Instructions:The Test Warning! There are more than one version of the test.
At the end of each problem a maximum point which one may get for a correct solutionof 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 apersonalized formula sheet which has your name on it. This should be submitted alongwith the test.
Time: 8:10-9:30 Part I13:40-15:00 Part IIThe multi-choice problems 1-15 must be answered on the original paper.
Limits: Part I: Maximum 46 of which 23 are G points and 23 are VG points.Part II: Maximum 20 VG-points and 5 MVG ¤ quality
In the multi-choice problems below just circle the correct answer (s) in the test paper andwrite the answer in the space provided as “Answer: _____________”
1. In which type of wave is the disturbance perpendicular to the direction of wave travel?
a) Longitudinal
b) Transitional
c) Transverse
d) Circular
Answer: Alternative (C): Transverse [1/0]
2. Mechanical wave motion in a medium transfers
a) only mass
b) only energy
c) both mass and energy
d) neither mass nor energy
Answer: Alternative (b): only energy [1/0] A mechanical wave can
only carry energy. No mass is carried by a mechanical wave.
3. The driver of a car hears the siren of a police car which is moving away from her. If the
actual frequency of the siren is 2000 Hz, the frequency heard by the driver may be
a) 2200 Hz
b) 2100 Hz
c) 2000 Hz
d) 1900 Hz
Answer: Alternative (d): 1900 Hz [1/0] The frequency of a source
moving away from a stationary observer decreases as: ⎟ ⎠ ⎞
⎜⎝ ⎛ +
=′
vv
f f
s1
4. If the frequency of a sound wave in air at STP remains constant, its energy may beincreased by increasing its
a) velocity
b) wavelength
c) period
d) amplitude
Answer: Alternative (d): Amplitude [1/0]. The energy of a mechanical
wave is a function of its amplitude and frequency.
9. If the air temperature in a rooms is decreased, the fundamental frequency of the organpipes
9. If the air temperature in a rooms is decreased, the fundamental frequency of the organpipesa) will be decreased.a) will be decreased.
b) will not be effected.b) will not be effected.
c) will be increased.c) will be increased.d) be equal to the frequency of the second harmonics.d) be equal to the frequency of the second harmonics.
Answer: Alternative (a): will be decreased. [1/0]
Any decrease in the air temperature will decrease the velocity of the
sound in the air due to the fact that: smT v / 60.0331 ⋅+≈ where T is
the air temperature in centigrade. Lowering velocity, on the other
hand will result in lower frequency:λ
v f = .
10. A standing wave pattern is produced when a guitar string is plunked. Whichcharacteristic of the standing wave immediately begins to decrease?a) speed.
b) frequency.
c) period.
d) amplitude.
Answer: Alternative (d): amplitude. [1/0]
The amplitude of the mechanical wave decreses due to dissipative
forces like friction that damps the energy of the wave which is directly
proportional to the amplitude. This means after a while, even though
still the string is vibrating and producing sound, but it is producing
such a small amount of energy that is too weak to hear.
11. In which direction will segment S move as the traveling wave passes through it?
a) right only
b) down only
c) up only
d) down, then up, then down
e) up, then down, then up
f) right, then up, then down, then up
g) right, then down, then up, then down
S
v
Rope
Answer: Alternative (d): down, then up, then down again. [1/0]
12. The diagram below represents a ropealong which two pulses of equalamplitude,
12. The diagram below represents a ropealong which two pulses of equalamplitude, A , approach point S . As thetwo pulses pass through point S , themaximum vertical displacement of the
rope at point S will bea) A2
b) A
c) 2 A
d) 0
Answer: Alternative (a): A2 . [1/0] Due to the superposition of the
waves.
13. The diagram below represents a standing wave. Point S on the standing wave is
a) a node resulting from constructive interference.
b) a node resulting from destructive interference.
c) an antinode resulting from constructive
interference.
d) an antinode resulting from destructive interference.
