8/3/2019 Suggested+Solutions+V1+Test+FyBNVC06+Ch11 12+Waves Sound
http://slidepdf.com/reader/full/suggestedsolutionsv1testfybnvc06ch11-12waves-sound 1/19
Suggested Solutions V1 FyBNVC06 Ch11-12 Mechanical waves – Sound NV-College
Physics B: FyBNVC06Mechanical Waves, SHO, Sound
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
Pass (G): minimum 20 pointsVG: minimum 40 points, minimum 12 VG pointsMVG: minimum 45 points, minimum 24 VG points, ¤ MVG-quality work.
© [email protected] Not for Sale . Free to use for educational purposes. 1/19
8/3/2019 Suggested+Solutions+V1+Test+FyBNVC06+Ch11 12+Waves Sound
http://slidepdf.com/reader/full/suggestedsolutionsv1testfybnvc06ch11-12waves-sound 2/19
Suggested Solutions V1 FyBNVC06 Ch11-12 Mechanical waves – Sound NV-College
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.
© [email protected] Not for Sale . Free to use for educational purposes. 2/19
8/3/2019 Suggested+Solutions+V1+Test+FyBNVC06+Ch11 12+Waves Sound
http://slidepdf.com/reader/full/suggestedsolutionsv1testfybnvc06ch11-12waves-sound 3/19
Suggested Solutions V1 FyBNVC06 Ch11-12 Mechanical waves – Sound NV-College
5. An opera singer’s voice is able to break a thin crystal glass when the singer’s voice and
the vibrating glass have identical
a) speed
b) amplitude
c) frequency
d) wavelength
Answer: Alternative (c): frequency [1/0] The thin crystal breaks due
to the resonance phenomenon, i.e. it starts vibrating at the frequency
of high pitched opera singer, and due to the fact that crystal glass are
rigid, it shatters.
6. The mass of harmonic oscillator of period s0.1 is increased by a factor of nine, i.e.
12 9mm = . The period of the new harmonic oscillator isa) s03.0
b) s0.1 .
c) s0.3
d) s0.9
Answer: Alternative (c): s0.3 [1/0]k m
T ⋅= π 2 is used. The period is
proportional to the square root of the mass. If mass increases, then period would
increase but only proportional to the square root of the mass: mT ∝ .
7. Sound waves are fastest in
a) solids.
b) liquids.
c) gases.
d)
vacuum.Answer: Alternative (a): solids. [1/0]
Why? Explain below: The speed of sound is greatest in solids. Solids, in
general are denser than liquids and liquids have higher density than
gases. Atoms are closer to each other and therefore are able to kick each
other faster in liquid and solids than gases: ρ / Bv = . Note that solids and
specially liquids are much less compressible and have larger elastic
moduli, B . [0/1]
© [email protected] Not for Sale . Free to use for educational purposes. 3/19
8/3/2019 Suggested+Solutions+V1+Test+FyBNVC06+Ch11 12+Waves Sound
http://slidepdf.com/reader/full/suggestedsolutionsv1testfybnvc06ch11-12waves-sound 4/19
Suggested Solutions V1 FyBNVC06 Ch11-12 Mechanical waves – Sound NV-College
8. Of the following procedures, which ones will higher the pitch of a guitar string?
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: Alternatives (b), (c) and (f). [2/0]
This is due to the fact that:λ v
f = and Lm
F v T
/ = . Therefore if we
increase the velocity of the propagation of wave v , the frequency will
be increased. This can be achieved by either increasing the tension T F
or by lowering the density of the string, i.e. by decreasing Lm
. This
can be easily achieved by using lighter string. The string therefore
may be thinner, or is made of lighter material for example plastic
instead of metal. Changing the length is more complicated. This is
due to the fact that the length of the wire appears in two places
simultaneously.
If we consider the fundamental frequency, L L 22 1
1 =⇔= λ λ
m LF
mF L
L LmF
L Lvv
f T T T
⋅=⋅====
21
21
21
211 λ
Therefore, shortening of the string will increase the frequency! Note
that this is identical conclusion that one may get it directly from
LmF
v T
/ = , but reasoning solely based on
LmF
v T
/ = is incomplete and
possibly misleading.
