe neck and the quantity, K, represents thestiffness of the volume of air in the cavity of theresonator as shown in the figure.

Problem 9.8

(a) Helmholtz resonator - cylindrical cavity, 100mm radius, 200mm high.Neck 40mm long with a radius of 30mm. Volume of cylinder, V = π × (0.1)2 × 0.2 = 6.28 × 10-3 m3

Area of neck, A = π × (0.03)2 = 2.83 × 10-3 m2.Length of neck, R = 0.04m + end corrections.

so R = 0.0742m.From equation 9.38 in the text:


Q 'ρcRs

R /AV '413.661000

0.0742

2.83 × 6.28 × 10&6

1/2

' 26.7

¢p 2 ¦ ' 4 × 10&10 × 108 ' 0.04Pa 2

Thus, f0 = 844.6/2π = 134Hz.

(b) The quality factor may be calculated using equation 9.40 in the text.Thus:

(c) If we define effectiveness as the frequency range within 3dB of themaximum, then the bandwidth of effectiveness is obtained using equation7.22 in the text as . Thus we can expectΔf ' f /Q ' 135/26.7 ' 5.0Hzthe device to be effective between 132 and 137Hz.

(d) Incident plane wave of 80dB Lp and 135Hz.

Power dissipated = = at resonance.¢p 2 ¦*Zs*

¢p 2 ¦Rs

The mean square sound pressure is:

Thus the power dissipated = 0.041000

' 40µWatts

(e) Power in a plane wave = IA = ¢p 2 ¦Aρc

' 40 × 10&6

Thus area of plane wave, A = 413.6 × 40 × 10&6 /0.04 ' 0.4136m 2

(f) The sabine absorption of the resonator, = 0.414m2Sα

Cross-sectional area of resonator = . Thus the cross-π × 0.12 ' 0.031m 2

sectional area of the resonator volume is 13 times smaller than its Sabineabsorption area. Thus to make the wall look anechoic, we would need1/0.414 = 2.4 resonators per square metre.

(g) From equation 1.4 in the text, an ambient temperature variation from -5EC to 45EC corresponds to a speed of sound change from

to , which is a range1.4 × 8.314 × 268/0.029 1.4 × 8.314 × 318/0.029from 328 to 357m/s. From equation 9.38, it can be seen that theresonance frequency of the resonator will vary from 808 to 880 rad/secwhich corresponds to a range from 129 to 140Hz which is outside the3dB range discussed in part (c). The problem could be overcome by


pt ' pi % pr ' Aiej(ωt % kx) % Are

j(ωt & kx)

ut ' ui % ur '1ρc

Ai ej(ωt % kx) & Ar ej(ωt & kx)

ρcS

Ai % Ar

Ai & Ar

' ZL ' Zc

Ai % Ar

Ai & Ar

pt

Sut

' Zi ' Zc

Ai ejkL % Ar e&jkL

Ai ejkL & Ar e&jkL

ZL

Lx0

Zi

using two additional types of resonator with resonance frequencies of131.5 and 137.5Hz respectively at 20EC.

Problem 9.9

(a) The total acoustic pressureanywhere along the tube isthe sum of the incidentand end r e f l ec tedpressures and may bewritten in terms of twocomplex constants, Ai andAr, representing the complex amplitudes of the incident and reflected(from x = 0) waves respectively. Thus:

Using equations 1.6, and 1.7, the acoustic particle velocity may bewritten as:

At x = 0, . Thus:pt

Sut

' ZL

At x = L:

Expanding the exponents gives:


Zi ' Zc

Ai cos(kL) % jAi sin(kL) % Ar cos(kL) & jAr sin(kL)

Ai cos(kL) % jAi sin(kL) & Ar cos(kL) % jAr sin(kL)

' Zc

Ai % jAi tan(kL) % Ar & jAr tan(kL)

Ai % jAi tan(kL) & Ar % jAr tan(kL)

' Zc

Ai % Ar % jtan(kL)(Ai & Ar)

Ai & Ar % jtan(kL)(Ai % Ar)

Zi '

Ai % Ar

Ai & Ar

% jtan(kL)

1 %Ai % Ar

Ai & Ar

jtan(kL)

' Zc

ZL

Zc

% jtan(kL)

1 %ZL

Zc

jtan(kL)

Zi ' Zc

ZL % jZc tan(kl)

Zc % jZL tan(kl)

Rearranging gives:

Rearranging gives:

(b) (i) Zc 'ρcS

'414

0.2 × 0.2' 10,400 MKS Rayls

(ii) At x = 0, pr/pi = 0.5; thus Ar/Ai = 0.5 and

Ai % Ar

Ai & Ar

'1 % 0.51 & 0.5

' 3

Thus, ZL = 3Zc = 31,000 MKS Rayls

(iii) Zi ' 1035031050 % j10350tan(200π × 10.29 /343)10350 % j31050tan(200π × 10.29 /343)

' 103503105010350

' 31000MKS Rayls


κ2n ' (ω/c)2 & [(nxπ /Lx)

2 % (nyπ /Ly)2 ]

(ω/c)2 ' [ (nxπ /Lx)2 % (nyπ /Ly)

2 ]

cn 'ω2

ωc

2

&nxπ

Lx

2

&nyπ

Ly

2

1/2

ω2,2 ' 3434πLy

2

%2πLy

2 1/2

'343Ly

16π2 % 4π2 1/2'

4819Ly

cn

c0

fnx,nyf

Problem 9.10

(a) The wavenumber is:

(i) Cut-on when = 0. That is when:κ2

(ii) Phase speed is defined as . Thus:cn ' ω /κn

This equation is shown sketched in the figure below, where ω =2πf.

(b) Drop in Lp over a distance of Ly/4 for 2,2 mode.Cut-on frequency is given by:

Excitation frequency = 0.75ω2,2 = 3614/Ly rad/sec. The wavenumber isthen:


κ2n '

3614343Ly

2

&16π2 % 4π2

L 2y

ΔLp ' 20log101

e&jκ2,2Ly /4

' 20log101

e&9.294/4' 20.2dB

Wt ' 0.5Rep2v(

2 ' 0.5ReZcv2v(

2 ' 0.5ReZc*v2*2

τ 'Wt

Wi

'0.5ReZc*v2*

2

0.5ReZc*v*2' *

v2

v*2

R % jX 'Zs

Zc

vZc

Zs v1 Zc v2

Thus, at the excitation frequency. The acoustic pressureκ2,2 ' ± j9.293

Ly

is related to the distance along the duct as . Thusp(x) % e&jκnx

. Setting x1 = 0 and x2 = Ly/4, we can write the followingp(x1)

p(x2)'

e&jκ2,2x1

e&jκ2,2x2

for the reduction in sound pressure level as the wave travels from x1 to x2.

Problem 9.11

(a) The transmitted power is given by:

A similar expression can be derived for the incident power. Thus thetransmission coefficient is given by:

The characteristic impedance is defined as Zc = ρc/A. Thus:

The circuit equations are v ' v1 % v2 and v1Zs ' v2Zc

from which:


v ' v2 1 %Zc

Zs

and *v2

v*2 '

1

*1 %1

Rs % jXs

*2

τ ' *v2

v*2 '

1

*1 %1

Rs % jXs

*2

'*Rs % jXs*

2

*Rs % jXs % 1*2

'R 2

s % X 2s

(Rs % 1)2 % X 2s

*Rp*2 ' 1 &

R 2s % X 2

s

(Rs % 1)2 % X 2s

'(Rs % 1)2 % X 2

s & R 2s & X 2

s

(Rs % 1)2 % X 2s

*Rp*2 '

2Rs % 1

(Rs % 1)2 % X 2s

The transmission coefficient can then be written as:

(b) The power reflection coefficient, , is related to the transmission*Rp*2

coefficient, τ, by . Thus:*Rp*2 ' 1 & τ

Rearranging gives:

Problem 9.12

(a) The Insertion Loss is calculated using equation 9.54 as it is a constantvolume velocity source. Assuming that the damping term of Equation(7.33) is negligible, the resonance frequency of the muffler is defined by:

. IL ' 10log10 1 &2π × 20ω0

2 2

' 10

Thus ω0 = 61.6rad/sec. To be a little conservative, use ω0 = 60rad/sec.The required chamber volume is then found using equation 9.38 in the


V '3432 × 0.01767

0.3 × 652' 1.92m 3

0PV '0m

MRT

0m 'MP 0VRT

'0.029 × 101.4 × 103 × 2.894

8.314 × 288' 3.55kg/s

0V '0mRTMP

'3.55 × 8.314 × 623

0.029 × 12 × 106' 0.0529m 3 /s

U '0.0529 × 4

π × 0.12' 6.735m/s

ρ '0m0V'

3.540.0529

' 66.9kg/m 3

text. The cross-sectional area of the inlet pipe is. Thus the required chamber volume is:A ' π × 0.152 /4 ' 0.01767m 2

(b) The attenuating device could be made smaller by using a low pass filteras described on pp.433-438 in the text.

Problem 9.13

Low pass acoustic filter (see figure 9.11 in text). Head loss 4 velocity.heads due to tube inlets and exits, so total pressure drop, Δp = 2ρU2. We nowmust calculate the flow speed, U. Assume that the choke tube diameter is thesame as the inlet and exit pipes and equal to 0.1m. Flow rate at STP =250,000m3/day = 2.894m3/sec. The mass flow rate can be calculated from theUniversal Gas Law. Thus:

and:

At operating conditions, T = 623EK and P = 12 × 106 Pa. Thus:

The velocity is thus:

The gas density is:

Thus the pressure drop, Δp is:


Δp ' 2 × 67.2 × 6.7352 ' 6.1kPa

cg ' γRT /M ' (1.3 × 8.314 × 623/0.029)1/2 ' 482m/s

f0 'cg

2π

Ac

Rc

1V1

%1V2

1/2

'4822π

π × 0.12

4 × 1.81 % 1

1/2

' 7.2Hz

which is much less than 0.5% of 12MPa, so it is OK.

Speed of sound in the gas is:

Following the design procedure on pages 437 and 438 of the text, we have

1. f0 = 0.6 × 10 = 6Hz

2. Try V1 = V2 = 1m3

4. Choke tube diameter = 0.1m and length = 1.8m (assuming the chambersare cylinders 1m long and 0.64m diameter).

7. Resonance frequency is:

which is close enough to 6Hz for now.

Choke tube x-sectional area = π × 0.01/4 = 0.007854m2

Equation 9.71 in the text may be used to calculate the Insertion Loss for themuffler. Substituting the values into this equation gives IL = 30dB which istoo much. Thus try changing the volumes to 0.75m3 and the choke tube lengthto 1.6m. This gives an Insertion Loss of 17.6dB which is too small. Trychanging the volumes to 0.9m3 and the choke tube length to 1.7m. This givesan Insertion Loss of 25.9dB which is too large. Try changing the choke tubelength to 1.4m. This gives an Insertion Loss of 20.8dB which is OK.

