Dalgalar II

7/31/2019 Dalgalar II

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University of Cape Town

Department of PhysicsPHY2014F

Vibrations and Waves

Andy BufflerDepartment of Physics

University of Cape [email protected]

Part 2

Coupled oscillators

Normal modes of continuous systemsThe wave equation

Fourier analysis

covering (more or less)

French

Chapters 2, 5 & 6


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Problem-solving and homework

Each week you will be given a take-home problem set to

complete and hand in for marks ...

In addition to this, you need to work through the followingproblems inFrench, in you own time, at home. You will not

be

asked to hand these in for marks. Get help from you friends, the

course tutor, lecturer, ... Do not take shortcuts.Mastering these problems is a fundamental aspect

of this course.

The problems associated with Part 2

are:

2-2, 2-3, 2-4, 2-5, 2-6, 5-2, 5-8, 5-9, 6-1, 6-2, 6-6, 6-7, 6-10,

6-11, 6-14

You might find these tougher: 5-4, 5-5, 5-6, 5-7


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The superposition of periodic motions

Two superimposed vibrations of equal frequency

1 1 0 1cos( )x A t = +

2 2 0 2cos( )x A t = +

combination can be written as

0cos( )x A t = +

Using complex numbers:0 1( )

1 1

j tz A e

+=0 2( )

2 2

j tz A e

+=1 2z z z= +

{ }0 1 2 1( ) ( )1 2j t jz e A A e + = +

A A2

A1

0 1t +

2 1

2 1 =

Phase differenceThen

( )2 2 1sin sinA A = and

2 2 2

1 2 1 2 2 12 cos( )A A A A A = + +

French

page 20


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If we add two sinusoids of slightly different frequency and we observe beats

x1x2

x1

+x

2

t

t

Superposed vibrations of slightly different frequency: Beats

2cos t

1 2

cos cost t +1 2cos

2

t

1 2

2beatT

=

1cos t

1 2 1 21 2cos cos 2cos cos

2 2t t t t

+ + =

1

2

French

page 22


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Combination of two vibrations at right angles

1 1 1cos( )x A t = +2 2 2cos( )y A t = +

???

Write and1 0cos( )x A t=

Consider case where frequencies are equal and let initialphase difference be

2 0cos( )y A t = +

Case 1 : 0 = 1 0cos( )x A t=

2 0cos( )y A t=

2

1

Ay x

A= Rectilinear motion

2 = 1 0cos( )x A t=

2 0 2 0cos( 2) sin( )y A t A t = + =

Case 2 :

2 2

2 2

1 2

1x yA A

+ = Elliptical path inclockwise direction

French

page 29


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Case 3 : = 1 0cos( )x A t= 2

1

Ay xA

=

Combination of two vibrations at right angles 2

2 0 2 0cos( ) cos( )y A t A t = + =

3 2 = 1 0cos( )x A t=2 0 2 0cos( 3 2) sin( )y A t A t = + = +

Case 4 :

2 2

2 2

1 21

x y

A A + = Elliptical path inanticlockwise direction

4 = 1 0cos( )x A t=

2 0cos( 4)y A t = +

Case 5 :

Harder to see use a graphical approach


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Superposition of simple

harmonic vibrations at

right angles with an initialphase difference of 4


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Superposition of two perpendicular simple

harmonic motions of the same frequency forvarious initial phase differences.


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Abbreviated construction for the superposition of

vibrations at right angles

see French page 34.


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Perpendicular motions with different frequencies: Lissajous figures

See French page 35.

Lissajous figures for

with various

initial phase differences.

2 12 =

= 0 4 3 42


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2 1:

1:1

1:2

2:3

1:3

3:4

5:6

4:5

3:5

= 0 4 3 42

Lissajous figures


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Coupled oscillators

When we observe two weakly coupled identical oscillators

A and B, we see:

t

t

xA

xB

these functions arise mathematically from the addition of twoSHMs

of similar frequencies

so what are

these two SHMs?

These two modes are known as normal modes which are states

of the system in which all parts of the system oscillate with SHM

either in phase or in antiphase.