© [email protected] Not for Sale . Free to use for educational purposes. 4/19
8/3/2019 Suggested+Solutions+V1+Test+FyBNVC06+Ch11 12+Waves Sound
http://slidepdf.com/reader/full/suggestedsolutionsv1testfybnvc06ch11-12waves-sound 5/19
Suggested Solutions V1 FyBNVC06 Ch11-12 Mechanical waves – Sound NV-College
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]
© [email protected] Not for Sale . Free to use for educational purposes. 5/19
8/3/2019 Suggested+Solutions+V1+Test+FyBNVC06+Ch11 12+Waves Sound
http://slidepdf.com/reader/full/suggestedsolutionsv1testfybnvc06ch11-12waves-sound 6/19
Suggested Solutions V1 FyBNVC06 Ch11-12 Mechanical waves – Sound NV-College
© [email protected] Not for Sale . Free to use for educational purposes. 6/19
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 , ?=
T F , ?= L
sm Amk vkAmv / 60.003.0
05.020
21
21 22 =×==⇔= smv / 60.0=
-2,0
-1,5
-1,0
-0,5
0,0
0,5
1,0
1,5
2,0
0,25 0,75 1,25 1,75 2,25S
Rope
v v
S
8/3/2019 Suggested+Solutions+V1+Test+FyBNVC06+Ch11 12+Waves Sound
http://slidepdf.com/reader/full/suggestedsolutionsv1testfybnvc06ch11-12waves-sound 7/19
Suggested Solutions V1 FyBNVC06 Ch11-12 Mechanical waves – Sound NV-College
© [email protected] Not for Sale . Free to use for educational purposes. 7/19
15. How long the block in the problem above, is in contact with the spring.a) s14.3
b) s314.0
c) s21085.7 −×
d) None: _____________
Answer: Alternative (c): s21085.7 −× [1/0]
Why? Show the details of your calculations in the space provided below: [0/2]
ssk m
T 314.010
05.022005.0
22 ==⋅=⋅=⋅= π π π π
The block was in contact for only a quarter of a period:
sst contact 21085.70785.0
44−×==== sT 314.0 [0/2]
8/3/2019 Suggested+Solutions+V1+Test+FyBNVC06+Ch11 12+Waves Sound
http://slidepdf.com/reader/full/suggestedsolutionsv1testfybnvc06ch11-12waves-sound 8/19
Suggested Solutions V1 FyBNVC06 Ch11-12 Mechanical waves – Sound NV-College
© [email protected] Not for Sale . Free to use for educational purposes. 8/19
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]
Suggested solution:
Data: mkgmkgmg Lm
/ 100.6 / 1060.0 / 60.0 43 −− ×=×== , mcm L 330.00.33 == ,
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.
8/3/2019 Suggested+Solutions+V1+Test+FyBNVC06+Ch11 12+Waves Sound
http://slidepdf.com/reader/full/suggestedsolutionsv1testfybnvc06ch11-12waves-sound 9/19
Suggested Solutions V1 FyBNVC06 Ch11-12 Mechanical waves – Sound NV-College
© [email protected] Not for Sale . Free to use for educational purposes. 9/19
Second Method:
B
A A B
A
B
B
A
B
A
f f
L L L L
Lv Lv
f f
Lvv
f v
f ⋅=⇔==⇔==⇔=
2
221
1 λ λ
mmmmm f f
L L B
A A B 37037.0293.0330.0293.0
495440
330.0 ==−=⇔=⋅=⋅= 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 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.
8/3/2019 Suggested+Solutions+V1+Test+FyBNVC06+Ch11 12+Waves Sound
http://slidepdf.com/reader/full/suggestedsolutionsv1testfybnvc06ch11-12waves-sound 10/19
Suggested Solutions V1 FyBNVC06 Ch11-12 Mechanical waves – Sound NV-College
© [email protected] Not for Sale . Free to use for educational purposes. 10/19
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 .