Thus, the final design is for 2 volumes, each of 0.9m3 with a 0.1m diameterchoke tube, 1.4m long connecting them.

In practice, conservatism would usually dictate sticking with the 30dB designif it is practical.


26 '343

2 × ππ ×0.052

3 × 4 × Rc

(0.03&1 % 0.03&1 )1/2

compressor

12

43

L

Problem 9.14

As the tubes are short, theymay be treated as lumpedelements and the designprocedure outlined onpages 437 and 438 in thetext may be used. Theimpedance of the threetubes is three times that fora single tube.

The design frequency is 40Hz. Thus the required resonance is 0.65 × 40 =26Hz. To simplify matters, use the largest allowable chamber volumes andsmallest allowable choke tube diameter. Using equation 9.80, we obtain:

Thus and Rc = 192mm.26R1/2c ' 11.403

The filter thus consists of 2 volumes, each 0.03m3, connected by three 0.05mdiameter tubes which are approximately 0.2m long.

Problem 9.15


v ' v2 % v3 % vL

v2Z2 ' v3(Z3 % Z4) ' VLZL

v3 ' vL

ZL

Z3 % Z4

v2 ' vL

ZL

Z2

v ' vL 1 %ZL

Z3 % Z4

%ZL

Z2

IL ' 20log10*1 %ZL

Z3 % Z4

%ZL

Z2

*

ZL 'ρcA

'413.6 × 4

π × 0.022' 1.317 × 106

Z2 ' Z4 ' & jρc 2

Vω' & j

1.206 × 3432

0.2 × 2π × 10' & j1.129 × 104

Z1

Z2

v2

v3vL

ZL

Z4

Z3

v

(a) The equivalent acoustical circuit is shown above.

(b) The Insertion Loss of the muffler is given by equation 9.64 in the text.The circuit equations are:

Thus the Insertion Loss is given by:

(c) For no reflections from the pipe exit:

The volume impedances are:


Z1 ' Z3 ' jρcA

tan2πLλ

' j413.6 × 4

π × 0.022tan

6π343

' j7.24 × 104

IL ' 20log10*1 %1.317 × 106

j(7.24 × 104 & 1.129 × 104 )%

1.317 × 106

& j1.129 × 104*

' 20log10*1 & j21.54 % j116.7*

. 39.6dB

v ' v1 % v2 % v3 % v4

v1Za ' (v2 % v3 % v4) (Zb /2) % (v3 % v4) (Zb /2) % v4Zg

v2(Zc % Zd) ' (v3 % v4) (Zb /2) % v4Zg

v3Ze ' v4Zg

c

d

b

ea

f g

v

Zf Zb/2 Zb/2

v2 v3Zc

Zd

Za

v1 Ze Zg v4

and the tube impedances are:


Problem 9.16

(a) Equivalent circuit diagram.

(b) System equations:


Zi ' Z3 %1

(1 /Z2) % (1 /Z1)' Z3 %

Z1Z2

Z1 % Z2

'Z3Z1 % Z3Z2 % Z1Z2

Z1 % Z2

Z3 '4jρc

πd 23

tan(kL3); Z2 ' &4jρc 2

πd 22 L2ω

0.5m 0.5m0.5m

0.4m

10mm10mm

Z1 Z2Z3 Zi

Zi

Z3

Z2 Z1

(c) Inductive (with resistive part) impedances are Zb, Zc, Zf and Zg.Capacitative impedances are Za, Zd and Ze.

Problem 9.17

(a) Impedance looking into the tube is:

(b)


Z1 ' &4jρc

πd 21

cot(kL1)

Zi '

&(4jρc)2

π2d 21 d 2

3

cot(kL1)tan(kL3) &(4jρc)2c

π2d 22 d 2

3 L2ωtan(kL3)

%(4jρc)2c

π2d 21 d 2

2 L2ωcot(kL1)

&4jρcπ

cot(kL1)

d 21

%c

d 22 L2ω

&cot(kL1) tan(kL3)

d 21 d 2

3

&c tan(kL3)

d 22 d 2

3 L2ω%

ccot(kL1)

d 21 d 2

2 L2ω' 0

&ω × 8 × 10&2 cot(ω /686) tan(ω /686)

& 343 × 10&4 tan(ω /686) % 343 × 10&4 cot(ω /686) ' 0

&ω % 0.429(& tan(ω /686) % cot(ω /686)) ' 0

and:

Thus:

(c) Resonance occurs when inductive impedance = capacitative impedance,or Zi = 0.

Eliminating (4jρc/π)2 from previous expression for Zi and setting theresult = 0, we obtain:

Substituting for k = ω/343, L1 = L2 = L3 = 0.5, d1 = d3 = 0.01 and d2 = 0.4,we obtain:

Simplifying gives:

Solving by trial and error:


reciprocatingcompressor

Z1Z2

Z3 ZL

Z4

Z5

Z4

Z3Z1

Z5

Z2

ZL

reciprocatingcompressor

v

v1

v2v3

ω Value ofexpression

ω Value ofexpression

1102018

29319.4-5.3-1.6

1717.3

17.1517.1586

0.3-0.280.017

-0.000045

Thus the resonance frequency = 17.1586/2π = 2.7Hz.

(d) At resonance, there will be a pressure maximum at the closed end and amaximum particle velocity at the open end.

Problem 9.18

(a)

(b) Constant volume velocity source. Thus . The Insertionv ' v1 % v2 % v3

Loss is given by:


IL ' 20log10*vv3

*

Z2v1 ' Z3(v2 % v3) % v2(Z4 % Z5)

v1 'Z3v2

Z2

%Z3v3

Z2

%Z4v2

Z2

%Z5v2

Z2

v2(Z4 % Z5) ' v3ZL; thus v2 'v3ZL

Z4 % Z5

v1 ' v3

Z3ZL

Z2(Z4 % Z5)%

Z3

Z2

%Z4ZL

Z2(Z4 % Z5)%

Z5ZL

Z2(Z4 % Z5)

IL ' 20log10*(v1 % v2 % v3)

v3

*

' 20log10/00001 %

ZL

Z4 % Z5

%Z3ZL

Z2(Z4 % Z5)%

Z3

Z2

/0000%

Z4ZL

Z2(Z4 % Z5)%

Z5ZL

Z2(Z4 % Z5)

Examining pressure drops, we may write:

Thus:

Also:

Substituting for v2 in the expression for v1 gives:


(c) The result of (b) is no longer valid if the dimensions of the chambersrepresented by impedances Z2 and Z5 exceed one quarter of a wavelengthof sound.


1Zs

'1Zv

%1Za

Zv ' &jρc 2

Vωand Za ' j

ρωRA

1Zs

' jVω

ρc 2& j

πa 2

ρωR

' jVω

ρc 2& j

3π2a

8ρω 2 & 2.5aπ

4V

1/3

ZaZv

cross-sectionalarea, , of holeA

Volume, Vv

Zv Za

Problem 9.19

(a) Load seen by speaker (ignoring external air load), Zs given by:

where:

and R is the effective length of the orifice. From equation 9.16, the endcorrection for the side of the hole in free space is 8a/3π. If we assumethat the enclosure is cylindrical of diameter D, then the end correction for

the side of the hole in the enclosure is and the totalR0 '8a3π

(1 & 2.5a /D)

effective length of the hole is then . If the cylinderR '8a3π

(2 & 2.5a /D)

diameter is equal to its length, then andD ' (4V /π)1/3

. R '8a3π

2 & 2.5aπ

4V

1/3

Thus:


Zs ' & j16ρωc 2 (1 & 1.153aV &1/3)

16Vω2 (1 & 1.153aV &1/3) & 3π2ac 2

16Vω2 (1 & 1.153aV &1/3) ' 3π2ac 2

ω0 '3π2ac 2

16V (1 & 1.153aV &1/3)

Thus:

(b) At low frequencies, the second term in the denominator will be largerthan the first and so the phase of the impedance will be +j which meansthat the acoustic pressure leads the acoustic particle velocity by 90E. Asthe particle displacement also leads the velocity by 90E, the particledisplacement at the orifice and the acoustic pressure will be in phase. Asthe box is small compared to a wavelength, the acoustic pressure will bein phase with the displacement of the cone; thus the out flow from theorifice will be in the same direction as the cone motion.

At higher frequencies, the second term in the denominator will becomesmaller than the first and the phase of the impedance will be -j, a 180Eshift from the lowerfrequency case. Thus inthis case the out flow fromthe orifice will be in theopposite direction to thecone motion and will thusreinforce the out flow atthe cone as shown in thefigure.

The crossover frequency is thus when the two terms in the denominatorare equal, which is when:

or:

(c) We need to solve the above equation for V, given that ω0 = 2π × 100.Rearranging the above equation and substituting values for variables, weobtain:


V '3π2ac 2

16ω20 (1 & 1.153aV &1/3)

'3π2 × (0.01 /π)1/2 × 3432

16 × 4 × π2 × 104(1 & 1.153 × (0.01 /π)1/2V &1/3)

'0.0311

1 & 0.0651V &1/3

ab

1m3d

1.5m3f

2m3

0.2m

L

0.2m dia0.1m long

0.2m dia0.1m long

0.2m dia10m long

c

e

Equivalent acoustical circuit

Za Zc Ze

ZLZfZdZb

p

vb vd vf vL

Solving by trial and error gives V = 0.039m3.

Problem 9.20

Referring to the figures above and using equation 9.55 in the text for aconstant pressure source, we may write the following Insertion Loss andacoustical circuit equations:


IL ' 20log10*p

vL ZL

* (1)

p ' (vb % vd % vf % vL )Za % vb Zb (2)

vb Zb ' vd Zd % Zc (vd % vf % vL ) (3)

vd Zd ' vf Zf % Ze(vf % vL ) (4)

vf Zf ' vL ZL (5)

vd ' vf

Zf

Zd

%Ze

Zd

%vL Ze

Zd

(6)

vd ' vL

ZL

Zd

%Ze ZL

Zd Zf

%Ze

Zd

(7)

vb ' vd

Zd

Zb

%Zc

Zb

% vf

Zc

Zb

% vL

Zc

Zb

(8)

vb ' vL

Zd

Zb

%Zc

Zb

ZL

Zd

%Ze ZL

Zd Zf

%Ze

Zd

%Zc ZL

Zb Zf

%Zc

Zb

(9)

pZL vL

'Za % Zb

ZL

Zd

Zb

%Zc

Zb

ZL

Zd

%Ze ZL

Zd Zf

%Ze

Zd

%Zc ZL

Zb Zf

%Zc

Zb

% Za

ZL

Zd

%Ze ZL

Zd Zf

%Ze

Zd

%ZL

Zf

% 1

Using eq. (4), we can write:

Using eqs. (5) and (6), we can write:

Using eq. (3):

Using eqs. (5), (7) and (8):

Using eqs. (2), (5), (7) and (9):