French

page 121


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t

t

A B

xA

xB

Coupled oscillators


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The double mass-spring oscillator

m

Axk

m

Bx k

Individually we know that andA Amx kx= B Bmx kx=

For both oscillators:0

k

m

=

Now add a weak coupling force:

m

Axk

m

Bx kck

For mass A: ( )A A c B Amx kx k x x= +

or where ,2 20 ( )A A B Ax x x x= + 2

0k

m = 2 ck

m =


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2 2

0 ( )A A B Ax x x x= +

The double mass-spring oscillator 2

For mass A:

For mass B:2 2

0( )

B B B Ax x x x=

two coupled differential equations how to solve ?Adding them:

22

02( ) ( )

A B A B

dx x x x

dt+ = +

Subtract B from A:2

2 2

02( ) ( ) 2 ( )

A B A B A Bd x x x x x xdt

=

Define two new variables:1 A Bq x x= +

2 A Bq x x=

Then and2

21 0 12

d qqdt =

2

2 22 0 22 ( 2 )d q

qdt = +

called normalcoordinates


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The two equations are now decoupled

221

12 s

d qq

dt

=

Write 222

22 f

d qq

dt=

2 2

0s =

2 2 2

02

f = +

s

= slow

f

= fast

which have the solutions:

1 1cos( )

sq C t = +

2 2cos( )fq D t = +

Since and 2 A Bq x x=

We can write and

1 A Bq x x= +

( )

1

1 22A

x q q= +( )

1

1 22B

x q q=


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1 2

cos( ) cos( )2 2A s f

C Dx t t = + + +

Then1 2cos( ) cos( )

2 2B s f

C Dx t t = + +

SoxA

andxB

have been expressed as the sum and difference of

two SHMs

as expected from observation.

C,D, 1

and 2

may be determined from the initial conditions.

whenxA

=xB

,then q2

= 0

there is no contribution from the

fast mode and the two masses move in phase

the coupling spring

does not change length and has no effect on the motion 0s =

whenxA

= xB

,then q1

= 0

there is no contribution from the

slow mode

the coupling spring gives an extra force

each mass

experiences a force giving( )2c

k k x +

2 2

02= +

2 2 cf

k k

m

+

=


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symmetric mode antisymmetric

mode mixed mode


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We now have a system with two natural frequencies, and

experimentally find two resonances.

Frequency

Amplitude

F h


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Pitch and bounce oscillator

d

Axk

m

Bx

L

k

Two normal modes (by inspection):

Bouncing

A Bx x=

Restoring force = 2kx2

2 2

d x

m kxdt = Pitching

A B

x x= Ax

Bx Centre of mass stationary

2

bounce

2k

m =

I =

21 12 12

( )kd d mL = 2

26kdmL

=2

2

pitch 26k dm L

=

French

page 127


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N= 2

( )1 0 01

2 sin2 2 1

= =

+

( )2 0 022 sin 3

2 2 1 = =+


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N

= 4

N

= 3

French


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N-coupled oscillators

fixed fixedTension T

l

1 2 3 p1 p p+1 N

consider transverse displacements that are small.

Each bead has mass m

1 2 3 p1 p p+1 N

1p

p

Transverse force on pth particle:1sin sinp p pF T T = +

1 1p p p py y y yT T

l l

+ = +

for small

y

French

page 136


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N-coupled oscillators 2

2

1 12p p p p p

p d y y y y yF m T Tdt l l

+

= = +

( )

2

2 2

0 0 1 12 2 0p

p p p

d y

y y ydt + + =

where ,20

T

ml = p

= 1, 2

N

a set ofN

coupled differential equations.

Normal mode solutions: sinp py A t=

Substitute to obtainN

simultaneous equations

( ) ( )2 2 20 0 1 12 0p p pA A A + + =

or2 21 1 0

2

0

2p p

p

A AA

+ +=

i 3


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From observation of physical systems we expect sinusoidalshape functions

of the form sinpA C p=

Substitute into

2 2

1 1 02

0

2p p

p

A A

A

+

+=

And apply boundary conditions and0 0A = 1 0NA + =

and

n

= 1, 2, 3,

N

(modes)

( )02 sin 2 1nn

N

= +

find that1

n

N

=

+

There areN modes: sin sin sin1

pn pn n n n

pny A t C t

N

= =

+

N l d ill t 4


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02

n

0 1 2 3 N+1 n

( )02 sin 2 1nn

N

= +


For smallN:

N l d ill t 5


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Forn


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N

coupled oscillators haveN

normal modes and henceN

resonances

response



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Continuous systems

Continuous systems


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Continuous systems

Consider a string stretched between two rigid supports

x = 0 x =L

tension T

String has mass m

and mass per unit length m L=

Suppose that the string is disturbed in some way:

y

x

The displacementy

is a function ofx

and t : ( , )y x t

TNormal modes of a stretched string


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T

T

+

y

x xx x+

Consider the forces on a

small length of string

Restrict to small amplitude disturbances then is small andcos 1 = sin tan yx

= = =

The tension T

is uniform throughout the string.