Answer: Alternative (b): S is a node resulting from destructive
interference. [1/0]
14. A g0.50 block of wood is pressed against a horizontal spring of force constantm N / 0.20 . The spring is compressed cm0.3 and then the block is released. The
wooden block lies on a frictionless horizontal desk. The block leaves the spring at:a) sm / 00.6 .
b) sm / 60.0
c) sm / 06.0
d) sm / 12
e) None: ____________ m/s
Answer: Alternative (b): smv / 60.0= [1/0]
Why? Show the details of your calculations in the space provided below: [0/2]
Suggested solution:Data: kggm 050.00.50 == , m N k / 0.20= mcm A 03.00.3 == ; Problem: ?=v , ?=
16. The A string of a violin has a linear density of mg / 0 and an effective length of 60.cm0.33 .
a) Find the tension required for its fundamental frequency to be note A, i.e.: Hz440 [0/2]
b)
If the string is under this tension, how far from one end should it be pressed againstthe fingerboard in order to have it vibrate at a fundamental frequency of note B,i.e.: Hz495 ? [0/2]
Problem: Note A: Hz f 440= , ?=T F , ?= L Note B: Hz f 4951
=
Velocity Velocity of propagation of the waveλ
λ ⋅== f
T v
Wave on astring
The velocity of a wave v on a stretchedstring of mass m and length L under whichis under tension T F .
LmF v T
/ =
The fundamental frequency of a string is 1 f which is related to the lengthof the string and therefore the wave length of the first harmonic through:
21λ = L ⇔ L21
=λ
( ) ⎟ ⎠ ⎞
⎜⎝ ⎛ ⋅⋅=⇔
⎪⎩
⎪
⎨
⎧
⋅=
⎟
⎠
⎞⎜
⎝
⎛ ⋅=
⇔
⎪⎩
⎪
⎨
⎧
⋅=
=⇔= L
m
f F f v
L
mvF
f v
v
Lm
F
Lm
F v
T
T T T
222
/ / λ λ λ
( ) ( ) ⎟ ⎠ ⎞
⎜⎝ ⎛ ⋅⋅=⎟
⎠ ⎞
⎜⎝ ⎛ ⋅⋅=⇔
⎪⎩
⎪⎨
⎧
=
⎟ ⎠ ⎞
⎜⎝ ⎛ ⋅⋅=
Lm
f L Lm
L f F
L
Lm
f F T
T 21
221
1
211 42
2λ
λ
( ) ( ) ( ) N N Lm
f LF T 6.5059.50100.644033.044 42221
2 ≈=×⋅⋅=⎟ ⎠ ⎞
⎜⎝ ⎛ ⋅⋅= −
Answer: The tension required for violin’s fundamental frequency to be
note A is N Lm f LF T 6.504 2
12 ≈⎟
⎠ ⎞⎜
⎝ ⎛ ⋅⋅=
( )mm
LmF
f Lm
f
F L
Lm
f LF T T BT 293.0
100.659.50
49521
21
44 4
121
22 =×⋅
=⎟ ⎠ ⎞
⎜⎝ ⎛ ⋅
=⎟ ⎠ ⎞
⎜⎝ ⎛ ⋅⋅
=⇔⎟ ⎠ ⎞
⎜⎝ ⎛ ⋅⋅= −
mmm L L B 37037.0293.0330.0 ==−=−=l mm37=l Answer: If the string is under this tension, 37 mm from one end (or 29 cmfrom the other end) should it be pressed against the fingerboard in order
to have it vibrate at a fundamental frequency of note B.
Answer: If the string is under this tension, 37 mm from one end (or 29 cmfrom the other end) should it be pressed against the fingerboard in orderto have it vibrate at a fundamental frequency of note B.
Third Method:
mmmm L L
mm
Lm Hz
sm f vv
f
sm L
mF
v
B A
B B
B B
T
37037.0293.0330.0
293.02
586.02
586.0495
/ 4.290
/ 4.2901060.06.50
3
==−=−=
===⇔===⇔=
=⋅
== −
l
λ λ
λ
Answer: If the string is under this tension, 37 mm from one end (or 29 cmfrom the other end) should it be pressed against the fingerboard in orderto have it vibrate at a fundamental frequency of note B.