IL ' 20log10/0000

Za % Zb

ZL

Zd

Zb

%Zc

Zb

ZL

Zd

%Ze ZL

Zd Zf

%Ze

Zd

%Zc ZL

Zb Zf

%Zc

Zb

/0000% Za

ZL

Zd

%Ze ZL

Zd Zf

%Ze

Zd

%ZL

Zf

% 1

Za 'jρcAa

tan(k Ra) % Ra ; Zb ' & jρc 2

Vbω; Zd ' &j

ρc 2

Vdω

Zc ' jρcAc

tan(k Rc ) % Rc ; Ze ' jρcAe

tan(k Re ) % Re ; ZL 'ρcAL

Ra 'ρcAa

ωc

tDawa

2Aa

1 % (γ & 1)5

3γ% 0.288

ωc

d log10

4Aa

πh 2a

% εω2

c 2

Aa

2π% M

t ' 2µ / (ρω) ' 0.00550/ ω ; (µ ' 1.8 × 10&5 kgm &1 s &1)

where the impedances are defined as:

From equation 9.29, p.417:

Rc and Re are defined similarly to Ra, except that subscripts c and e aresubstituted for subscript a respectively.ρc = 1.206 × 343 = 414; Ae = Ac = Aa = π × 0.22/4 = 0.00314m2.Pipe end corrections, R0 = (8 × 0.1)(1 - 1.25 × 0.2/1)/(3 × π) = 0.064m

hc = he = largest of t or 0.05; ha = largest of t or tube inlet radius.wa = 10m, we = wc = 0.1m, γ = 1.4, Vb = 1, Vd = 1.5, Vf = 2m3

Assume that M is small enough to neglect.εa ' εc ' εe ' 0


12

Re Δ pZ0v (

1 '12

Re Z0 v1 v (

1 '*v1*

2

2Re Z0

v1(Z1 % Z0) ' v2 Z2 ' Δ p

v1 'Δ p

Z1 % Z0

W0 '*v1*

2

2Re 6Z0 > ' *Δ p*2

2

Re 6Z0 >*Z1 % Z0*

2

p

pZ1Z1

Z2Z2 Z0Z0v2 v1 v2 v1

Problem 9.21

(a) Fan and plenum equivalent circuit diagram is shown in the figure below.

Power flow through Z0 is the radiated sound power and is equal to:

v1 is an rms quantity and is the pressure drop across Z0. EquatingΔ pZ0

circuit pressure drops gives:

At low frequencies, ReZ0 = 0, and so W0 = 0

(b) For Z2, use the analysis leading up to equation 9.35 in the text. For Z1 usethe analysis leading to equation 9.14 for the imaginary part and equation9.29 for the real part.

The expressions are valid over the frequency range from zero up to wherethe neck diameter or plenum dimensions approach 0.2 of a wavelength.

(c) The resistive component consists of viscous friction losses due to airparticles vibrating back and forth against the edge of the inlet.

(d) Varying the fan position along the duct will have the effect of adding aninductive impedance between the fan and plenum and will also changeimpedance, Z1. Thus the radiated sound power will vary.


cg 'γRTM

'1.3 × 8.314 × (273 % 900)

0.03

1/2

' 650m/s

25 ' 10log10 1 & (ω /ω0 )2 2 ;

Thus, ω0 'ω

18.78'

2π × 50

18.78' 72.5rad /s

11.3

&18 × 0.1314.2V

&2 × 0.01

A 2 × 6502

72.52 A 1.5 V

10 π × 8 × 10&3 × 0.01> 1

0.77 &5.73 × 10&3

V&

4.734 × 10&8

A 23.707 × 106 A 1.5 V > 1

Amin '3 × 4.734 × 10&8

0.77

1/2

' 4.3 × 10&4 m 2

Vmin '3 × 5.73 × 10&3

0.77' 0.022m 3

[0.769 & 0.143 & 0.051]4.41 ' 2.53 , which is a bit large

Problem 9.22

The speed of flow, U0 = 0.1m/s. The speed of sound in the gas is:

Following the design procedure on p.432 in the text, the desired resonancefrequency is given by:

Using equation 9.60 in the text, the design equation is:

Rewriting gives:

To begin, set each term in brackets = (0.77/3). Thus:

However, this will not satisfy the minimum requirement for A1.5V. Thusincrease A and V by a factor of 2. Thus try V = 0.04m3 and A = 9.6 × 10-4m2 (d = 35mm).

Design equation becomes:


[0.769 & 0.143 & 0.095]2.78 ' 1.48 , which is OK

R '7.06 × 10&4

0.0465072.5

2

' 1.42m

0.3m 0.6m

Try reducing d to 30mm; A = 7.06 × 10-4 and the design equation is

The required tail pipe length, R, is:

Design summary (to achieve 25dB at 50Hz)Volume = 40litresTail pipe diameter = 30mmtail pipe length = 1.42m

Could use a smaller tail pipe length and larger tail pipe diameter if a largervolume were chosen.

Problem 9.23

Use maximum allowableOD = 0.6m.Inside diameter= 0.3m = 2hLiner thickness,R = 0.15mThus R/h = 1.0M = -17/343 = -0.05At 500Hz,2hλ

'0.3 × 500

343' 0.437

From Figure 9.15 in the text, for M = 0, R/h = 1.0 and 2h/λ = 0.437, the figurewith the highest attenuation is the top right figure corresponding to R1R/ρc =2. For a square duct lined on 4 sides (equivalent to a circular duct), theattenuation would be 6.3dB per length of lined duct equal to the duct radius.

From Figure 9.17, top right figure, the attenuation for M = -0.1, is7.1 dB perlength of duct equal to the duct radius. As M = -0.05, use an attenuation rateof 0.5(6.3 +7.1) = 6.7dB per length of duct equal to the duct radius.

From figure 9.22 in the text, (expansion ratio = (0.6/0.3)2 = 4), the requiredliner attenuation for an overall attenuation of 15dB is 10.8dB.Thus the required length of liner = 0.15 × 10.8/6.7 = 0.24m


Sλ

' π ×0.32

4

1/2

×500343

' 0.39

450mm 300mm

75mm

Additional noise reduction if direction of sound is distributed equally in alldirections at the duct inlet (for example, if the duct were venting a machineenclosure) can be found from figure 9.21, p.459 in the text, where:

From the figure, the additional attenuation is 7dB.

Problem 9.24

The total attenuation is made up of! entrance losses (figure 9.21, p.459)! exit losses (table 9.5, p.464, numbers in brackets)! liner attenuation (figure 9.15, p.449)

Mach no., M = 0.0. Area of open duct = 75 × 300 × 10-6 = 0.0225, σ = 0. Tomaximise the attenuation choose the curve corresponding to R/h = 4 and R1R/ρc= 4 in figure 9.15 which corresponds to M = 0.0. For the large dimension, 2h= 0.3 and duct length is equal to 3h, and for the small dimension, 2h = 0.075and the duct length is equal to 12h. The following table may now begenerated.


2h1 /λ

short

2h2 /λ

long

attenuation (dB) - lining only

short sides long sides all sides

500100020004000

0.4370.8751.7493.499

0.1090.2190.4370.875

7.54.62.40.9

19.124.229.524.0

26.630.231.924.9

The total loss may now be calculated with the aid of the following table.


Octave bandcentre

frequency(Hz)

S /λ(S=0.3×0.075)

Entranceloss(dB)

Exit loss (dB)

D '4Aπ

' 0.169

liningloss(dB)

totalloss(dB)

500100020004000

0.2190.4370.8751.749

3.17.39.6

10.0

4.21.60.60

26.630.231.924.9

33.939.142.134.9

(a) It is clearly possible to achieve 30dB or more in each of the octave bands,1 2 & 4kHz. In fact a thinner liner would most likely be adequate.

(b) The best attenuation possible at 500Hz is 34dB.

Problem 9.25

Dissipative muffler - 3 attenuations: inlet, outlet and lined section (noexpansion loss as it is mounted on an enclosure). Try beginning with thesmallest allowed cross section of duct. Thus, S = 0.25 m2

and and S /λ ' f × S /c ' 1.458 × 10&3 f 2h /λ ' 0.5 f /c ' 1.458 × 10&3 f

frequency(Hz)

inletSλ

'2hλ loss

outletloss

totalatten.

needed

Lineratten

needed

125 0.18 2 5.5 9 1.5

1000 1.46 10 0 15 5

2000 2.91 10 0 15 5

If we use the largest outer cross section allowed, the ratio of liner thicknessto half airway width is 1.0. Using curve 3 in Figure 9.16, assuming a flowspeed of M=0.1, and using R1R/ρc = 8, we obtain the following attenuationsfor 0.25 m length of duct lined on all 4 sides.

125 Hz 1.2 dB1000Hz 3.4 dB2000 Hz 1.0 dBIt is clear that the critical frequency is 2000 Hz and to satisfy the requirementof 5 dB at this frequency, we need a length of duct equal to 5 × 0.25/1.0 =


45o

100mm

200mm

300mm

a

b

a

b

1.25 m. Note that many other solutions would be acceptable as well.

Problem 9.26

(a) From the figure to the right, usingsimilar triangles, it can be seen thatthe ratio, R/h is 100/200 = 0.5 (asone side only of the airway islined). It may also be assumed thatthe flow speed is small enough toignore.

(b) For the ratio R/h = 0.5, anacceptable value of R1R/ρc is 2.Thus the required flow resistance ofthe liner is R1 = 2 × 413.6/0.0707 =11,700 MKS Rayls.

(c) h = 2R = 200//2 = 141.4mm. Ductcross-sectional area = (0.2//2) ×0.4 = 0.0566m2. The effective ductlength, L = 300/2 = 424mm = 3h.Wavelength, λ = 343/f. The exitloss is obtained from Table 9.5 inthe text and the inlet loss is fromfigure 9.21 in the text (assuming diffuse field input). The lined duct lossis from figure 9.15 in the text (curve 2, top right figure). The results aresummarised in the following table.


r

Octavebandcentre

frequency(Hz)

2h/λ

linedductloss(dB)

length, 3h

/S/λInletLoss(dB)

ExitLoss(dB)

TotalAtten.(dB)

631252505001000200040008000

0.0520.1030.2060.4120.8251.653.306.60

0.10.62.16.68.23.00.90.3

0.0440.0870.170.350.691.392.785.55

002

6.18.8101010

12.38.24.21.20.1000

14991418131110

Problem 9.27

(a) Insertion Loss is the difference in sound level at the end of the duct withand without the silencer in place. Transmission Loss is the difference insound pressure level measured at the inlet and outlet of the silencer.

(b)

Power transmitted down duct with no muffler = 1.Power transmitted with muffler = τ.Thus IL = -10log10τ

Power incident on muffler = 1.Power transmitted by muffler = τThus Transmission Loss, TL = -10log10τ = IL

(c) Dissipative attenuators absorb energy and contain surfaces lined withsound absorbing material. They are cost effective for high frequencynoise.