Net horizontal force is zero: cos( ) cos 0T T + =

Vertical force: sin( ) sinF T T = +

Thenx x x

y yF T Tx x

+ =

g

French

page 162

Normal modes of a stretched string 2


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x x x

y yF T T

x x+

=

Use( ) ( )dg g x x g x

dx x

+ =

Then2

2

yF T x

x=

or ( )

2 2

2 2

y y

x T xt x

=

giving

2 2

2 2

y y

x T t

=

Check: has the

dimensions

T21 v

Then is the speed at which awave propagates along

the string see later

v T =

Write2 2

2 2 2

1y y

x v t

=

One dimensional wave equation


: mass per unit length



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Look the standing wave (normal mode) solutions

Normal mode: all parts of the system move in SHM at the

same frequency

Write: ( , ) ( )cosy x t f x t=( )f x is the shape

function

substitute into wave equation

2 2

2 2( , ) ( ) cosy x t d f x tx dx

=

( )

22

2

( , )

( ) cos

y x t

f x tt

=

22

2 2

( ) 1cos ( )cos

d f xt f x t

dx v =

which must be true for all t

then

2 2

2 2 ( )d f

f xdx v

=

No a odes o a st etc ed st g 3



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2 2

2 2 ( )

d f

f xdx v

=

which has the same form as the eq. of SHM:2

2

02

d xx

dt=

has general solution:0sin( )x A t = +

Thus we must have: ( ) sinf x A x

v

= +

Apply boundary conditions: y

= 0 at x = 0 and x

=L

(0) 0f = and ( ) 0f L =

x

= 0,f

=0 : 0 sin 0A

v

= +

i.e.

0 sin Lv

=

0 =

x = L,f =0 : L nv

=i.e. n = 1,2,3,

g



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n

n v

L

=Write n = 1,2,3,( ) sin sinn n

x n v n xf x A A

v L L

= =

Therefore

shape function, or eigenfunction

x

= 0

x

=L

n = 1

n = 3n

= 2

n = 5n

= 4

( )1( ) sinf x A x L=( )2( ) sin 2f x A x L=

( )3( ) sin 3f x A x L=

( )4( ) sin 4f x A x L=

( )5( ) sin 5f x A x L=

1

v L =

3 3 v L =

2 2 v L =

5 5 v L =

4 4 v L =

g


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Normal modes of a stretched string

n

= 1

n

= 2

n

= 3

n = 4


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Full solution for our standing waves:

( , ) sin cosn nn x

y x t A tL

=

n

n v

L

=

The mode frequencies are evenly spaced:1n n =

n

0 1 2 3 n

3

2

1

(recall the beaded string)

This continuum approach breaks

down as the wavelength approaches

atomic dimensions

also if there is

any stiffness in the spring whichadds an additional restoring forcewhich is more pronounced in the

high frequency modes.



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All motions of the system can be made up from thesuperposition of normal modes

1

( , ) sin cos( )n n nn

n xy x t A t

L

=

= +

with n

n v

L

=

Note that the phase angle is back since the modes may

not be in phase with each other.

Whispering galleries


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Whispering galleries

best example is the inside

dome of St. Pauls cathedral.

If you whisper just inside the

dome, then an observer close to

you can hear the whisper

coming from the oppositedirection

it has travelled

right round the inside of the

dome.

Longitudinal vibrations of a rodFrench

page 170


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x x x+

x x + + + x +

1F 2F

section of massive rod

section is displaced and stretched

by an unbalanced force

Average stress =x

Average strain = Y

x

Y: Youngs modulus

stress at = (stress atx) +x x+ (stress)

xx

page 170

Longitudinal vibrations of a rod 2


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If the cross sectional area of the rod is

1F Yx

=

2

2 2F Y Y x

x x

= +

and

2

1 2 2F F Y x ma

x

= =

2 2

2 2Y x xx t

=

or2 2

2 2

x Y t

=

2 2

2 2 2

1

x v t

=

Yv =

Longitudinal vibrations of a rod 3


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Apply boundary conditions: one end fixed and the other free

x =L :x

= 0 : i.e. 0 =

n

= 1,2,3,

( , ) ( )cosx t f x t =Look for solutions of the type:

(0, ) 0t =

0F Yx

= =

( ) sinf x A xv

= +

where

then cos 0L

v

=

or ( )12L nv

=

( ) ( )1 12 2n

n v n Y

L L

= =The natural angular frequencies


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x = 0 x =Ln

= 1

n = 3

n

= 2

n

= 5

n

= 4

12

Y

L

=

232

YL

=

3

5

2

Y

L

=

Normal modes for different boundary conditions


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45n

= 1 n

= 2 n

= 3

Simplysupported

Clampedone end

Free bothends

Clampedboth ends

The elasticity of a gasl

French

page 176


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A , p Bulk modulus:dp

K VdV=

Kinetic theory of gases: Pressure

2 21

rms rms3 3

m

p v vAl= =If then

21rms2k

E mv=2

3

kEpA l

=

Now move piston so as to compress the gas work done on

gas:

kW pA l E = =

Then ( )2

2 2 2 5( )

3 3 3 3

kk

E l l lp E pA l p p

A l A l A l l

= = =

adiabatic

5

3

pK V p

V

= =

giving

and1.667K p

v = =

Sound waves in pipesFrenchpage 174


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A sound wave consists of a series of compressions and

rarefactions of the supporting medium (gas, liquid, solid)

In this wave individual molecules move longitudinally with

SHM. Thus a pressure maximum represents regions in which themolecules have approached from both sides, receding from the

pressure minima.

wave propagation

p g

Longitudinal wave on a spring


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Standing sound waves in pipes


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Pressurep

p0

Flow

velocity u

0

x

x

x

x

0 :t=

2 :t T=p

u

Standing sound waves in pipes 2


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Consider a sound wave in a pipe. At the closed end the flow

velocity is zero (velocity node, pressure antinode).At the open end the gas is in contact with the atmosphere,

i.e.p

=p0

(pressure node and velocity antinode).

p

u

p0

0

pressure node

velocity antinode

pressure antinode

velocity node

Open end Closed end

Standing sound waves in pipes 3


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Pipe closed at

both ends

Pipe open at

both ends

Pipe open at

one end

n n vL =

2 2

n nvL

f

= =

2

nvf

L=

( ) ( )2 1 2 1

4 4

n n vL

f

= =

( )2 1

4

n vf

L

=

( )2 12

nn v

L =

Sound


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Audible sound is usually a longitudinal compression wave in air towhich the eardrum responds.

Velocity of sound (at NTP) ~ 330 m s-1

By considering the transport of energy by a compression wave,

can show that 2 2 22 mP f Avs =

whereA

is cross sectional area of air column andsm

is

maximum displacement of air particle in longitudinal wave

Then intensity = 2 2 22 mP

f vsA

= unit: W m-2

Sound 2


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The human ear detects sound from ~10-12

W m-2

to ~1 W m-2

use a logarithmic scale forI :

10

0

10logI

I

=

decibels

whereI0

= reference intensity

= 10-12

W m-2

Intensity level or loudness:

Musical sounds


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Waveforms from real musical instruments are complex, and maycontain multiple harmonics, different phases, vibrato, ...

Pitch is the characteristic of a sound which allows sounds to be

ordered on a scale from high to low (!?). For a pure tone, pitch

is

determined mainly by the frequency, although sound level may

also change the pitch. Pitch is a subjective sensation and is a

subject in psychoacoustics.

The basic unit in most musical scales is the octave. Notes judged

an octave apart have frequencies nearly (not exactly) in the ratio2:1. Western music normally divides the octave into 12 intervals

called semitones ... which are given note names (A through G withsharps and flats) and designated on musical scales.

Musical sounds ...2


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Timbre is used to denote tone quality or tone colour of asound and may be understood as that attribute of auditorysensation whereby a listener can judge that two sounds are

dissimilar using any criteria other than pitch, loudness or

duration. Timbre depends primarily on the spectrum of the

stimulus, but also on the waveform, sound pressure and

temporal characteristics of the stimulus.