17. At a recent concert, a dB meter registered dB120 when placed m0.4 in front of aloudspeaker on the stage.a) What was the power output of the speaker, assuming uniform spherical spreading of
the sound and neglecting absorption in the air? [0/2]b) How far away would the intensity level be a somewhat reasonable dB80 ? [0/2]
Suggested solution:Data: Sound wave dB120= β , mr 0.4= ,Problem: ?=P , ?=r to the intensity dB80= β a) Answer: The power output of the speaker, assuming uniform spherical
spreading of the sound and neglecting absorption in the air was:wP 200≈ .
dB120= β ⇔ dB I I
120log100
=⎟⎟
⎠
⎞⎜⎜
⎝
⎛ = β ⇔ 12log
0
=⎟⎟
⎠
⎞⎜⎜
⎝
⎛ I I
⇔ 12
0
10= I I
⇔
212120
12 / 0.1101010 mw I I =⋅=⋅= − [0/1]
2 / mw AP
I = ⇔ w I r w I AP ⋅⋅=⋅= 24π ⇒
( ) wwwP 2001.2010.10.44 2 ≈=⋅⋅= π [0/1] wP 200≈
using: ⎟⎟
⎠
⎞⎜⎜
⎝
⎛ =
0
log10 I I
β , a x =log ⇔ a x 10= , 2120 / 10 mw I −= , 2 / mw
AP
I = and
the surface area of a sphere of radius r : 24 r A ⋅= π .
b) Answer: mr 400≈ away from the speaker the intensity level is asomewhat reasonable dB80 .
18. In a car engine, each piston moves up and down in an approximation of SHO. A typicalpiston has a mass of g.500 , a total travel (stroke) of cm0.12 and a frequency of oscillation of Hz60 . Find the maximum force it experiences. [0/4]
Suggested solutions:Data: kggm 500.0.500 == ; Hz f 60= Using:
( )
( )
( )
Aa
xt Adt
xd dt dv
a
t Adt dx
v
t A x
f mk
xmk
dt xd
kxma
xk F
amF ⋅=⇔
⎪
⎪
⎪
⎩
⎪⎪
⎪
⎨
⎧
⋅−=⋅⋅−===
⋅⋅==
⋅⋅=
⇔
⎪
⎪
⎪
⎩
⎪
⎪
⎪
⎨
⎧
⋅=
=
−=
−=
⇔
⎪⎩
⎪⎨
⎧
⋅−=
⋅=2
max
222
22
2
2
sin
cos
sin
2
ω
ω ω ω
ω ω
ω
π ω
ω
( ) A f mF A f m AmmaF ⋅=⇔⋅⋅=⋅== 22max
22maxmax 42 π π ω
Noting that in the piston the amplitude is half of the total travel distance:mcmcm A 060.00.6
20.12 ===
( ) ( ) kN N A f mF 3.47.4263060.060500.044 2222max
≈=⋅××=⋅= π π The maximum force experienced by the piston is kN A f mF 3.44 22
max≈⋅= π
Second Method:
( ) ( ) m N m f k
mk
f mk
f 061715.06044421 222222 =⋅⋅=⋅⋅=⇔=⋅⇔= π π π π
kN N F Ak F 3.42634212.006171maxmax ≈=⋅=⇔⋅=
The maximum force experienced by the piston is kN A f mF 3.44 22max
≈⋅= π
Third Method:Conservation of mechanical energy requires that:
2
2max2
max2
21
21
Amv
k mvkA =⇔=
Using: A f v ⋅= π 2max
( ) m f A
A f mk ⋅=⋅= 222
2
42 π π
( ) ( ) m N m f k 061715.06044 2222 =⋅⋅=⋅⋅= π π
kN N F Ak F 3.42634212.0
06171maxmax≈=⋅=⇔⋅=
The maximum force experienced by the piston is kN A f mF 3.44 22max
19. An ambulance is rushing in a straight highway a seriously injured patient to hospital athkm / .120 . It emits siren at Hz400 .