dB I I
80log100
=⎟⎟
⎠
⎞⎜⎜
⎝
⎛ = β ⇔ 8log
0
=⎟⎟
⎠
⎞⎜⎜
⎝
⎛ I I
⇔ 8
0
10= I I [0/1]
241280
8 / 10101010 mw I I −− =⋅=⋅=
2 / mw AP
I = ⇔ 2m I P
A = ⇔ 22
4m
I P
r ⋅
=π
⇔ m I
Pr ⎥
⎦
⎤
⎢⎣
⎡
⋅=
π 4
mr ⎥
⎦
⎤
⎢
⎣
⎡
⋅
= −4
104
06.201
π
⇔ mr 400= [0/1] mr 400≈
8/3/2019 Suggested+Solutions+V1+Test+FyBNVC06+Ch11 12+Waves Sound
http://slidepdf.com/reader/full/suggestedsolutionsv1testfybnvc06ch11-12waves-sound 11/19
Suggested Solutions V1 FyBNVC06 Ch11-12 Mechanical waves – Sound NV-College
© [email protected] Not for Sale . Free to use for educational purposes. 11/19
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
≈⋅= π
8/3/2019 Suggested+Solutions+V1+Test+FyBNVC06+Ch11 12+Waves Sound
http://slidepdf.com/reader/full/suggestedsolutionsv1testfybnvc06ch11-12waves-sound 12/19
Suggested Solutions V1 FyBNVC06 Ch11-12 Mechanical waves – Sound NV-College
© [email protected] Not for Sale . Free to use for educational purposes. 12/19
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]
Suggested Solutions:
Doppler EffectPhysical Quantity Definition FormulaVelocity of sound smv / 343= Observedfrequency
Source moving at sv towardstationary observer ⎟
⎠ ⎞
⎜⎝ ⎛ −
=′
vv
f f
s1
Observedfrequency
Source moving ats
v away fromstationary observer ⎟
⎠ ⎞
⎜⎝ ⎛ +=′
vv
f f
s1
Observedfrequency
Observer moving towardstationary source ⎟
⎠ ⎞
⎜⎝ ⎛ +⋅=′
vv
f f o1
Observedfrequency
Observer moving away fromstationary source ⎟
⎠ ⎞
⎜⎝ ⎛ −⋅=′
vv
f f o1
a)
smhkmvs / 3
1000036
0010120 / .120 =
////×== [0.5]
Hz
vv
f f
s
443
3433.33
1
400
1
=⎟ ⎠ ⎞
⎜⎝ ⎛ −
=⎟ ⎠ ⎞
⎜⎝ ⎛ −
=′ [0.1] Answer: Hz f Guard 443=′ [0.5]
b)
smhkmvO / 250036
001090 / .90 =
////×==
The frequency of the siren heard by a police officer moving at hkm / .90 inthe opposite direction as the police car approaches the ambulance:
( )( )
( )( )
Hzvvvv f
vv
vv
f v
v f f S
OO
S
O 4753.33343
2534340011
1 =−+⋅=
−+⋅=⎟
⎠ ⎞⎜
⎝ ⎛ +⋅
⎟ ⎠ ⎞
⎜⎝ ⎛ −
=⎟ ⎠ ⎞⎜
⎝ ⎛ +⋅′=′′ [0/2]
The frequency of the siren heard by a police officer moving at hkm / .90 inthe opposite direction as the police car moves away from the ambulance:
( )( )
( )( )
Hzvvvv
f v
v
vv
f v
v f f
S
OO
S
O 3383.33343
253434001
11 =
+−
⋅=+−
⋅=⎟ ⎠ ⎞
⎜⎝ ⎛ −⋅
⎟ ⎠ ⎞
⎜⎝ ⎛ +
=⎟ ⎠ ⎞
⎜⎝ ⎛ −⋅′=′′ [0/2]
Answer: Moving toward Hz f 475=′′ . Moving away: Hz f 338=′′
8/3/2019 Suggested+Solutions+V1+Test+FyBNVC06+Ch11 12+Waves Sound
http://slidepdf.com/reader/full/suggestedsolutionsv1testfybnvc06ch11-12waves-sound 13/19
Suggested Solutions V1 FyBNVC06 Ch11-12 Mechanical waves – Sound NV-College
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.
© [email protected] Not for Sale . Free to use for educational purposes. 13/19
8/3/2019 Suggested+Solutions+V1+Test+FyBNVC06+Ch11 12+Waves Sound
http://slidepdf.com/reader/full/suggestedsolutionsv1testfybnvc06ch11-12waves-sound 14/19
Suggested Solutions V1 FyBNVC06 Ch11-12 Mechanical waves – Sound NV-College
© [email protected] Not for Sale . Free to use for educational purposes. 14/19
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.
8/3/2019 Suggested+Solutions+V1+Test+FyBNVC06+Ch11 12+Waves Sound
http://slidepdf.com/reader/full/suggestedsolutionsv1testfybnvc06ch11-12waves-sound 15/19
Suggested Solutions V1 FyBNVC06 Ch11-12 Mechanical waves – Sound NV-College
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.
© [email protected] Not for Sale . Free to use for educational purposes. 15/19
8/3/2019 Suggested+Solutions+V1+Test+FyBNVC06+Ch11 12+Waves Sound
http://slidepdf.com/reader/full/suggestedsolutionsv1testfybnvc06ch11-12waves-sound 16/19
Suggested Solutions V1 FyBNVC06 Ch11-12 Mechanical waves – Sound NV-College
© [email protected] Not for Sale . Free to use for educational purposes. 16/19
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=
On the other hand if 2
3λ
δ = . The implication is:
m123.03
185.023
2 =×== δ λ
Hzmsmv
f 2781123.0
/ 343 ===λ
This is well beyond the expected values!
