Suggested+Solution++Test+FyBNVC09+Ch11 12+Waves Sound

8/3/2019 Suggested+Solution++Test+FyBNVC09+Ch11 12+Waves Sound

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Suggested Solutions FyBNVC09 Ch11-12 Mechanical wavesSound NV-College

[email protected] for Sale. Free to use for educational purposes. 1/20

Physics B

FyBNVCO09 Ch11, and 12 Mechanical Waves and Sound

Instructions:

Time: 140 minutes: 8:10-10:30

The Test At the end of each problem a maximum point which one may get for a correct solution

of the problem is given. (2/3/) means 2 G points, 3 VG points and an MVG quality.

Tools Approved formula sheets, ruler, and graphic calculator. You may use one page of a

personalized formula sheet which has your name on it. This should be submitted along

with the test.

Grade limits: There are 13 problem, 3 of which are MVG type. The test gives a maximum of 35

points of which 16 are VG points.

Lower limits for examination grade

Pass (G): 12 points

Pass with distinction (VG): 23 points of which at least 6 VG-points.

Pass with special distinction (MVG): 28 points of which at least 12 VG-pointsand most of MVG-qualities.

Problems number 10, 12, and 13 are heavily graded and are of greatest importance for

both VG and MVG. You may choose to solve these problems before solving the others.Waves Sting Pipe Pipe Bottle Standing W Earthquak Pitch Doppler Intensity Level

Problem 1 2 3 4 5 6 7 8 9

G 2 2 2 2 2 1 1 1 1

VG 1 1 1 1 1 2

G

VG

Waves Interference Doppler Organ Pipe Standing W Mechanical Waves

Problem 10 11 12 13 35 Total min G min VG min MVG

G 2 2 3 19 Total 12 23 28

VG 2 2 4 3 16 VG points 6 12

MVG M235 M12345 M12345 M12345 M12345G

VG

MVG

Improvement is required in deeper understanding , and/or a proper treatment of

Waves: Velocity of propagation, frequency, wave length, amplitudeWaves on strings, tension, length, mass per unit length Intensity, Intensity level Interference. Standing waves.

Standing waves on strings, double-end-fixed. Fundamental frequency, Overtones Standing waves on tubes, double-end-open. Fundamental frequency, Overtones Standing waves on tubes, one-end-closed. Fundamental frequency, Overtones Doppler Effect. Units Significant figures

Comments:


2/20



In the multi-choice problems below just circle the correct answer (s) in the test paper.

1. If a violin string vibrates at Hz588 as its second harmonic, what are the frequencies

of first three harmonics? [1/0]

a. Hzf 5881 , Hzf 17612 , Hzf 47013 ,

b. Hzf 2941 , Hzf 5882 , Hzf 8823 ,

c. Hzf 2941 , Hzf 8823 , Hzf 47015 ,

d. Hzf 1471 , Hzf 2942 , Hzf 5883 ,

e. None above, they are:

Why? Show a sample of your calculations. [1/0]

Answer:

Alternative ______

[1/0]Suggested solution: Answer: Alternative B

Hzf 5882 ; ...,5,4,3,2,12620 nHznfnfn

Hzf 2942

5881 First harmonic or Fundamental frequency

Hzf 5882 Second harmonic or First overtone frequency

Hzf 88232943 Third harmonic or second overtone frequency

2. What is the length of an organ pipe whose

fundamental frequency is Hz440 at C20 . How

long is the pipe, if it is open at both ends? At

C20 , the speed of sound in air is sm/343 .

a. cm5.19 ,

b. cm9.38 ,

c. cm0.78 ,

d. None above, it is: _________________

Why? Show your calculations. [1/0]

Answer:

Alternative__________ [1/0]

Suggested solution: Answer: Alternative b[1/0]

mf

VL

L

Vf

Vf 389.0

4402

343

22 00

1

0

[1/0] Answer: cmL 9.38


3/20



3. If the air temperature in a room is increased, the fundamental frequency of the organpipes

a) will be increased.

b) will not be effected.

c) will be decreased.

d) will be equal to the frequency of the second harmonics.