a) What is the frequency of the siren heard by a stationary hospital guard? [2/0]b) What is the frequency of the siren heard by a police officer moving at hkm / .90 in
the opposite direction as the police car approaches the ambulance and as it movesaway from the ambulance? [0/4]
20. When a Hz440 tuning fork is struck simultaneously with the playing of the note A ona guitar, 4 beats per second are heard. After the guitar string is tightened slightly toincrease its frequency, the beat frequency increases to 6 beats per second. What is thefrequency of the guitar string after it is tightened? What should we do to tune the guitarstring to exactly 440 Hz? [0/3]
Suggested solutions:The original frequency of the guitar string was either Hz436 or Hz444 .This is due to the fact that 4 beats per second was heard initially, i.e.:
21 f f f B −= 4404 −=
G f
Hz f G 436= or Hz f G 444= If the original guitar frequency was Hz f G 436= the tightening of thestring, i.e. increasing the tension in the string, would result in decreasingof the beat frequency. This is due to the fact that increasing tension
directly increases the velocity of the wave in the string: Lm
F v T
/ = which in
turn increases the frequency:λ v
f = .
Therefore, due to the fact that the beat frequency increases from 4 to 6when the tension in the string is increased, we may conclude that theoriginal frequency of the guitar was rather Hz f G 443= and we mustslightly decrease the tension in the string by loosening it a little.
21. A cm.75 long guitar string of mass g80.1 is placed near a tube open in one end, andalso of the same length cm.75 . How much tension should be in the string if it is toproduce resonance (in its fundamental mode) with the third harmonics in the tube?
[0/4/¤]Suggested solutions:
Data: mcm L 75.0.75 == ; Fundamental mode: String:2
1String Lλ = ,
kggm 31080.180.1 −×== ; One end open pipe: Third harmonics:4
3 3Pipe Lλ ⋅= ;
Tension ?=T F , if PipeString f f 31
= ; Speed of sound in air (SPT) smv / 343=
Hz Hz L L
v f
PipePipe 343
75.043433
43433
34343
33
=××=×===
λ [0/1]
Hz f f PipeString 34331 == [0/1]
m L L StringStringString 50.175.0222 1
1 =×=⋅=⇔= λ λ
smv f vv
f / 5.5145.134311=×=⇔⋅=⇔= λ
λ , [0/1]
⎟ ⎠ ⎞
⎜⎝ ⎛ ⋅=⇔=⇔=
Lm
vF v Lm
F Lm
F v T
T T 22
/ /
( ) N F T 3.635
75.0
1080.15.514
32 =
⎟⎟
⎠
⎞⎜⎜
⎝
⎛ ×⋅=
−
[0/1/¤] N F T 635≈
Answer: The tension should be N F T 635≈ in the string if it is to produceresonance (in its fundamental mode) with the third harmonics in the tube.
Second method:
⎪
⎪
⎩
⎪⎪
⎨
⎧
⋅=⇔=
⎟ ⎠ ⎞
⎜⎝ ⎛ ⋅=⇔=⇔=
λ λ
f vv
f
Lm
vF v Lm
F Lm
F v T
T T 22
/ / ⇔ ( ) ⎟
⎠ ⎞
⎜⎝ ⎛ ⋅⋅=
Lm
f F T 2λ
Third harmonics in one end open Pipe: Hz Hz
L Lv
f Pipe
Pipe 34375.04
343343433
34343
33
=××=×===
λ Hz f f PipeString 34331
==
Fundamental frequency of the string:
m L L StringString 50.175.0222 1
1 =×=⋅=⇔= λ λ
;
( ) ( ) N Lm
f F T 3.63575.01080.1
50.13433
22 =⎟⎟
⎠
⎞⎜⎜
⎝
⎛ ×⋅×=⎟
⎠ ⎞
⎜⎝ ⎛ ⋅⋅=
−
λ N F T 635≈
Answer: The tension should be N F T
635≈ in the string if it is to produceresonance (in its fundamental mode) with the third harmonics in the tube.