One subjective rating scale for timbre (von Bismarck, 1974)

dull

compact

full

colourful

sharp

scattered

empty

colourless

Two dimensional systems Frenchpage 181


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the membrane has mass per unit area ,

and a surface tension S

which gives a force Sl

perpendicular to a length l in the surface

y

x

y

x

The forces on the shaded

portion are

Sx

Sx

SySy

Consider an elastic

membrane clamped

at its edges

Two dimensional systems 2


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If the membrane is displaced

from thez

= 0 plane then a

cross section through the

shaded area shows:

+

z

x xx x+

Sy

Sy

looks exactly like the case of the stretched string.2

2zS y xx The transverse force on the element will be

And if we looked at a cross section perpendicular to this

the transverse force will be2

2

zS x y

y



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The mass of the element is .x y

Thus2 2 2

2 2 2

z z zS y x S x y x y

x y t

+ =

or2 2 2

2 2 2

z z z

x y S t

+ =

a two dimensional wave equation

with the wave velocity Sv

=

L k f l d l i f h f



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( , , ) ( ) ( )cosz x y t f x g y t=2 2

2 2( )cos

z d fg y t

x dx

=

( )2

2

2( ) ( ) cos

zf x g y t

t =

Look for normal mode solutions of the form:

2 2

2 2( )cos

z d gf x t

y dy

=

2 2

2 2( )cos ( )cosd f d g

g y t f x tdx dy + =2

2( ) ( )cosf x g y t

v

2 2 2

2 2 2

1 1d f d g

f dx g dy v

+ = i.e.

In a similar fashion to the 1D case, find

1

1( ) sinnx

n xf x A

L

=

and2

2( ) sinny

n yg y B

L

=



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then1 2 1 2

1 2,( , , ) sin sin cosn n n n

x y

n x n yz x y t C tL L

=

1 2

22

1 2,n n

x y

n v n v

L L

= +

where the normal mode frequencies are

e.g. for a membrane having sides 1.05L and 0.95L

then1 2

2 21 2

,1.05 0.95

n n

n nv

L

= +

Normal modes of a rectangular membrane


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up

down1,1

2,1 2,2

3,1 3,2

Normal modes of a circular membrane


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1,0 2,0 3,0

1,1 2,1 2,2


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Modes of vibration of a 38 cm cymbal. The first 6 modes

resemble those of a flat plate ... but after that the resonances

tend to be combinations of two or more modes.

Normal modes of a circular drum


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Chladni plates


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Soap films


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Holographic interferograms

of the top and bottom plates

of a violin at several resonances.


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of a classical guitar top

plate at several resonances.


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showing the vibrations of a 0.3 mm

thick trombone driven acoustically at 240 and 630 Hz.


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Time-average hologram interferograms

of inextensional

modes in a C5

handbell

Normal modes of a square membrane

v2n area per point =

2v


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one point per

normal mode

1 2

2 2

1 2,

2 2n n

n v n vf

L L

= +

12

v

L

22

v

L

32

v

L

4 2

v

L

0

12

v

L 2 2

v

L 3 2

v

L 4 2

v

L 5 2

v

L00 1 2 3 4 5

0

1

2

3

4

4,3f

1n

Normal modes having the same frequency are said to be degenerate

2L

Normal modes of a square membrane for large n1 and n2


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1

n

2ndf

f

area =

2

2

v

L

area per mode =

14(2 )f df

Number of modes with

frequencies betweenf

and (f+ df) =

2

14

2(2 )

Lf df

v

2

2

2 L f df

v

=

Three dimensional systems

Consider some quantity which depends on x y z and t

French

page 188


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Consider some quantity

which depends onx

, y , z

and t

,

e.g. the density of air in a room.

In three dimensions:2 2 2 2

2 2 2 2 2

1

x y z v t

+ + =

which can be written:2

2

2 2

1

v t

=

The solutions for a rectangular enclosure:

1 2 3

22 2

31 2, ,n n n

x y z

n vn v n vL L L

= + +

and for a cube: 1 2 32 2 2

, , 1 2 3n n n

v

n n nL

= + +

How many modes are there with frequencies in the range

Three dimensional systems 2


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f and (f+ df) ?Set up an imaginary cubic lattice with spacing 2v L2n

1n

3n

df

f

and consider positive frequencies only.