Reactive mufflers change the radiation impedance "seen" by the sound


TL ' &10log10AR%

Acosθ

πr 2

R 'Sα

(1 & α)

' 2(3.32 × 2.82 % 3.32 × 2.95 % 2.82 × 2.95) × 0.1 /0.9

' 6.106m 2

3.32m

2.82m

source (tonal noise or system resonances) and reflect energy back to thesource. They can also dissipate energy through viscous losses atentrances and exits of small air passage ways. Reactive mufflers are costeffective for low frequency broadband noise. They can also be tuned toattenuate tonal noise at one or more frequencies.

Active mufflers act in a similar way to reactive mufflers but theimpedance change, sound reflection or absorption is provided by a soundsource such as a loudspeaker. These mufflers are cost effective for lowfrequency tonal noise problems and in many cases they are preferred toreactive mufflers because of their relatively small size and installationconvenience. They are also preferred in dirty environments wherereactive mufflers can become clogged.

Problem 9.28

(a) Room 3.32m long, 2.82m wide and 2.95m high. Doors at each end are2.06m high, 0.79m wide. Can treat it like a plenum chamber.

and the sound power attenuation or transmission loss is:α ' 0.1

A = 2.06 × 0.79 = 1.627m2, r = 3.32m

Thus:


TL ' &10log101.6276.106

%1.627

π × 3.322' 5.0dB

R ' 6.106 ×0.90.1

×0.50.5

' 54.95m 2

TL ' &10log101.62754.95

%1.627

π × 3.322' 11.2dB

TL ' &10log101.6276.106

' 5.7dB

TL ' &10log101.62754.95

' 15.3dB

W '¢p 2 ¦4ρc

A '4 × 10&10 × 108.5 × 1.627

4 × 413.6' 1.244 × 10&4 W

(b) Increasing to 0.5 gives:α

Thus,

That is, a 6dB improvement.

(c) If the direct line of sight were prevented, then the direct field termcontribution will be zero. Thus for case (a) above:

and for case (b) above,

That is, there is little difference in the first case where the reverberantfield contribution dominates.

(d) Sound power leaving the doorway of the first room is given by the soundintensity directed towards the door multiplied by the door opening area.Using equations 7.33 and 1.75 in the text, we obtain:

The sound power level is:Lw ' 10log10 W % 120 ' 81.0dB re 10&12 W

Thus the sound power incident on the second doorway is: 81.0 - 5.0 = 76.0dB.

Assuming no reflection from the doorway, this corresponds to a soundpressure level in the doorway given by equation 6.2 in the text as (where


Lp ' Lw & 10log10 A % 0.15

' 76.0 & 10log10(1.627) % 0.15 ' 74.0dB

Ns ' fd /c ' 500 × 1/343 ' 1.458

Lp ' Lw & K %DIM & AE (dB re 20µPa)

100m

90o

R2m

1

2

the correction for ρc … 400 has been included):

Problem 9.29

At a distance of 50m, the angular orientation from the stack axis of the linejoining the stack to the observer is approximately 90E. The directivity indexmay be obtained using figure 9.27 in the text. The Strouhal number is:

Thus from figure 9.27 in the text, DIM = -9.6dB.

The sound pressure level at the receiver is related to the sound power radiatedby the stack using equation 5.158 in the text which may be written as:

Lw = 135dB

K = 10log10(2πr2) = 10log10(2π × 1002) = 48dB (equation 5.161 in text)

AE = Aa + Ag + Am + Ab + Af (equation 5.165 in text)

Aa = 0.2 - 0.3dB (table 5.3, p. 225 in text)

Ag = -3dB (attenuation of ground reflected wave = attenuation of directwave)

Am = (+3, -1)dB (table 5.10, p.243 in text)

Ab = Af = 0


Lp ' 135 & 48 & 9.6 & 0.2 % 3 & (%3, &1) ' 77 & 81dB re 20µPa

Thus:

If meteorological influences are ignored, Lp = 80dB re 20µPa.

10

Solutions to problems invibration isolation

Problem 10.1

(a) The difference in levels between the plant room and apartment for thetwo cases of the plant operating and the test source operation areobtained by subtracting appropriate rows in the table given in theproblem and are given in the table below


63 125 250 500

Equipment operating 25 36 38 40

Test source operating 34 37 38 39

Inspection of the above table indicates that in the 63Hz band, the problemis dominated by structure-borne noise whereas at higher frequencies, air-borne noise dominates. Thus improved vibration isolation will only helpthe low frequency noise problem and after allowing for the A-weightingcorrection of table 3.1, it can be seen that reducing the 63Hz problem willnot significantly reduce the A-weighted noise level in the apartment.

Problem 10.2

The single degree of freedom model gives inaccurate estimations of vibrationtransmission in the audio frequency range because it treats the spring andsupported mass as lumped elements and does not include the effects of wavetransmission along the spring. The wave transmission effects are negligibleat sub-audio frequencies but are often important mechanisms of vibrationtransmission in the audio frequency range.


Ts ' mL

0

12ρS

yL0x 2 dy '

12ρS

L 20x 2 L 3

3

'12

(ρSL )0x 2

3

'12

ms 0x2

Ttot ' Ts % Tm '12

m3

0x 2 %12

m 0x 2 '12

m %ms

30x 2

Tmax '12

m %ms

3(ωA)2

V '12

kx 2max '

12

kA 2

12

m %ms

3(ωA)2 '

12

kA 2

m

ydy Lk, ms

0

xProblem 10.3

(a) Assuming that the displacementof the end of the spring attachedto the mass is described by x(t).An intermediate point on thespring, a distance of y from thefixed base will have adisplacement of (y/L)x(t). Thetotal kinetic energy of the springis:

The total kinetic energy of the spring and mass is:

If x(t) = Asin(ωt), then and: 0x(t) ' ωAsin(ωt )

The maximum potential energy stored in the spring is:

where k is the spring stiffness.

Equating Tmax with Vmax gives:

Vibration isolation 305

ωn 'k

m %ms

3

f0 'ωn

2π'

12π

km % ms /3

E 'force /area

spring extension/spring length'

kx /Sx /L

'kLS

cL 'Eρ

'kLSρ

' fsλs ' 4fs L

fs '14

kSρL

' 0.25k

ms

M2ξ

Mx 2'

1

c 2L

M2ξ

Mt 2

ξ ' Aej (ωt & (ω /cL )x )

% Bej (ωt % (ω /cL )x % θ)

Thus:

and the resonance frequency f0 is thus:

(b) Effective Young's modulus, E, is given by:

Longitudinal wave speed is thus given by:

Thus the surge frequency is given by:

(c) The wave motion in the spring satisfies the one dimensional waveequation given by:

where ξ is the spring longitudinal displacement as a function of axiallocation x.

The solution for this equation is the same as for the acoustic case. Thatis:


ξ ' A ej (ωt & (ω /cL )x )

& ej (ωt % (ω /cL )x )

mM2ξ

Mx 2' &kL

MξMx

&mω2 ej (ωt & (ω /cL )L )

& ej (ωt % (ω /cL )L )

' jkLωcL

ej (ωt & (ω /cL )L )

% ej (ωt % (ω /cL )L )

mω2 ej (ω /cL )L )

& e& j (ω /cL )L )

' jkLωcL

e& j (ω /cL )L )

% ej (ω /cL )L )

ωLcL

tanωLcL

'ρSLm

'ms

m'

1N

One boundary condition is that at x = 0, ξ = 0. Substituting this into thesolution to the wave equation gives . Thus the wave equationBejθ ' &Asolution becomes:

The second boundary condition is that the inertia force of the mass at x= L is equal to the spring force. That is:

where k is the spring stiffness.

Substituting the wave equation solution into this gives:

which can be rewritten as:

As shown in part (b), kL = cL2ρS. Using this relation and rearranging the

above equation gives:

The surge frequency is 2π × ω, where ω is the solution of the abovetranscendental equation. As stated in the problem, the upper frequencybound will be 0.9 × this value.


ke 'k1 k2

k1 % k2

ω1 'k1 k2

(k1 %k2)m

ω2 'k1

m

ω1

ω2

'k2

k1 % k2

'1

1 % k1 /k2

m

k1

k2

0

x

MmMm

Mi

MfMf fmfm ffff

finfin

without isolator with isolator

Problem 10.4

As shown in the figure, let the framestiffness be represented by k2 and theisolator stiffness by k1. The effectivestiffness, ke is then given by:

Thus the resonance frequency is given by:

With a rigid frame, the resonance frequency is given by:

The ratio of the two (which is plotted in figure 10.7) is:

Problem 10.5

(a) The equivalent mobility electrical circuits are shown in the two figuresbelow.


fmMm ' ff Mf

fin ' fm % ff ' ff 1 %Mf

Mm

' ff

Mm %Mf

Mm

T1 'ff

fin

'Mm

Mm % Mf

fm ' ff

Mi % Mf

Mm

fin ' fm % ff ' ff 1 %Mi % Mf

Mm

' ff

Mm % Mi % Mf

Mm

T2 'ff

fin

'Mm

Mm % Mf % Mi

TF 'T2

T1

'Mm % Mf

Mm % Mf % Mi

(b) For the circuit without the isolator:

and:

The force transmissibility is the ratio of ff /fin, which is:

For the circuit with the isolator:

and:

The force transmissibility is the ratio of ff /fin, which is:

The force transmissibility with the isolator compared to that without theisolator is then T2/T1 and is:

which is the same as equation 10.31.


Problem 10.6

2b = 0.7 m; machine width = 0.9m2e = 1.0 m; machine depth =1.2m2h = 0.2 ma = 0.5 - h = 0.4 m; machine height = 2×(0.4-h) = 0.6mvertical dimension of mass, 2d = 2(a - h) = 0.6mk = 4000 N/m

Radius of gyration about vertical y-axis (eq. 10.18 in text):

δy ' [ (0.9/2)2 % (1.2/2)2 ] /3 ' 0.4330 m

Radius of gyration about horizontal x-axis (eq. 10.18 in text):

mδx ' [ (0.6/2)2 % (1.2/2)2 ] /3 ' 0.3873

Radius of gyration about horizontal z-axis (eq. 10.18 in text):

mδz ' [ (0.6/2)2 % (0.9/2)2 ] /3 ' 0.3122

f0 '12π

km

'1

2π4000 × 4

50' 2.847Hz

Now the rocking modes will be calculated, first in the x-y plane (about the z-axis) and then in the z-y plane (about the x-axis):

Wx ' (δz /b) kx /ky '0.31220.35

1/4 ' 0.446

Mx ' a /δz ' 0.4/0.3122 ' 1.281

From figure 10.5 in the text, Ωa = 0.38 and Ωb = 1.19. Thus:

andfa ' 0.38 × 2.847 × 0.35/0.3122 ' 1.21Hz

fb ' 1.19 × 2.847 × 0.35/0.3122 ' 3.80Hz

Wz ' (δx /e) kz /ky '0.3873

0.51/4 ' 0.3873

Mz ' a /δx ' 0.4/0.3873 ' 1.033

From figure 10.5 in the text, Ωa = 0.33 and Ωb = 1.09. Thus:


TF 'Mm % Mf

Mm % Mf % Mi

TF '0.1 % 0.2

0.1 % 0.2 % 1' 0.231

k2

k1

'm1 m2

(m1 % m2 )2

2π f2 'k2

m2

ζ2 '3(m2 /m1 )

8(1 % m2 /m1 )3

andfc ' 0.33 × 2.847 × 0.5/0.3873 ' 1.21Hz

fd ' 1.09 × 2.847 × 0.5/0.3873 ' 4.01Hz

The resonance frequency of the rotational mode is calculated using equation10.17 in the text and is:

fy '1π

0.352 × 4000 % 0.52 × 4000

50 × 0.4332' 4.01Hz

Problem 10.7

From equation 10.31, the increase in force transmission is:

From the question, Mf = 0.2Mi = 2Mm. Thus:

which corresponds to a reduction in force transmission by a factor of 4.3.