22. A source emits sound of wavelength m34.2 and m86.2 in air ( C T °= 20 ) [0/4/¤]
a) How many beats per second will be heard?b) How far apart in space are the regions of maximum intensity?
Suggested solutions:Data: m34.21 =λ , m86.22 =λ , smv / 343= ; ?= B f ; ?= Bλ
Hz Hzv
f 14758.14634.2
343
11
≈===λ
[0/1]
Hz Hzv
f 12093.11986.2
343
22
≈===λ
[0/1]
Hz Hz f f f B 2765.2693.11958.14621≈=−=−= [0/1]
mm f v
B B 1387.12
65.26343 ≈===λ
Answer: 27 beats per second will be heard: Hz f f f B 1521 ≈−= . The regionsof maximum intensity are m B 13≈λ apart in space. [0/1/¤]
Corrections :
Beat frequency must be actually small, possibly less than 5 Hz, otherwisewe may not notice or detect (hear) it. The wavelengths emitted by thesource must, therefore, be much closer to each other, may be:
m34.21=λ , m46.22
=λ , smv / 343= ; ?= B f ; ?=
Bλ
Hz Hzv
f 14758.14634.2
343
11
≈===λ
Hz Hzv
f 14043.13946.2
343
22
≈===λ
Hz f f f B 15.743.13958.14621=−=−=
mm f v
B B 4897.47
15.7343 ≈===λ
Answer: 7 beats per second will be heard: Hz f f f B 721≈−= . The regions
of maximum intensity are m B 48≈λ apart in space.
This is much better! Beat frequency must be actually small, possibly about5 Hz, otherwise we may not notice it.
23. Michael hears a pure tone coming from two sources that seems to be in the Hz1000500 − range. The sound is loudest at points equidistance from two sources. In
order to determine exactly what the frequency is, Michael moves about and finds that thesound level is minimum at a point cm5.18 farther from one source than the other. Whatis the frequency of the sound? [0/3/¤]
Suggested solutions:Data: mcm 185.05.18 ==δ , smv / 343= ; ?= f ;Due to the fact that at equidistance points the sound is loudest, we mayconclude that the sources are in harmony, and oscillate at the samefrequency. On the other hand destructive interference occurs at the pointswhere the path difference is an odd numbers of half of the wavelength:
( )2
12λ
δ += n . [0/1]
The waves from the distinguished sources reach the point completely outof phase and they end up completely annihilating killing each other.
If we assume the path difference is2λ
δ = : [0/1]
2λ
δ = ⇔ ( ) m37.0185.022 =⋅== δ λ
Hzv
f 92737.0
343 ===λ
This is within the expected limits. [0/1/¤] Answer: The sources oscillate at identical frequency of Hz f 927=
24. A highway overpass was observed to resonate as one full loop when a small earthquakeshock the ground vertically at Hz0.4 . The highway department put up a support at thecenter of the overpass, anchoring it to the ground.
a) What resonance frequency would you now expect for the overpass?b) It is noted that earthquakes rarely do significant shaking above Hz0.5 or Hz0.6 .
Did the modification do any good? Why? Explain. [0/3/¤]Suggested solutions:Data: Hz f 0.4= ; L2=λ
Hz Lv
Hz Lvv
f 0.80.42
=⇔===λ
[0/1]
Now the resonance frequency of the new situation is related to the “new
length” of the bridge,2 L
Lnew= :
L L
L L newnewnew
new=⋅==⇔=
222
2λ
λ [0/1]
The eventual earthquake vibrates the bridge at Hz Lvv
f new
new 0.8===λ
.
Answer: The expected resonance frequency of the “new bridge” is Hz f new 0.8= which is rare! The modification succeeded. [0/1/¤]
Note that the higher harmonics of the new situations are all out of rangeof the earthquake frequency: ( ) Hzn f nnew