8/3/2019 Suggested+Solutions+V1+Test+FyBNVC06+Ch11 12+Waves Sound
http://slidepdf.com/reader/full/suggestedsolutionsv1testfybnvc06ch11-12waves-sound 17/19
Suggested Solutions V1 FyBNVC06 Ch11-12 Mechanical waves – Sound NV-College
© [email protected] Not for Sale . Free to use for educational purposes. 17/19
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
⋅= 0.8
8/3/2019 Suggested+Solutions+V1+Test+FyBNVC06+Ch11 12+Waves Sound
http://slidepdf.com/reader/full/suggestedsolutionsv1testfybnvc06ch11-12waves-sound 18/19
Suggested Solutions V1 FyBNVC06 Ch11-12 Mechanical waves – Sound NV-College
© [email protected] Not for Sale . Free to use for educational purposes. 18/19
25. a) Show that if the tension in a stretched string is changed by T F Δ , the frequency of thefundamental is changed by
f F F
f T
T ⋅Δ=Δ2
[0/4/¤]
b) By what percent must the tension in a piano string be increased or decreased to raisethe frequency from Hz436 to Hz440 ? [0/2]
Suggested Solutions:
Wave on astring
The velocity of a wave v on a stretchedstring of mass m and length L underwhich is under tension T F .
LmF
v T
/ =
Velocity Velocity of propagation of the waveλ
λ ⋅== f
T v
λ ⋅== f Lm
F v T
/ . Square it: 22
/ λ ⋅= f
LmF T
Multiply both sides of the equation by Lm
: Lm
f F T ⋅⋅= 22 λ [0/1]
Differentiate w.r.t. f : f Lm
df dF T ⋅⋅⋅= 22 λ [0/1]
Multiply both sides by df : df f
L
mdF T
⋅⋅⋅⋅= 22 λ
Using Lm
f F T ⋅⋅= 22 λ ⇔
22
f F
Lm T =⋅λ and substituting it in the equation above:
df f F
df f f F
df f Lm
dF T T T
⋅⋅=⋅⋅⎟⎟
⎠
⎞⎜⎜
⎝
⎛ ⋅=⋅⋅⎟
⎠ ⎞
⎜⎝ ⎛ ⋅⋅= 222 2
2λ ⇔ df f F
dF T T
⋅⋅= 2 [0/1]
Multiply both sides byT F
f ⋅2
:
df f F
F f
dF F
f T
T T
T
⋅//⋅//
⋅/⋅//
/=⋅⋅
222
⇔ df dF F
f T
T
=⋅⋅2
⇔T
T
dF F
f df ⋅
⋅=
2 [0/1]
Rearrange the terms and change the infinitesimal change in frequency andtension: df and T dF to small changes in them, i.e.: f df Δ→ and T T F dF Δ→ :
f F F
f T
T ⋅Δ=Δ2
QED. [0/0/¤]
f F F
f T
T ⋅Δ=Δ2
⇔
T
T
F F
f f
2
Δ=Δ
Answer: To increase the frequency from Hz436 to Hz440 , we must raise
the tension in the wire by %8.1≈Δ f f :
%8.1%82.10182.0110
2440
4222
≈===⋅=Δ⇔Δ=Δ⇔Δ=ΔT
T
T
T
T
T
F F
f f
F F
F F
f f [0/2]
8/3/2019 Suggested+Solutions+V1+Test+FyBNVC06+Ch11 12+Waves Sound
http://slidepdf.com/reader/full/suggestedsolutionsv1testfybnvc06ch11-12waves-sound 19/19
Suggested Solutions V1 FyBNVC06 Ch11-12 Mechanical waves – Sound NV-College
Second method:
λ ⋅== f Lm
F v T
/ ⇔ ( )2
/ λ ⋅= f
LmF T ⇔
( )( ) ( )2
22
2
2
22
22
2 f
f f f f f
f f F
F F
Lm
f f F F
L
m f F
T
T T
T T
T
Δ+Δ⋅+=Δ+=Δ+⇔
⎪
⎪
⎩
⎪
⎪
⎨
⎧
⎟ ⎠ ⎞
⎜⎝ ⎛ ⋅⋅Δ+=Δ+
⎟
⎠
⎞⎜
⎝
⎛ ⋅⋅=
λ
λ
Due to the fact that ( ) f f <<Δ 2 we may ignore this term:
2
2 2 f
f f f F
F F
T
T T Δ⋅+≈Δ+
2
211
f f f
F F
T
T Δ⋅+≈Δ+
2
2 f
f f F F
T
T Δ⋅
≈Δ
QED
© [email protected] Not for Sale . Free to use for educational purposes. 19/19