Volume of shell =21

8(4 )f df

3

2vL

Volume per mode =

Number of modes with

frequencies betweenf

and (f+ df) =

3

218

2(4 )

Lf df

v

3 2

34 L f dfv=

Number of modes with 24 V f df =

Three dimensional systems 3


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frequencies betweenf and (f+ df) 3v=

holds for any volume V

provided its dimensions are much

greater than the wavelengths involved.

need to multiply by a factor of 2 when dealing withelectromagnetic radiation (2 polarization states)

Ultraviolet catastrophe for blackbody radiation

Equipartition

theorem: in thermal equilibrium each mode has an

average energy in each of its two energy stores

Hence, energy density of radiation in a cavity:

12 Bk T

( )2

123

42 2

V f df df kT

c

=

or

2

3

8 fkTc

= f

experiment

!?

Planck was able to show, effectively by

assuming that energy was emitted an

absorbed in quanta of energy hf that the


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absorbed in quanta of energy hf

, that the

average energy of a cavity mode was not kT

but1hf kT

hf

e

where Plancks constant h

= 6.67

10-34

J K-1

Then

2

3

8

1hf kTf df hf

df c e

=

which agrees extremely well with experiment.

energy

density

no. of

modes inrangef

tof

+df

average

energy permode

Plancks law


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Introduction to Fourier methods

We return to our claim that any physically observed shape

French

page 189


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y p y y p

function of a stretched string can be made up from normal

mode shape functions.

x

( )f x

i.e.1

( ) sinnn

n xf x B

L

=

=

a surprising claim

?

first find

n

multiply both sides by

and integrate over the range x

= 0 to x = L

1

sin

n x

L

1 1

10 0( )sin sin sin

L L

n

n

n x n x n xf x dx B dxL L L

=

=

1 1( )sin sin sin

L L

n

n x n x n xf x dx B dx

=

Fourier methods 2


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10 0 nL L L=

If the functions are well behaved, then we can re-order things:

1 1

10 0

( )sin sin sin

L L

n

n

n x n x n xf x dx B dxL L L

= =

[n1

is a particular integer and n

can have any value between 1 and .]

Integral on rhs:

( ) ( )1 110 0

1sin sin cos cos2

L L

n n x n n xn x n x dx dxL L L L

+ =

Fourier methods 3

Both (n1 + n) and (n1 n) are


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integers, so the functions

on the intervalx

= 0 toL

must look like from which it is evident that

( )1cos

n n x

L

( )1

0

cos 0

L n n xdx

L

=

Except for the special case when n1

and n are equal then ( )1

cos 1n n x

L

=

and

( )1

0

cos

L n n xdx L

L

=

Fourier methods 4

Thus all the terms in the summation are zero, except for the


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single case when n1

= n i.e.

1 2n

LB=

( ) ( )1

1 11

0 0

1( )sin cos cos

2

L L

n

n n x n n xn xf x dx B dx

L L L

+ =

1

1

0

2 ( )sin

L

n n xB f x dxL L

=

i.e.

We have found the value of the coefficient for some

particular value ofn1

the same recipe must work for any

value, so we can write:

0

2( )sin

L

n

n xB f x dx

L L

=

Fourier methods 5


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The important property we have used is that the functions

and

are orthogonal over the intervalx

= 0 tox

=L.

i.e.

1sinn x

L

sinn x

L

1

0

sin sin

Ln x n x

dxL L

=

10 if n n

1if2

L n n=

Read French pages 195-6

Fourier methods 6

The most general case (where there can be nodal or


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g (

antinodal

boundary conditions atx

= 0 andx

=L) is

0

1( ) cos sin2

n n

n

A n x n xf x A BL L

=

= + +

where

0

2

( )sin

L

n

n x

B f x dxL L

=

0

2

( )cos

L

n

n x

A f x dxL L

=

Fourier methods 7

One of the most commonly encountered uses of Fourier methods is

the representation of periodic functions of time in terms of sine


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and cosine functions

Put2

T

=

This is the lowest frequency in

clearly there are

higher frequencies

by the same method as before, write

( )f t

0

1

2 2( ) cos sin

2n n

n

A nt nt f t A B

T T

=

= + +

0

1cos sin2

n n

n

AA n t B n t

== + +

where and ( )0

2( )sin

T

n f t n t dt

T

= ( )0

2( )cos

T

nA f t n t dt

T

=

T( )f t

t

Waveforms of ...


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a flute

a clarinet

an oboe

a saxophone

Fast Fourier transform experiments, 10 March 2008


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Odd functions

An odd periodic function ( ) ( )f t f t =

h


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where 2 2T t T

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Dalgalar II