Why? Explain and show a sample of your calculations. [1/0]

Answer:

Alternative_______ [1/0]

Suggested answer: Answer: Alternative a:According to smTv /60.0331 , velocity of sound is an increasing linear

function of the temperature, and frequency of the wave is alsoproportional to the velocity of it in the medium. Since the change in thelength of the pipe is minimal in the temperature range of interest, thefundamental wavelength of the sound is independent of the temperature.Therefore, the frequency of the sound increases (higher pitch) as afunction of the temperature:

HzL

Tvf

4

60.0331

0

0

if it is open in one end.

HzL

Tvf

2

60.0331

0

0

if it is open in both-ends.

Note that, the incremental increase in the length of the pipe due to itsthermal expansion is negligible in the room temperature range.


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4. What resonant frequency would you expect from blowing an empty soda

bottle that is cm4.16 deep? What would the resonant frequency be if it

was %75 filled?

a. Hz262 , and Hz786 respectively.

b. Hz1046 , and Hz1834 respectively.

c. Hz523 , and Hz0902 respectively.

d. Hz953 , and Hz8582 respectively.

e. None above, they are:

Answer: Alternative ______ [1/0]

Why? Show your calculations in the space below. [1/1]

Suggested solution:

Alternative c: Hz523 , and Hz0902 respectively. [1/0]

mL 164.0 . One end closed pipe: Hzm

sm

L

vf 523

164.04

/343

40

; [1/0]

Since %75.0 of the m164.0 deep bottle is filled, only %25.0 of it is empty.

Therefore:

mm

L 041.04

164.0 HzHz

sm

L

vf 09020912

041.04

/343

40

[0/1]


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5. The velocity of waves on a string is sm/0.94 . If the fundamental frequency of the

standing waves is Hz349 , how far apart are two adjacent nodes?

a. m08.1

b. cm9.53

c. cm9.26 ,

d. cm5.13

e. None above its ___________



Suggested solution: Answer: Alternative d cm5.13 [1/0]

cmmf

vvf 9.26269.0

349

0.94

[1/0]

Nodes of all waves are always2

apart:

cmmm 5.13135.02269.0

2 [0/1]
http://www.walter-fendt.de/ph14e/stwaverefl.htm


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6. In an earthquake, it is noted that a footbridge oscillates at fundamental frequency of

period s25.2 . What other possible resonant periods of motion are there for this

bridge?

a) ...,3,2,125.2 nsn

b) ..,3,2,14

9

ns

n

c) ...,3,2,19

4

ns

n

d) ...,3,2,125.2

2ns

n.

e) None above. It is ..

Answer:

Alternative ______ [1/0]


Suggested solution: Answer: Alternative b i.e.: ...,3,2,14

9

ns

nTn

Tf

1 ; ...,3,2,10 nfnfn ;

...,3,2,1

1

111

0

0

0

0

nn

T

Tnfnf

Tn

n

HzHzHzf9

4

425.2

4

25.2

10

; ...,3,2,1

9

4

5.2

1 nHz

nHznfn

...,3,2,14

9

9

4

11

ns

ns

nfT

n

n [1/1]

Tacoma Narrows Bridge: Construction,

Oscillation, Collapse:

Simulations of bridge collapse under

earthquakes:
http://www.youtube.com/watch?v=gHgQALH9-7M&feature=relatedhttp://www.youtube.com/watch?v=gHgQALH9-7M&feature=relatedhttp://www.youtube.com/watch?v=gHgQALH9-7M&feature=relatedhttp://www.youtube.com/watch?v=qIiuDUgvZpYhttp://www.youtube.com/watch?v=qIiuDUgvZpYhttp://www.youtube.com/watch?v=qIiuDUgvZpYhttp://www.youtube.com/watch?v=qIiuDUgvZpYhttp://www.youtube.com/watch?v=qIiuDUgvZpYhttp://www.youtube.com/watch?v=qIiuDUgvZpYhttp://www.youtube.com/watch?v=IqK2r5bPFTM&feature=relatedhttp://www.youtube.com/watch?v=gHgQALH9-7M&feature=relatedhttp://www.youtube.com/watch?v=gHgQALH9-7M&feature=related


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7. Of the following procedures, which one (s) will lowerthe pitch of a guitar string? i.e. it

will make its frequency smaller.

a) Decrease string tension.

b) Increase string tension.

c) Shorten the string.

d) Lengthen the string.

e) Use a heavier string.

f) Use a lighter string.


Why? Use words, formulas and logic in the space below to

show how you arrived in your conclusion. [0/1]

Suggested solution: Answer: Alternative a, d, and e:

usingLm

Fv T

/ , L21 , and

L

vvf

vf

200

.

If the velocity of the propagation of the wave in the stringLm

Fv T

/ is

decreasedthe frequency

vf will decrease. This may be achieved,

either by decreasing the tension TF (alternative a) or by using a heavierstring (alternative e). The low pitch (frequency) also may be achieved bylengthening the wire (alternative d) which in turn increases the

wavelength of the sound:Lm

F

LL

vvf T

/2

1

200

.


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8. The siren of a police car at rest emits a sound wave at Hz6001 . What frequency will

you hear, if you are at rest and the police car is moving away from you at hkm/108 .

a. I will hear the siren at Hz4701 .

b. I will hear the siren at Hz7531 .c. I will hear the siren at Hz7401 .

d. I will hear the siren at Hz4601 .

e. I will hear the siren at Hz4602 .

f. I will hear the siren at Hz1851 .

g. None above. I will hear the siren at.


Why? Use words, formulas and logic in the

space below to show how you arrived in your

conclusion. [0/1]

Suggested solution: Alternative a

This is due to theDoppler effect

smv /303600

1000108 HzHz

v

v

ff

s

47014711

343

301

6001

1

Hzf 4711

Doppler

Effect Applet
http://www.walter-fendt.de/ph14e/dopplereff.htmhttp://www.walter-fendt.de/ph14e/dopplereff.htmhttp://www.walter-fendt.de/ph14e/dopplereff.htmhttp://library.thinkquest.org/19537/java/Doppler.htmlhttp://library.thinkquest.org/19537/java/Doppler.htmlhttp://library.thinkquest.org/19537/java/Doppler.htmlhttp://library.thinkquest.org/19537/java/Doppler.htmlhttp://library.thinkquest.org/19537/java/Doppler.htmlhttp://www.youtube.com/watch?v=76X1LtWkUE8http://www.astro.ubc.ca/~scharein/a311/Sim/doppler/Doppler.htmlhttp://www.walter-fendt.de/ph14e/dopplereff.htm


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9. The intensity level of the sound from a rock concert at a distance of m0.5 from the

main loudspeaker is dB130 which is the threshold of pain. How far away would the

intensity level be a somewhat reasonable dB100 ?

a. At m160 from the loudspeaker.

b. At m125 from the loudspeaker.

c. At m100 from the loudspeaker.

d. At m75 from the loudspeaker.

e. At m50 from the loudspeaker.

f. At m25 from the loudspeaker.

g. None above. At m from the

loudspeaker.

Answer:

Alternative ______ [1/0]

Why? Use words, formulas and logic in the space

below to show how you arrived in your conclusion.