Problem 10.8

Referring to figure 10.11 in the text and the discussion on pages 496, theoptimal absorber is characterised by:

The excitation frequency is 3000/60 = 50Hz and this should be equal to the


9.87 × 104 × m2

107'

1000m2

(1000 % m2 )2

k2 ' (100π)2 × m2 ' 19.7MN/m

ζ '3 × 200/1000

8(1 % 200/1000)3' 0.21

resonance frequency of the absorber mass and spring system.Thus or (100π)2 ' k2 /m2 k2 ' 9.87 × 104 × m2

Using the first equation and substituting the above for k2, and the given valuesfor k1 and m1, we obtain:

which results in a negative mass m2. The text indicates that the damping massshould be as large as possible; thus it seems impractical to try to satisfy thesecond of the above three design equations. Thus let the design mass m2

=20% of m1 = 200kg. Thus the required stiffness is given by:

The required damping may be obtained from equation 10.47 as:

Problem 10.9

Damping a vibrating surface will only reduce the resonant response. Thusdamping will only result in a reduction in sound radiation if the resonantmodes are contribution most to the radiated sound field. This is generally thecase when the surface is excited mechanically. However, if the surface isexcited by an acoustic wave on the side opposite that which is causing theradiation problem, then it is likely that the sound radiation will be dominatedby vibration modes which are being forced to vibrate at frequencies wellabove their resonances. In this case damping the surface will not reduce thesound radiation directly but may reduce it a little because adding mass to thevibrating structure will decrease its mobility for excitation by the incomingsound field. See pages 504-506 in the text.

Problem 10.10

(a) 100dB below 1 volt corresponds to a voltage of 1 × 10-100/20 Volts =


u 'pρc

'2 × 10&5 × 1080/20

413.6' 4.836 × 10&4 m/s

0urms ' 2π fu ' 2π × 1000 × 4.836 × 10&4 ' 3.04m/s 2

drms 'u

2π f'

4.836 × 10&4

2π × 1000' 0.078µm

10µVolts.

Accelerometer mass, m (grams), is approximately equal to sensitivity inmV/g. Smallest detectable acceleration (in g) is the smallest detectedvoltage in milli-volts divided by m. Thus smallest detectable accelerationis 10-2/m "g" which in metres/sec is 0.01 divided by the accelerometerweight in grams.

(b) The required relation can be determined by considering the mass loadingeffect of the accelerometer. To obtain results within 3dB of the correctlevel, the accelerometer mass must satisfy the requirement that

.m < 3.7 × 10&4(ρcL h 2 / f )

For steel, ρ = 7800 and cL = 5150/0.954. Thus the condition must be satisfied.mfu < 15,580 h 2

(c) From part (a), if the smallest acceleration to be detected is 0.01g, then thelightest accelerometer which can be used is 1 gram. Substituting m = 1and h = 1 in the relation of (b) above gives fu = 15.58kHz.

Problem 10.11

(a) For a large plate vibrating as a piston, sound will be radiated as planewaves with no near field. Thus at any point the pressure and acousticparticle velocity are related by p = ρcu. As the acoustic particle velocityadjacent to the plate is equal to the plate velocity, the velocity of the plateis given by:

The r.m.s. acceleration is thus given by:

(b) The r.m.s. displacement is given by:


(c) At high frequencies accelerations are generally large compared todisplacements, whereas the opposite is true at very low frequencies. Anaccelerometer is thus the best means of measuring the acceleration at1kHz, provided that the accelerometer did not significantly mass load theplate (see equation 10.53 in the text). For thin plates at high frequencies,a measurement of the sound pressure close to the plate may be the bestway of determining the plate response.

Problem 10.12

(a) Adding damping will reduce the sound radiated by a vibrating surface ifthe surface vibration modes which are excited are resonant as is usuallythe case if the structure or surface is excited mechanically (but notacoustically). A full discussion of this concept may be found on pages504-506 of the text.

(b) Adding stiffness to a vibrating surface will only decrease its soundradiation or increase its transmission loss at frequencies below the firstresonance frequency of the surface. If the surface is excited at resonanceby a tonal excitation source, then adding stiffness will increase itsresonance frequencies and if the carefully done will result in a reductionin radiated sound.

(c) Adding mass to a vibrating surface will reduce the sound radiation andincrease the panel transmission loss at frequencies above the firstresonance frequency of the surface and below the surface criticalfrequency.

Problem 10.13

Vibration velocities measured in octave bands on a diesel engine are listed inthe following table.


Octave bandcentre

frequency(Hz)

63 125 250 500 1k

rms vibration velocity(mm/s) 5 10 5 2 0.5

rms accelerationestimate (v2πf) (m/s2)

1.98 7.85 7.85 6.28 3.14

rms displacementestimate (v/2πf) (µm)

12.6 12.7 3.18 0.64 0.08

(a) Overall rms velocity = (52 + 102 + 52 + 22 + 0.52)1/2 = 12.4 mm/s(b) Overall velocity in dB re 10-6 mm/s = 20 log10(12.4/10-6) = 142dB re 10-6

mm/s(c) Estimate of the overall acceleration level in dB re 10-6 m/s2 = 20

log10(13.29/10-6) = 142dB re 10-6m/s2

(d) Estimate of the overall displacement level in dB re 10-6µm = 20log10(18.22/10-6) = 145dB re 10-6µm

Problem 10.14

(a) I would mount the accelerometer so that its axis was normal to the beamsurface and thus parallel to the beam displacement.

(b) Longitudinal waves would cause the accelerometer to vibrate normal toits axis and the cross-axis sensitivity of the accelerometer will result ina signal due to the longitudinal wave. For a properly selected and alignedaccelerometer the effect can be very small but for any otheraccelerometer the effect could be significant depending on the relativelevels of the bending and longitudinal displacements and the actual crossaxis sensitivity of the accelerometer.

Problem 10.15

(a) Using the relation, , we obtain f0 = 11.1Hzf0 '1

2πgd


TF '1 % (2×0.05×4.49)2

(1 & 4.492 )2 % (2 × 0.05 × 4.49)2' 0.057

Mi 'j2 π × 504.905e%6

' j6.405 × 10&5 m/s/N

TF ' /000 /000&3.183 × 10&6 & 2 × 10&5

&3.183 × 10&6 & 2 × 10&5 % 6.405 × 10&5' 0.567

4πζ

1 & ζ 2' 0.5

(b) We may use equation (10.14) in the text. The value of X is3000/(60×11.1) = 4.49. Thus:

Thus the reduction in transmitted vertical force is -20 log10(0.057) =25dB

(c) We may use equation (10.31) in the text. First we must calculate theisolator mobility. The overall spring stiffness is found by settingequations (10.2) and (10.3) in the text, equal. The result is k = 1000×9.81/0.002 = 4.905MN/m. The isolator mobility is thencalculated using to give:Mi '

jωki

The mobility of the supported mass is calculated using Mm = 1/j2πfm =-j3.183 × 10-6 m/s/N.

Using equation (10.31), we obtain:

Thus the reduction in transmitted vertical force is now -20 log10(0.567)= 5dB; thus there is an increase of 20dB.

Problem 10.16

The critical damping ratio is given by:


ζ '0.25

16π2 % 0.25' 0.04

Thus:

and η = 0.02.

This loss factor is about 10 to 20 times greater than would be expected froma sheet of steel, so one might conclude that the product would be effective.One application would be for lining of parts bins.

11

Solutions to problems inactive noise control

Problem 11.1

Acoustic mechanisms associated with active noise control include:1. Suppression of the primary source by changing its input impedance with

a control source2. Reflection of energy as a result of causing an impedance mismatch at the

control source3. Absorption of energy by the control source4. Local cancellation at the expense of increased levels elsewhere

Applications:

(a) Feasible, reference sensor would be tacho signal, control source shouldbe downstream of primary source and remote from turbulence generatingparts of the duct system, and error sensor should be downstream ofcontrol source (out of source near field). Mechanism involved issuppression of primary source by changing its radiation impedance.

(b) Not feasible if the source of noise is the grille. This is because it wouldbe difficult to obtain a causal reference signal. If the source of noise isupstream of the grille, then active control may be feasible. A referencesignal could be obtained from a microphone (with a turbulence filter)placed upstream. The control source would be placed at least 1.2mdownstream of the reference sensor (depending on the controller timedelay) and the error sensor would be placed 0.5 to 1m downstream of thecontrol source, and it may even work better if placed on the room side ofthe grille. Mechanism would be reflection and absorption of primaryenergy.

(c) Not feasible if global control is needed. Possible to establish zones oflocal noise reduction.

Active noise control318

(d) Not feasible due to difficulty in obtaining a causal reference signal.

(e) Feasible, reference sensor would be tacho signal on rotating shaft relatedto noise producing machine, control sources should be in factory cornerif room is small, otherwise they should be near the locations where noisereduction is needed. Mechanism involved is suppression of primarysource by changing its radiation impedance for small room and localcancellation for large room.

(f) Feasible, reference sensor would be tacho signal from aircraft engine,control sources would be placed in cabin ceiling or in seat headrests, anderror sensors should be as close as possible to the passengers.Mechanism involved is suppression of primary source by changing itsradiation impedance, although local cancellation may dominate in somecases

(g) Not generally feasible due to complexity of radiated sound field.

(h) Feasible. Feedback system is needed. Control sources could be actuatorson the fuselage skin or loudspeakers in the passenger headrests.Microphones would be best located in passenger head rests.

(i) Feasible. Reference signal would be derived from electricity mainssignal, control sources could be shakers on the tank or loudspeakerssurrounding the transformer and very close to it. Error sensors wouldneed to surround the transformer and be located further away than thecontrol sources. Mechanism is suppression of primary noise by changingthe radiation impedance of the transformer tank.

