Suggested+Solution++Test+FyBNVC09+Ch11 12+Waves Sound
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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:
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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
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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 ___________
Answer: Alternative ______ [1/0]
Why? Show your calculations in the space below. [1/1]
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]
Why? Show your calculations in the space below. [0/1]
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.
Answer: Alternative ______ [1/0]
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.
Answer: Alternative ______ [1/0]
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]
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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/ -
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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 -
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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.
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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 -
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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]
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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 -
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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
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Suggested Solutions FyBNVC09 Ch11-12 Mechanical wavesSound NV-College
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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 -
8/3/2019 Suggested+Solution++Test+FyBNVC09+Ch11 12+Waves Sound
19/20
Suggested Solutions FyBNVC09 Ch11-12 Mechanical wavesSound NV-College
[email protected] for Sale. Free to use for educational purposes.19/20
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.
http://groups.physics.northwestern.edu/vpl/waves/wavetypes.htmlhttp://groups.physics.northwestern.edu/vpl/waves/wavetypes.htmlhttp://groups.physics.northwestern.edu/vpl/waves/wavetypes.htmlhttp://www.walter-fendt.de/ph14e/stlwaves.htmhttp://www.walter-fendt.de/ph14e/stlwaves.htmhttp://groups.physics.northwestern.edu/vpl/waves/wavetypes.htmlhttp://groups.physics.northwestern.edu/vpl/waves/wavetypes.htmlhttp://groups.physics.northwestern.edu/vpl/waves/wavetypes.htmlhttp://groups.physics.northwestern.edu/vpl/waves/wavetypes.htmlhttp://groups.physics.northwestern.edu/vpl/waves/wavetypes.htmlhttp://www.walter-fendt.de/ph14e/stlwaves.htmhttp://groups.physics.northwestern.edu/vpl/waves/wavetypes.html -
8/3/2019 Suggested+Solution++Test+FyBNVC09+Ch11 12+Waves Sound
20/20
Suggested Solutions FyBNVC09 Ch11-12 Mechanical wavesSound NV-College
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.