[0/2]

Answer: Alternative e, i.e.: At mr 160 away from the speaker the

intensity level is a somewhat reasonable dB100 .

dBI

I135log10

0

11

13log

0

1

I

I 13

0

1 10I

I

21213

0

13

1 /10101010 mwII

dBII 100log10

0

22

10log

0

2

I

I 10

0

2 10I

I

221210

0

10

2 /10101010 mwII [0/1]

Due to the fact that it is the same loudspeaker that generates thesound, the power P for both cases is the same:

2/mw

A

PI 2m

I

PA 22

4m

I

Pr

1

2

2

1

2

2

2

1

4

4

I

I

I

P

I

P

r

r

mmrrI

I

r

r16015810005

1000

1

1000

1

10

1052

2

21

2

2

1

[0/1]


10/20


[email protected] for Sale. Free to use for educational purposes.10/20

10. Explain the phenomena of interference in words, formulas and necessary figures. Whatare conditions for interference to take place? What are constructive interference and

destructive interference? In the figure below identify points where constructive

interference, as well as points where destructive interference takes place. [2/2/M235]

Suggested solution:When two waves approach eachother they simply pass each other.If the waves are share identicalfrequency, f , wavelength, , and

amplitude, A , they may produceconstructive interference ordestructive interferences as theymeet. If two waves are in phase,and their crests (peaks) and their

through (minimum) come to agiven point at the same time, theiramplitude are added to each otherand produce a wave of amplitudeA2 at the point.

The phenomenon is calledconstructiveinterference.At points where crests of one wavecome at the same time as throughof the other wave reach, their

amplitude cancel each other andinterference is destructive.In the accompanying figure, twosources that are connected to thesame main vibrator produce twowaves of identical frequency, wavelength and amplitude.

The dark points in the figure aretroughs and bright lines represent

crests. Notice the grey area; theyrepresent destructive interference,where waves from two sourcescompletely kill each other.Phenomenon of interference isbased on mathematicalsuperposition of two sinusoidalfunctions that have the same wavenumber.
http://www.walter-fendt.de/ph14e/interference.htmhttp://www.walter-fendt.de/ph14e/interference.htmhttp://www.walter-fendt.de/ph14e/interference.htmhttp://www.walter-fendt.de/ph14e/interference.htmhttp://www.walter-fendt.de/ph14e/interference.htmhttp://www.falstad.com/ripple/


11/20



If the path difference between twowaves reaching the point is:

...,3,2,1,0 nn

the interference is constructive:...,3,2,1,0sin nnd

If the path difference between twowaves reaching the point is:

...,3,2,1,02

12 nn

the interference is destructive:

...,3,2,1,02

12sin nnd

Principle of superposition: Interference of two waves: interference patternapplets:

Interference of water waves. Applets:
http://zonalandeducation.com/mstm/physics/waves/interference/waveInterference3/WaveInterference3.htmlhttp://zonalandeducation.com/mstm/physics/waves/interference/waveInterference3/WaveInterference3.htmlhttp://zonalandeducation.com/mstm/physics/waves/interference/twoSource/TwoSourceInterference1.htmlhttp://zonalandeducation.com/mstm/physics/waves/interference/twoSource/TwoSourceInterference1.htmlhttp://www.ngsir.netfirms.com/englishhtm/Interference.htmhttp://www.ngsir.netfirms.com/englishhtm/Interference.htmhttp://www.ngsir.netfirms.com/englishhtm/Interference.htmhttp://zonalandeducation.com/mstm/physics/waves/interference/twoSource/TwoSourceInterference1.htmlhttp://zonalandeducation.com/mstm/physics/waves/interference/twoSource/TwoSourceInterference1.htmlhttp://en.wikipedia.org/wiki/File:Interference_of_two_waves.svghttp://zonalandeducation.com/mstm/physics/waves/interference/waveInterference3/WaveInterference3.html


12/20



M V G - q u a l i t y The student in solving the problemshows highest (MVG) qualitysolution in problem 12 by

M2 Analyses and interprets the results,

concludes and evaluates if they arereasonable.

analyzing and interpreting

interference by using principle ofsuperposition.Path difference is discussed.Constructive interference:

...,3,2,1,0sin nnd

Destructive interference:

...,3,2,1,02

12sin nnd

M3 Carries out mathematical proof, oranalyses mathematical reasoning.

using words, figures, and formulasto define conditions for

constructive interference anddestructive interference areanalyzed, and interpreted.