φ 'f (k (ct ± r ))

r

12

Errata in the 3rd edition ofEngineering Noise Control

p xi, Change “Noise Reduction Index (NRI)” to “Noise ReductionCoefficient (NRC)”

p xv, change “FWHA” to “FHWA”

p xviii In line 19, change “Noise Reduction Index” to “Noise ReductionCoefficient”

p16, In line 3, change the equation to (1 /hf) E /ρ > 2

p16, line 10, change DP = 1.346E to DP = 1.099E

p16, Change Eq. (1.3) to

p18, In Eq. (1.5), change "332" to "331"

p27, Change Eq. 1.40a to

p29, 3 lines above Eq, (1.50), change "1.36" to "1.41"

p34, Change the reference just above Eq. (1.69) to "Fahy, 1995" p35, First line under Eq. (1.67), change "1.65" to "1.64"

p41, 4 lines above Section 1.10., replace "pet" with "per"

p45, 2 lines under Eq. (1.89) and in Eq. (1.90), remove the subscript, " t "from pt.

DC 'DF

1 %DF

EW

2Rt

%ρw

ρν2

Errata for third edition text book320

DND ' 2(L

)

Aeq,8h & 90) /L

Ta ' 8×2&(91.2 & 90.0) /3 ' 8/20.39 ' 6.1 hours

p51, Table 1.3, line 3, replace "U " with "u"

p51, Table 1.3, line 5, replace "Zd " with "ZA "

p51, Heading 1.12.2, replace "Z " with "Zs"

p72, line immediately below the figure, add "is the" after the word,"ordinate"

p76, Line 13, change "sound" to "sounds"

p87, 2 lines above Example 2.1, the text should read, "Figure 2.10(b) is analternative representation of Figure 2.10(a)"

p111, line 4, change "1252" to "61252".

p134, 3rd line, replace H with H )

p142, The number "3" and "0.3" should be replaced by "3.01" and "0.301"respectively in Equations (4.37) to (4.41) inclusive

p143, Replace equation 4.43 and the 2 lines preceding it with:The daily noise dose (DND), or "noise exposure", is defined as equalto 8 hours divided by the allowed exposure time, Ta with LB set equalto 90. That is:

p143, Replace the sentence following equation (4.42) with: "If the numberof hours of exposure is different to 8, then to find the actual allowedexposure time to the given noise environment, the "8" in Equation(4.42) is replaced by the actual number of hours of exposure."

p144, 3rd equation down should be:

p147, Replace Figure 4.6 with the more accurate figure below.

Errata for third edition text book 321

impact and steady state (equal energy)

5 dB / doublingsteady state

impulse180

170

160

150

140

130

120

110

100

90

0.1 1 10 10 10 10 10 10 10 10 102 3 4 5 6 7 8 92 5 5 5 5 5 5 5 5

8-hour dB(A)equivalent

B-duration x number of impulses (ms)

Pe

ak s

ound

pre

ssur

e le

vel (

dB r

e 2

0 P

a)

p147, 4 lines under Figure 4.6, change "1414" to "1474".

p149, 5th and 6th lines from the top, change "645" to "60645" in four places.

p150, 13 lines from the bottom, change Figure 4.6 to Figure 4.7.

p153, First line after the headings in Table 4.6, change "0.06" to "0.6".

p157, Fig 4.9 caption, add "MAF" = minimum audible field.

p160, On y-axis, change label from "dB re 20 mPa" to "dB re 20 µPa"

p165, First paragraph in section 4.9, replace "1995" with "1995, 1999".

p176, Line above Eq. (5.6), change "r" to "r = a"

p176, In Eq (5.6), change "r" to "a"

p177, In Eq. (5.7), change "r" to "a"

p179, 2 lines under figure 5.2, replace "(x,y)" with "O" and label the observeras O in Figure 5.2

p192, 2 lines above Eq. (5.71), add "each of which has a radius of ai"immediately after "sources"


p192, 2 lines above Eq. (5.72), change "a" to "ai"

p192, Line above Eq. (5.72), change "ka" to "kai"

p192, In Eq. (5.72), change "a" to "ai" in 5 places

p192 Eq. 5.71 and below, change Q to in 4 placesQ

p192 last line add "amplitude" immediately after "velocity"

p193 Eq 5.73 and below change Q to in 2 placesQ

p225, In Table 5.3 caption, change "Sutherland et al., 1974" to "Sutherlandand Bass, 1979"

p226, 13 lines above Eq. (5.171), change "2613" to "9613".

p226, Paragraph beginning "Note that ISO" only applies to overall A-Weighted calculations and should be deleted here. The paragraphfollowing this one should also be deleted as the meteorological effectsshould not be taken into account in two separate places - either theyshould be included in the barrier calculations or calculated separatelybut not both.

p229, Interchange the 63 Hz and 2000 Hz labels on the curves in Fig. 5.19.

p232, Eq. 5.181, change "-0.09" to "-0.9"

p236, In Eq. (5.188) change "10.3" to "10.0"

p241, Table 5.9, -3.0<<<+0.5 should be replaced with -3.0<<<-0.5

p244, ISO 9613-2 procedures for calculating ground effects and shieldingeffects are based on an assumption of downwind propagation from thesound source to the receiver. Thus the only correction term (Equation(5.193)) that is offered by ISO for meteorological effects is a term toreduce the A-weighted calculated sound pressure level for long timeaverages of several months to a year. Thus section 5.11.12.4.should bedeleted and replaced with the paragraph above.

p251, In Figure 6.1, in the centre on the right hand side replaceγ ' 1/κ with γ ' κ


p253, 2 lines above section 6.6, change "1989" to "1995".

p259, The equation numbered "6.12" should be numbered "6.11"

p264, The equation numbered "6.25" should be numbered "6.24"

p264, 2 lines below Eq. 6.20, replace S1 with 1/S1

p264, 3 lines below Eq. 6.20, replace S2 with 1/S2

p267, The first equation should be numbered "6.26"

p267, In Fig 6.3, there are two curves labelled "4". The lower curve shouldbe labelled "5"

p292, 3 lines above Eq. 7.52, change to and add "at time t=0"¢p 2k (t) ¦ ¢p 2

k (0) ¦after "mode k"

p292, 2 lines above Eq. 7.52, change to ¢p 2k (t) ¦ ¢p 2

k (0) ¦p292, In Eq. 7.52, change to ¢p 2

k (t) ¦ ¢p 2k (0) ¦

p293, 3 lines above Eq. 7.55, change pk to pk(0)

p293 6 lines from the bottom, there should be a minus sign before loge

p294, 5 lines from the bottom, change (2000) to (2001)

p294, Eq. (7.59), replace 0.16VS

with 0.16V

S 2

p295, Eq. (7.64), multiply each of the three terms in brackets by -1

p295, 2 lines beneath Eq. (7.62), add "energy" before "reflection"

p295, 2 lines above Equation (7.64), change "2001" to "2000"

p296, lines 2 and 3, change "Sx , Sx and Sx " to , "Sx , Sy and Sz "

p301, In each of the top two lines of the table, add "(m2)" after Sα

p303, Section 7.7.2, change "NRI" to "NRC" in three places and change"Noise Reduction Index" to "Noise Reduction Coefficient" in twoplaces. Also change Eq. 7.76 to:


NRC 'α250 % α500 % α1000 % α2000

4(7.76)

p303, 2 lines from bottom, change "20 mm" to "20 µm"

p304, Caption of Figure 7.6, line 1, change "porous surface" to "rigidlybacked porous material" and in the last line, change "L" to R

p310, Immediately following Equation (7.88), add the following: "Note thatfor square, clamped-edge panels, the fundamental resonance frequencyis 1.83 times that calculated using Equation (8.21). For panels withaspect ratios of 1.5, 2, 3, 6, 8 and 10 the factors are 1.89, 1.99, 2.11,2.23, 2.25 and 2.26 respectively."

p310, Equation 7.85 should be: ξc 'ffc

1/2

p311, End of second full paragraph, change "Elbert" to "Elfert"

p329, Eq. (7.122), replace T60u with 1T60u

p330, 10th line, change "2000" to "2001"

p339, 12th line from the bottom, change "1973" to "1988"

p343, 5 lines above the figure, change "ASTM E90-66T" to "ASTM E413-87"

p347, replace the line immediately above section 8.2.4 and the last word inthe line above that with "contour value at 2000 Hz is increased by 1dB." and add "Note that IIC, Rw and STC values are all reported asintegers."

p352, 3 lines under Equation (8.36), change "below" to "above".

p353, change x-axis label to f (Hz) (log scale)"

p354, 2nd and 3rd lines from the bottom, replace "8.37" with "8.38"

p355, 2nd line after Eq. 8.44, replace fc2 /2 with fc1 /2


20 % 20log10 (2500/100) & 6 ' 42.0 dB

h ' 1 &f

fc1

2 2

1 &f

fc2

2 2

D '

2h

if f < 0.9 × fc1

π fc1

8 fη1η2

fc2

fif f > 0.9 × fc1

p355, 3rd line, replace "8.37" with "8.38"

p359, In Eq. 8.50, replace 10 log10 m1 with 20 log10 m1

p360, change x-axis label to "frequency (Hz) (log scale)"

p360, on the x-axis of the figure, change "0.5 fc2" to "0.5 fc1"

p360, first line of item (b) in the caption, change to "LineBpoint support ( fc2is the critical frequency of the point supported panel)"

p360, Under "Point B", item (a), replace "30log10 fc2" with "20log10 fc1 +10log10 fc2"

p360, Under "Point B", items (b) and (c), replace "40log10 fc2" with "20log10

fc1 + 20log10 fc2"

p360, Eq (a) under "Point C", add the term, "20 log10 (fc2 / fc1)" to the RHS ofthe equation

p360, last Eqn., change f l to fR

p361, replace Eq. 8.55 with:

p363, 6 lines from the bottom of the page, change the equation to:

p363, 4 lines from the bottom of the page, change "77" to "78" and "61" to"60" in 2 places

p363, last line, change "61" to "60" and "52" to "51"


p365, Section 8.2.6.2, 5 lines down, replace the sentence beginning with"Alternatively" with the following: "This mechanism can be consideredto approximately double the loss factor of the base panels.Alternatively, the panels could be connected together with a layer ofvisco-elastic material to give a loss factor of about 0.2."

p365, Section 8.2.6.2, 9 lines down, after the words "(0.3 to 0.6 m)", add thewords, "or connected with a layer of visco-elastic material or evennailed together".

p371, In the 500 Hz column, 7th number from the bottom, replace S1" with"51"

p379, 2 lines above "Example 8.4", change "Example 8.7" to "Example 8.8"

p380, replace the example table with the following table.

Octave band centre frequency (Hz)

63 125 250 500 1000 2000 4000 8000

TL from Table 8.2 30 36 37 40 46 54 57 59 from Table 7.1 0.013 0.013 0.015 0.02 0.03 0.04 0.05 0.06αw from Table 7.1 0.01 0.01 0.01 0.01 0.02 0.02 0.02 0.03α f

Si (m) 0.463 0.463 0.525 0.68 1.05 1.36 1.67 2.04α67 67 59 45.6 29.5 22.8 18.6 15.2SE /Siα i

10log10( ) 18 18 18 17 15 14 13 12SE /Siα iNR (dB) 12 18 19 23 31 40 44 47

p381, 3rd line down, Equation (8.75) should be (8.65), 6 lines downEquation (8.76) should be (8.66) and 8 lines down, Equation (8.6),should be (8.65).

p381, 4th Eq. in section 3, "30.5/30" should be "30.5/31"

p391, At the end of the paragraph above the figure, add the followingsentences. "When paths involving the ground reflected wave on thesource side are considered, the straight line distance, d, used inEquation (8.85) is the distance between the image source and thereceiver. The same reasoning applies to paths involving groundreflections on the receiver side."

p394, 3 lines following Eq. 8.98, replace "barier" with "barrier".

p395, replace the four equations for Ab with the following in the sameorder


Ab ' 15.8 % 20log10[5.8/4.5] ' 18.0 dB; AR ' 1.3 dB; Ab % AR ' 19.3 dB

Ab ' 19.8 % 20log10[7.2 /4] ' 24.9 dB; AR ' 2.6 dB; Ab % AR ' 27.5 dB

Ab ' 19.5 % 20log10[7.5/4.5] ' 23.9 dB; AR ' 5 dB; Ab % AR ' 28.9 dB

Ab ' 12.0 % 20log10[4.5/4] ' 13.0 dB

Ab ' 19.8 % 20log10[7.2/4] ' 24.9 dB

Ab % AR ' 19.3 dB

p395, 6 lines from the bottom, replace "4.6" with "4.7"

p395, Solution, item 1, last line, change "5.18" to "5.20".

p396, replace the two equations for Ab with the following in the sameorder.

p396, Item 3, lines 2 and 3, change the numbers to 19.3 dB, 19.3 dB, 27.5

dB, 28.9 dB, 28.9 dB, 13 dB, 24.9 dB and 24.9 dB

p396, Item 3, line 4, change "5.18" to "5.20".

p396, Item 3, line 4, change "10 dB" to "12 dB"

p399, Figure 8.19, replace r with R


R)s ' R θ cosα

h )

s ' Hb & R θ sinα

α '12

(π & θ) & β

β ' cos&1 (Hb /A)

θ ' ±cos&1 [1 & (A 2 /2R 2 ) ], *R* > A /2

N ' ± 2λ

X 2S % (hb & ZS )2 1/2

% X 2R % (hb & ZR )2 1/2

% b2% Y 2

1/2& d

Octave band centre frequency (Hz)63 125 250 500 1000 2000 4000 8000

0

10

20

30

40

50

Oct

ave

ban

d in

sert

ion

loss

Figure 8.21 Typical pipe lagging insertion loss for 50 mm glass-fibre, density 70-90 kg/m3, covered with a lead / aluminium jacket ofsurface density, 6 kg/m2. The I symbols represent variations inmeasured values for three pipe diameters (75 mm, 150 mm and 360mm).

p399, Replace Eq. (8.100) with:

p400, 1st paragraph, change "Figure 8.12" to "Figure 8.14"

p401, Eq. (8.107) should be:

p404, Figure 8.21 is missing (see following figure)


Cc ' 0.232ξc R /h (8.117)

Xc ' [41.6(m /h )1/2ξc (1 & 1/ξc )&1/4 ] & [258h / Rξc )] (8.116)

Xm ' [226(m /h)1/2ξc (1 & ξ 2c ) ] & [258h / (Rξc )] (8.119)

curveno

1234

0.010.10.51

h

p405, Replace Equations (8.116), (8.117) and (8.119) with the following:

p415, lines 6 and 7 under Eq 9.16, replace with, "the end correction. In thiscase, > = 0. For a"

p417, replace the text between Eqs. (9.25) and (9.26) with:"An alternative expression for the effective length, which may giveslightly better results than Equation (9.25), for grazing flow across theholes, and which only applies for flow speeds such that

, is (Dickey and Selamet, 2001)"uτ / (ωd ) > 0.03

p429, Move Equation (9.52) up one line.

p432, Item 5, line 1, Replace "Equation (8.48)" with "Equation (9.52)"

p439, line following Equation (9.81), replace µ with fm

p444, In Table 9.2, "19" should be "-19"

p453, 454, Replace the legend in the figures with

p459, Figure 9.21, x-axis label, change "S" to "A" and in the caption add"open" immediately before "duct".

p461, In the equation in the centre of the page, change "6" to "5"

p461, 4 lines below the equation in the middle of the page, change "5.5" to"7"


p461, 8 lines below the equation in the middle of the page, change "12.5"to "13"

p462, line 1, change "1.2" to "1.0"

p462, Figure 9.23 caption, last line, change "1992" to "1987"

p464, Replace Table 9.5 with the following:

Octave band centre frequency (Hz)Ductdiameter (mm) 63 125 250 500 1000 2000

150 18(20) 13(14) 8(9) 4(5) 1(2) 0(1)200 16(18) 11(12) 6(7) 2(3) 1(1) 0(0)250 14(16) 9(11) 5(6) 2(2) 1(1) 0(0)300 13(14) 8(9) 4(5) 1(2) 0(1) 0(0)400 10(12) 6(7) 2(3) 1(1) 0(0) 0(0)510 9(10) 5(6) 2(2) 1(1) 0(0) 0(0)610 8(9) 4(5) 1(2) 0(1) 0(0) 0(0)710 7(8) 3(4) 1(1) 0(0) 0(0) 0(0)810 6(7) 2(3) 1(1) 0(0) 0(0) 0(0)910 5(6) 2(3) 1(1) 0(0) 0(0) 0(0)1220 4(5) 1(2) 0(1) 0(0) 0(0) 0(0)1830 2(3) 1(1) 0(0) 0(0) 0(0) 0(0)

p470, Figure 9.27, caption, and Eq. (9.115), replace "D" with "d"

p471, Eq. (9.116) and (9.117) and 2 lines below Fig. 9.28, replace "D"with "d"

p476, 3rd and 6th line of the first paragraph, change "1979" to "1978"

p478, line above Equation (10.14), change "1979" to "1978"

p479, Figure 10.2, replace the lowest y-axis label (currently 0) with 0.02

p483, In Equation (10.18) and 2 lines above it, replace "e" with "q" toavoid confusion with the distance, e, between spring supports.

p484, In Figure 10.6, the force should be shown as acting on mass m2, not


mass m1.

p485 In Eqs. (10.25a,b), the left hand side should be squared.

p487, line above Equation (10.31), change "1986" to "1988"

p495, Equation (10.42), remove the symbol "d" from the right hand side.

P496, Equation (10.48), replace "d" with |F|/ k1

p496, Equation (10.47), the numerator on the RHS should be 3(m2 /m1)3

p498, 8 lines from the top of the page, change "1979" to "1978"

p513, Table 11.2, 3rd line in 2000 Hz column should be "25"

p513, Table 11.2, the 8000 Hz column should be replaced with 13, 15, 18,27, 35, 35, 26, 32, 32, 34, 42 and 44 respectively and the BFI columnfor the two tubeaxial entries should be " 7 "

p513, Remove the paragraph containing Equation (11.2) and remove"(11.2)" in the second to bottom line.

p514, last Equation, label (11.2)

p515, Example 11.1 table, replace "30" with "36"

p517, Equation (11.10), change to:Lw ' 72 % 13.5log10 kW (dB re 10&12 W)

p526, last line, change "8.8" to "8.3"

p528, replace the values in the table with the following.

0 72 77 80 81 80 76 69 6360 74 79 82 83 82 78 71 65120 61 66 69 70 69 65 58 52180 55 60 63 64 63 59 52 46


p535, 4 lines above Eq.(11.33), and 2 lines after Eq. (11.34), change "534"to "60534".

p536, 4 lines from the bottom, change "534" to "60534".

p541, Following Eq. 11.64, insert the statement, "If the second term inbrackets of Equation (11.64) exceeds 0.3, it is set equal to 0.3".

p542, line 3, change "534" to "60534".

p542, Immediately before Equation (11.67), add the following: "Note thatthe final spectrum levels must all be adjusted by adding orsubtracting a constant decibel number so that when A-weighted andadded together, the result is identical to the A-weighted overall levelsform Equations (11.65) and (11.66)."

p543, 1 line and 4 lines above Eq. (11.70), change "534" to "60534".

p544, Equation 11.73, second term on the right should have the "log10"removed and "17.27" replaced with "17.37", so it reads "-17.37(...........)"

p544, Replace the last paragraph with, "The octave band external soundpressure levels may be calculated using Equations (11.73) and(11.76) with octave band sound power levels used in Equation(11.76) instead of overall sound power levels."

p552, The constant in Equation (11.89) should be "55", not "53".

p558, Replace the paragraph following Table 11.29 with the following:"The road surface or condition correction is taken as zero for eithersealed roads at speeds above 75 km/hr or gravel roads. For speedsbelow 75 km/hr on impervious sealed roads, the correction is -1 dB.For pervious road surfaces, the correction is -3.5 dB. For concreteroads with deep random grooves greater than 5 mm in width, thecorrection is, Ccond = 4 - 0.03P where P is the percentage of heavyvehicles."

p559, Replace the nine lines following Eq. 11.102 with the following:"Low barriers such as twin beam metal crash barriers can have less


SELref ' SELv % 10 log10 N % C2

effect than soft ground. So if these are used with any proportion, Pd

, of soft ground, their effect should be calculated by looking at thelower noise level (or the most negative correction) resulting from thefollowing two calculations:

$ Soft ground correction (0 < Pd < 1.0), excluding the barriercorrection; and

$ hard-ground correction (Pd = 0) plus the barrier correction.”

p560, Remove the sentence beginning 12 lines from the bottom of the page,"Note that the two values for $ must add up to 180E "

p561, In the heading and first line, change "FWHA" to "FHWA"

p561, 6 lines from the bottom, add "Menge, et al.," before "1998".

p562, 4 lines under Equation (11.108), add "Menge, et al.," before "1998".

p563, 5th line in first paragraph, and 3 lines under Equation (11.109),replace "1995" with "U.K. DOT, 1995a".

p563, 3 lines under Equation (11.111), replace "1995" with "U.K. DOT,1995a,b".

p563, p564, Replace the last two lines of page 563 and the top three lines ofpage 564 with the following:"Note that different vehicle types must be considered as separatetrains. For any specific train type consisting of N identical units,the quantity SELref is calculated by adding 10log10N to SELv. Inaddition the track correction, C2 from Table 11.32 must also beadded so that:

p564, The second entry of "Freight vehicles, tread braked, 2 axles" shouldactually be "Freight vehicles, disc braked, 4 axles"

p565, Lines 1 and 3, change SEL to SELref.


p565, table 11.32, add ,C2, after "Correction" in the column 2 label.

p567, In Equation (11.121), remove the minus sign

p568, Add equation numbers, 11.122 and 11.123 to the equations at the topof the page.

p580, 10 lines above Equation (12.1), change "1985" to "1986".

p609, line 2 in the table for fresh water, change "988" to "998".

p609, line in the table for iron, Young’s Modulus = 206, density =7,600,= 4910, E /ρ

0 = 0.0005 and < = 0.27.

p609, line in the table for Nylon, move the "6.6" next to "nylon" andYoung’s Modulus = 2, density =1,140, = 1,320.E /ρ

p609, line in table for lead, loss factor = 0.015

p609, line in table for concrete, loss factor = 0.005 - 0.02

p610, the last column of numbers is the density and the 2nd last column isYoung’s modulus.

p617, In figure captions, change "C.6" to "C.5" and "C.5" to "C.6".

p621, Change number of Eq. 1.36 to C.24.

p622, In Equation (C.29), replace ZN with ZN /Dc

p623, In Equation (C.30), replace 2 with $ in three places.

p645, Missing references.Allard, J.F. and Champoux, Y. (1989). In situ two-microphonetechnique for the measurement of acoustic surface impedance ofmaterials. Noise Control Engineering Journal, 32, 15-23.