M5 The presentation is structured, andmathematical-physical language iscorrect.

presenting well-structured, clearsolutions. The mathematical-physical language is correct.


13/20



11. An ambulance carrying a seriously injured patient moves down a highway at a constant

velocity of hkm/126 . Its siren emits a sound at Hz.700 . Determine the frequency

heard by a car driver moving in the opposite direction at constant velocity of

hkm/0.90 as the car approaches the ambulance. [0/2]

Suggested solution:Data: Hzf 700 ,

smhkmvv sambulance /0.356003

1000126/126 ,

smhkmvv ocar /0.256003

100090/90 ,

This is doubleDoppler effectVelocity of sound in air smv /343 using smTv /60.0331 and 20T :

If the driver of the car (observer) was at rest and the ambulance (source)was approaching her, she would experience a higher pitch according to

Doppler effect:

v

v

ff

s

1

.

On the other hand if the ambulance (source) was stationary emitting

sound of frequency

vv

ff

s

1

and the car was approaching it, the driver of

the car would experience

v

vff o1 .

s

o

s

o

s

o

s

o

o

vv

vvf

v

v

v

vv

v

v

v

f

v

vv

v

f

v

v

v

vf

v

vff

1

1

1

1

1 [Extra M1235]

As the car approaches the ambulance the observed frequency of the sirenis:

HzHzfvv

vvf

s

o 83670035343

25343

[0/2]

Answer: The driver of the car experiences siren of frequency 836 Hz as itapproaches the ambulance that is sending a siren of 700. Hz.
http://www.walter-fendt.de/ph14e/dopplereff.htmhttp://www.walter-fendt.de/ph14e/dopplereff.htmhttp://www.walter-fendt.de/ph14e/dopplereff.htmhttp://www.walter-fendt.de/ph14e/dopplereff.htmhttp://www.youtube.com/watch?v=JZTKhswn684


14/20


15/20



12. A particular organ pipe can resonate at Hz735 , Hz0291 , and Hz3231 , but not

any other frequencies between Hz735 and Hz3231 . [2/4/M1235]

a) Is this open or closed organ. Why? Show details of your calculation and explainyour reasoning.

b) What is the fundament frequency of this pipe?

Suggested solution:Data: Overtones: Hz735 , Hz0291 ,

and Hz3231 , but nothing in between.

Both-ends-open pipe mth overtone

1mffm

1111 1 ffmfmff mm

Therefore if the pipe is an open tubeit is expected that mth overtone is aninteger factor of the fundamentalfrequency: 1mffm

Hzfn 7351 , nth overtone Hzfn 0291 , and (n+1)

th overtone

Hzfn 32311 ,

Hzff nn 294029132311

Hzff nn 29473502911

As demonstrated above if the pipe is a both-ends-open tube it is requiredthat:

...,3,2,1n294

294294

1

1111

n

Hzfnf

HzfHzfffff

n

nnnn [0/1]

Therefore, if Hzfnfn n2941 the pipe is both-end-open

But integer5.2294

735

294

1 Hzfn ; integer5.3

294

0291

294

Hzfn and

integer5.4294

3231

294

1 Hzfn

Therefore, the pipe cannot be a both-ends-open pipe.[0/1/Part of M1235]


16/20



One-End-Closed Tube or Pipe: Both-Ends-Open Tube or Pipe:

One-end-closed pipe (2m+1)

th

overtone 1212 112

112

mf

f

fmfm

m

1111212 21212 ffmfmff mm [0/1]