Barron, M. (1993). Auditorium acoustics and architectural design.E&FN Spon: London.


p646, Missing references.Beranek, L. L. (ed.) (1988). Noise and Vibration Control. Revisededition. Washington D.C: Institute of Noise Control Engineering.

Beranek, L.L. (1996). Concert and Opera Halls. How They Sound.Acoustical Society of America: New York.

Berglund, B., Lindvall, T. and Schwela, D.H. (1995). CommunityNoise. Stockholm: Stockholm University and Karolinska Institute.

Berglund, B., Lindvall, T. and Schwela, D.H. Eds. (1999).Guidelines for Community Noise. Geneva: World HealthOrganization.

p647, Missing references.Bragg, S.L. (1963). Combustion noise. Journal of the Institute ofFuel, Jan., 12B16.

Broner, N. and Leventhall, H.G. (1983). A criterion for predictingthe annoyance due to lower level low frequency noise. Journal ofLow Frequency Noise and Vibration, 2, 160B168.

p648, Missing references.Cazzolato, B.S. (1999). Sensing systems for active control of soundtransmission into cavities. PhD thesis, Adelaide University, SouthAustralia.

Cazzolato, B.S. and Hansen, C.H. (1999). Structural radiation modesensing for active control of sound radiation into enclosed spaces.Journal of the Acoustical Society of America, 106, 3732B3735.

Chapkis, R.L. (1980). Impact of technical differences betweenmethods of INM and NOISEMAP. In Proceedings of Internoise '80,pp. 831B834.

Chapkis, R.L., Blankenship, G.L. and Marsh, A.H. (1981).Comparison of aircraft noise-contour prediction programs. Journalof Aircraft. 18, 926 B 933.

p649, Missing references.Davy, J.L. (1993). The sound transmission of cavity walls due tostuds. In Proceedings of Internoise '93, pp. 975B978.

Davy, J.L. (1998). Problems in the theoretical prediction of soundinsulation. In Proceedings of Internoise '98, Paper #44.


Davy, J.L. (2000). The regulation of sound insulation in Australia.In Proceedings of Acoustics 2000. Australian Acoustical SocietyConference, Western Australia, November 15-17, pp. 155-160.

Delaney, M.E., Harland, D.G., Hood, R.A. and Scholes, W.E.(1976). The prediction of noise levels L10 due to road traffic. Journalof Sound and Vibration, 48, 305-25.

p650, Missing references.Dutilleaux, G., Vigran, T.E. and Kristiansen, U.R. (2001). An in situtransfer function technique for the assessment of acoustic absorptionof materials in buildings. Applied Acoustics, 62, 555-572.

Edge, P.M. Jr. and Cawthorn, J.M. (1976). Selected methods forquantification of community exposure to aircraft noise, NASA TND-7977.

Fahy, F.J. (2001). Foundations of Engineering Acoustics. London:Academic Press.

Fahy, F.J. and Walker, J.G. (1998). Fundamentals of Noise andVibration. London: E&FN Spon.

FHWA (1995). Highway Traffic Noise Analysis and AbatementGuide. U.S. Dept. of Transportation, Federal HighwayAdministration, Washington, D.C.

Fitzroy, D. (1959). Reverberation formula which seems to be moreaccurate with nonuniform distribution of absorption. Journal of theAcoustical Society of America, 31, 893-97.

Fleming, G.G., Burstein, J., Rapoza, A.S., Senzig, D.A. and Gulding,J.M. (2000). Ground effects in FAA's integrated noise model. NoiseControl Engineering Journal, 48, 16B24.

p652, Missing references.Hidaka, T., Nishihara, N. and Beranek, L.L. (2001). Relation ofacoustical parameters with and without audiences in concert hallsand a simple method for simulating the occupied state. Journal of theAcoustical Society of America, 109, 1028B1041.

Add Wosal, E-M. to the authors of the Hodgson (2002) paper.

Howard, C.Q., Cazzolato, B.S. and Hansen, C.H. (2000). Exhauststack silencer design using finite element analysis. Noise ControlEngineering Journal, 48, 113-120.


p653, Missing referenceJean, Ph., Rondeau, J.-F. and van Maercke, D. (2001). Numericalmodels for noise prediction near airports. In Proceedings of the 8th

International Congress on Sound and Vibration, Hong Kong, 2-6July, pp. 2929B2936.

p654, the reference, "Landau, L.D. and Lifsltitz, E.W." should be "Landau,L.D. and Lifshitz, E.W."

p654, Missing references.Kuo, S.M. and Morgan, D.R. (1996). Active noise control systems.New York: John Wiley.

Kurze, U.J. and Anderson, G.S. (1971). Sound attenuation bybarriers. Applied Acoustics, 4, 35B53.

Larson, K.M.S. (1994). The present and future of aircraft noisemodels: a user's perspective. In Proceedings of Noise-Con '94, pp969B 974.

Lee, J-W., Hansen, C.H., Cazzolato, B. and Li, X. (2001). Activevibration control to reduce the low frequency vibration transmissionthrough an existing passive isolation system. In Proceedings of the8th International Congress on Sound and Vibration, Hong Kong, 2-6July.

Li, K.M. (1993). On the validity of the heuristic rayBtraceBbasedmodification to the WeylBVan der Pol formula. Journal of theAcoustical Society of America, 93, 1727B1735.

Li, K.M. (1994). A high frequency approximation of soundpropagation in a stratified moving atmosphere above a porousground surface. Journal of the Acoustical Society of America, 95,1840B1852.

Li, K.M., Taherzadeh, S. and Attenborough, K. (1998). An improvedrayBtracing algorithm for predicting sound propagation outdoors.Journal of the Acoustical Society of America, 104, 2077B2083.

p655, Missing references.Maidanik, G. (1962). Response of ribbed panels to reverberantacoustic fields. Journal of the Acoustical Society of America, 34,809B826.


Menge, C.W., Rossano, C.F., Anderson, G.S. and Bajdek, C.J.(1998). FHWA Traffic Noise Model, Version 1.0, Technical Manual.U.S. Dept. Transportation, Washington, D.C.

p656, Missing references.Neubauer, R.O. (2000). Estimation of reverberation times in non-rectangular rooms with non-uniformly distributed absorption usinga modified Fitzroy equation. 7th International Congress on Soundand Vibration, Garmisch-Partenkirchen, Germany, July, pp.1709B1716.

Neubauer, R.O. (2001). Existing reverberation time formulae - acomparison with computer simulated reverberation times. InProceedings of the 8th International Congress on Sound andVibration, Hong Kong, July, 805-812.

Nilsson, A. (2001). Wave propagation and sound transmission insandwich composite plates. In Proceedings of the EighthInternational Congress on Sound and Vibration, Hong Kong, July,pp. 61B70.

p657, Missing references.Parkins, J.W. (1998). Active minimization of energy density in athreeBdimensional enclosure. PhD thesis, Pennsylvania StateUniversity, USA.

Passchier-Vermeer, W. (1968). Hearing Loss Due to Exposure toSteady State Broadband Noise. Report No. 36. Institute for PublicHealth Eng., The Netherlands.

Passchier-Vermeer, W. (1977). Hearing Levels of Non-NoiseExposed Subjects and of Subjects Exposed to Constant Noise DuringWorking Hours. Report B367, Research Institute for EnvironmentalHygiene, The Netherlands.

Plovsing, B. (1999). Outdoor sound propagation over complexground. In Proceedings of the Sixth International Congress onSound and Vibration, Copenhagen, Denmark, 685B694.

p658, Missing references.Price, A.J. and Crocker, M.J. (1969). Sound transmission throughdouble panels using Statistical energy analysis. Journal of theAcoustical Society of America, 47, 154B158.


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Raspet, R., L'Esperance, A. and Daigle, G.A. (1995). The effect ofrealistic ground impedance on the accuracy of ray tracing. Journalof the Acoustical Society of America, 97, 683B693.

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Saunders, R.E., Samuels, S.E., Leach, R. and Hall, A. (1983). AnEvaluation of the U.K. DoE Traffic Noise Prediction Method.Research Report ARR No. 122. Australian Road Research Board,Vermont South, VIC., Australia.

Sendra, J.J. (1999). Computational Acoustics in Architecture.Southampton: WIT Press.

p659, In the Shepherd reference, change "1985" to "1986"

p660, Missing references.Soom, A. and Lee, M. (1983). Optimal design of linear and nonlinearvibration absorbers for damped systems. Journal of Vibration,Acoustics, Stress and Reliability in Design, 105, 112B119.

Steele, C. (2001). A critical review of some traffic noise predictionmodels. Applied Acoustics, 62, 271-287.

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Takagi, K. and Yamamoto, K. Calculation methods for road trafficnoise propagation proposed by ASJ. In Proceedings of Internoise‘94. Yokohama, Japan, pp.289B294.

p660, Tse reference, change "1979" to "1978".


p661, Missing references.U.K. DOT (1988). Calculation of Road Traffic Noise. Departmentof Transport. London: HMSO.

U.K. DOT (1995a). Calculation of Railway Noise. Department ofTransport. London: HMSO.

U.K. DOT (1995b). Calculation of Railway Noise. Supplement 1.Department of Transport. London: HMSO.

Watters, B.G., Labate, S. and Beranek, L.L. (1955). Acousticalbehavior of some engine test cell structures. Journal of theAcoustical Society of America, 27, 449B456.

Wiener, F.M. and Keast, D.D. (1959). Experimental study of thepropagation of sound over ground. Journal of the Acoustical Societyof America, 31, 724.

Yoshioka, H. (2000). Evaluation and prediction of airport noise inJapan. Journal of the Acoustical Society of Japan (E), 21, 341B344.

Zaporozhets, O.I. and Tokarev, V.I. (1998). Aircraft noise modellingfor environmental assessment around airports. Applied Acoustics, 55,99B127.

p661, In the Zinoviev reference, replace "In print" with " 269, 535-548."

p663, after last line, add, "ANSI S3.6 B 1997. Specification forAudiometers."

p667, line 1, replace "E90-99" with "E90-02".

p715, Change "Noise Reduction Index" to "Noise Reduction Coefficient".

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