Therefore, if the pipe is a one-end-closedtube, as demonstrated above, thedifference between two consequentovertones is 11212 2fff mm . Therefore, it

is expected that the fundamentalfrequency of the pipe to be

11212 229402913231 fHzff mm , which implies: HzHzf 1472

2941

Checking the requirement integer12121

12112

m

f

ffmf mm

5147

735

147

12 nf

; 7147

0291

147

12 nf

and 9147

3231

147

32 nf

. [1/0]

Therefore, the pipe is a one end closed tube with the fundamental

frequency Hzf 1471 , and three consecutive overtones Hzf 7355 ,

Hzf 02917 , and Hzf 32319

. [0/1/Part of M1235]
http://www.walter-fendt.de/ph14e/stlwaves.htmhttp://www.walter-fendt.de/ph14e/stlwaves.htmhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://groups.physics.northwestern.edu/vpl/waves/wavetypes.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://groups.physics.northwestern.edu/vpl/waves/wavetypes.htmlhttp://www.walter-fendt.de/ph14e/stlwaves.htm


17/20



M V G - q u a l i t y The student in solving the problemshows highest (MVG) qualitysolution in problem 12 by

M1 Formulates and develops the problem,uses general methods with problem

solving.

generalizing the problem andcomparing standing in both one-

end closed as well as two-ends-open tubes (or pipes.)

M2 Analyses and interprets the results,concludes and evaluates if they arereasonable.

Analyzing and interpretingstanding waves in both one-endclosed as well as two-ends-opentubes (or pipes.) Longitudinal andtransversal waves are definedandcompared.

M3 Carries out mathematical proof, or analysesmathematical reasoning.

using words, figures, and formulasto compare and formulas for

standing waves in both one-endclosed as well as two-ends-opentubes (or pipes.)

M5 The presentation is structured, andmathematical-physical language iscorrect.

presenting well-structured, clearsolutions. The mathematical-physical language is correct.

13. The accompanying figure

illustrates a standing wave in astring at Hz735 . It is produced

by a frequency generator. The

string in the figure is cm0.51 .

[3/3/M1235]

a) Calculate the wavelength ofthe wave.

b) Calculate the velocity of thewave in the string.

c) Explain in words insufficient details,formulas,

andfigures what a standing

wave is and how it may be

produced.

d) Which type of wave is the one in the picture? Longitudinal or transversal? Whatare differences between longitudinal and transversal waves?

e) What will happen if the frequency of the generator is gradually changed? Will any

other standing wave occur if the frequency range of Hz200 to Hz5001 ? If

so, which ones and why? Explain in words in sufficient details, formulas, and

figures.

Suggested solution:Data: Hzf 7353 , mcmL 510.00.51


18/20



The standing wave in the figure is the third harmonic (second overtone).The wavelength of the standing wave may be found using the fact that:

mmLLL 340.0510.03

2

3

223

2

3333 [0/1] m340.03

The velocity of propagation of the wave in the string may be calculated,using the fact that:

smsmsmfT

v /.250/9.249/735340.0

[1/0] smv /.250

Astanding waveis produced when an incoming wave is mixed with thereflected wave according to theprinciple of superposition. This is due toconstructive interference and destructive interference. Nodes are due todestructive interference, and anti-nodes are due to constructiveinterference. Even though the wave, for example in the string of stringinstrument, goes back and forth, some points in the string seemstationary (nodes) and others may seem they are all the time in

maximum displacement from the equilibrium (anti-node). Standing wavesmay be produced only if both waves have identical wavelength andamplitude. Standing wave in a string or pipe may occur in more than onefrequency. As illustrated in the figures below:

Standing waves ontwo-end fixed strings

1

2

2

fnf

Ln

nL

n

nn

Standing waves on two-ends-open pipes:

1

2

2

fnf

Ln

nL

n

nn

Standing waves on one-end-closed pipes:

112

12

12

4

12

fnf

nL

n

n

Standing wave on a string

Standing waveson one end-closedTube

Standing waves ontwo-ends open tube

[0/1/part of M2]
http://www.walter-fendt.de/ph14e/stwaverefl.htmhttp://www.walter-fendt.de/ph14e/stwaverefl.htmhttp://www.walter-fendt.de/ph14e/stwaverefl.htmhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.google.se/imgres?imgurl=http://www.cdli.ca/courses/phys2204/unit04_org01_ilo11/les-11-l-d-fig01.gif&imgrefurl=http://www.cdli.ca/courses/phys2204/unit04_org01_ilo11/b_activity.html&usg=__AIvU-KQKWW3cijy11dtj1SqIP40=&h=727&w=623&sz=9&hl=sv&start=44&zoom=1&tbnid=khl7lHuxC9KY7M:&tbnh=134&tbnw=115&ei=tzHlTZXUEM-G-waxpLWUBw&prev=/search%3Fq%3Dstanding%2Bwave%2Bon%2Ba%2Bstring%26um%3D1%26hl%3Dsv%26sa%3DN%26biw%3D1021%26bih%3D474%26tbm%3Disch&um=1&itbs=1&iact=rc&dur=203&page=6&ndsp=9&ved=1t:429,r:0,s:44&tx=53&ty=58http://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.mta.ca/faculty/science/physics/suren/Harmonics/Harmonics.htmlhttp://www.walter-fendt.de/ph14e/stwaverefl.htm


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The wave produced in the stringis a transversal wave.Atransversal wavepropagatesin the direction perpendicular to

the direction of vibration ofatoms of the medium; anexample is the wave producedin a string instrument, or waterwaves.Atoms of the medium of alongitudinal wavevibrates in thesame direction as the directionof its propagation. Sound waveis a longitudinal wave.

[1/1/Part of M2]

Transversal waves versus Longitudinalwaves applets

The fundamental frequency of the wave in the string is Hz245 :

HzHzffff 2453

735

3

13 3113

Therefore, standing waves are produced in the following frequencies, inthe range of interest: [0/1/part of M3]

Fundamental frequency (firstHarmonic)

Hzf 2451

Second harmonic (Firstovertone)

HzHzf 49024522

Third harmonic (Secondovertone)

HzHzf 73524533

Fourth harmonic (Thirdovertone)

HzHzf 98024544

Fifth harmonic (Fourthovertone)

HzHzf 225124555

Sixth harmonic (Fifth overtone) HzHzf 470124566

M V G - q u a l i t y The student in solving the problem13 shows highest (MVG) quality by

M1 Formulates and develops the

problem, uses general methodswith problem solving.

Generalizing the problem andcomparing standing waves in stringsand one-end closed as well as two-ends-open tubes (or pipes.)

M2 Analyses and interprets the results,concludes and evaluates if they arereasonable.

Analyzing and interpreting standingwaves in strings and pipes.

M3 Carries out mathematical proof, oranalyses mathematical reasoning.

using words, figures, and formulas tocompare and formulas for standingwaves in strings and tubes.

M5 The presentation is structured, and

mathematical-physical language iscorrect.

presenting well-structured, clear

solutions. The mathematical-physicallanguage is correct.
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M V G - q u a l i t y 10 11a 11b 12c 12eM1 Formulates and develops the problem, uses

general methods with problem solving.

M2 Analyses and interprets the results, concludes

and evaluates if they are reasonable.M3 Carries out mathematical proof, or analysesmathematical reasoning.

M4 Evaluates and compares different methodsand mathematical models.

M5 The presentation is structured, andmathematical-physical language is correct.

M V G - q u a l i t y M1 Formulates and develops the problem, uses

general methods with problem solving.

M2 Analyses and interprets the results, concludesand evaluates if they are reasonable.

M3 Carries out mathematical proof, or analysesmathematical reasoning.

M4 Evaluates and compares different methodsand mathematical models.

M5 The presentation is structured, andmathematical-physical language is correct.

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