sm pdf chapter13-1
TRANSCRIPT
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361
Mechanical WavesCHAPTER OUTLINE
13.1 Propagation of aDisturbance
13.2 The Wave Model13.3 The Traveling Wave13.4 The Speed of Transverse
Waves on Strings13.5 Reflection and
Transmission of Waves13.6 Rate of Energy Transfer
by Sinusoidal Waves onStrings
13.7 Sound Waves13.8 The Doppler Effect13.9 Context
ConnectionSeismicWaves
ANSWERS TO QUESTIONS
Q13.1 To use a slinky to create a longitudinal wave, pull a few coils backand release. For a transverse wave, jostle the end coil side to side.
Q13.2 From vT
=
, we must increase the tension by a factor of 4.
Q13.3 It depends on from what the wave reflects. If reflecting from a lessdense string, the reflected part of the wave will be right side up.
Q13.4 The section of rope moves up and down in SHM. Its speed is always changing. The wave continues onwith constant speed in one direction, setting further sections of the rope into up-and-down motion.
Q13.5 As the source frequency is doubled, the speed of waves on the string stays constant and thewavelength is reduced by one half.
Q13.6 As the source frequency is doubled, the speed of waves on the string stays constant.
Q13.7 Higher tension makes wave speed higher. Greater linear density makes the wave move moreslowly.
Q13.8 As the wave passes from the massive string to the less massive string, the wave speed will increase
according to vT
=
. The frequency will remain unchanged. Since v f= , the wavelength must
increase.
Q13.9 Amplitude is increased by a factor of 2 . The wave speed does not change.
Q13.10 Sound waves are longitudinal because elements of the mediumparcels of airmove parallel andantiparallel to the direction of wave motion.
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362 Mechanical Waves
Q13.11 We assume that a perfect vacuum surrounds the clock. The sound waves require a medium for themto travel to your ear. The hammer on the alarm will strike the bell, and the vibration will spread assound waves through the body of the clock. If a bone of your skull were in contact with the clock,you would hear the bell. However, in the absence of a surrounding medium like air or water, nosound can be radiated away. A larger-scale example of the same effect: Colossal storms raging on theSun are deathly still for us.
What happens to the sound energy within the clock? Here is the answer: As the sound wavetravels through the steel and plastic, traversing joints and going around corners, its energy isconverted into additional internal energy, raising the temperature of the materials. After the soundhas died away, the clock will glow very slightly brighter in the infrared portion of theelectromagnetic spectrum.
Q13.12 The frequency increases by a factor of 2 because the wave speed, which is dependent only on themedium through which the wave travels, remains constant.
Q13.13 When listening, you are approximately the same distance from all of the members of the group. Ifdifferent frequencies traveled at different speeds, then you might hear the higher pitchedfrequencies before you heard the lower ones produced at the same time. Although it might be
interesting to think that each listener heard his or her own personal performance depending onwhere they were seated, a time lag like this could make a Beethoven sonata sound as if it werewritten by Charles Ives.
Q13.14 He saw the first wave he encountered, light traveling at 3 00 108. m s . At the same moment,
infrared as well as visible light began warming his skin, but some time was required to raise thetemperature of the outer skin layers before he noticed it. The meteor produced compressional wavesin the air and in the ground. The wave in the ground, which can be called either sound or a seismicwave, traveled much faster than the wave in air, since the ground is much stiffer againstcompression. Our witness received it next and noticed it as a little earthquake. He was no doubtunable to distinguish the P and S waves. The first air-compression wave he received was a shockwave with an amplitude on the order of meters. It transported him off his doorstep. Then he couldhear some additional direct sound, reflected sound, and perhaps the sound of the falling trees.
Q13.15 For the sound from a source not to shift in frequency, the radial velocity of the source relative to theobserver must be zero; that is, the source must not be moving toward or away from the observer.The source can be moving in a plane perpendicular to the line between it and the observer. Otherpossibilities: The source and observer might both have zero velocity. They might have equalvelocities relative to the medium. The source might be moving around the observer on a sphere ofconstant radius. Even if the source speeds up on the sphere, slows down, or stops, the frequencyheard will be equal to the frequency emitted by the source.
Q13.16 Wind can change a Doppler shift but cannot cause one. Both vo and vs in our equations must be
interpreted as speeds of observer and source relative to the air. If source and observer are movingrelative to each other, the observer will hear one shifted frequency in still air and a different shifted
frequency if wind is blowing. If the distance between source and observer is constant, there willnever be a Doppler shift.
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Chapter 13 363
Q13.17 Let t t ts p= represent the difference in arrival times of the two waves at a station at distance
d v t v ts s p p= = from the hypocenter. Then d tv vs p
= FHG
IKJ
1 1
1
. Knowing the distance from the first
station places the hypocenter on a sphere around it. A measurement from a second station limits itto another sphere, which intersects with the first in a circle. Data from a third non-collinear stationwill generally limit the possibilities to a point.
SOLUTIONS TO PROBLEMS
Section 13.1 Propagation of a Disturbance
P13.1 Replace x by x vt x t = 4 5. to get y
x t=
+
6
4 5 32
.a f
P13.2
FIG. P13.2
Section 13.2 The Wave Model
Section 13.3 The Traveling Wave
P13.3 f= =40 0 4
3
. vibrations
30.0 sHz v = =
42542 5
cm
10.0 scm s.
= = = =v
f
42 531 9 0 319
43
.. .
cm s
Hzcm m
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P13.4 Using data from the observations, we have = 1 20. m and f=8 00
12 0
.
. s. Therefore,
v f= =FHG
IKJ= 1 20
8 00
12 00 800.
.
..m
sm sa f .
P13.5 (a) Let u t x= +10 3 4
du
dt
dx
dt= =10 3 0 at a point of constant phase
dx
dt= =
10
33 33. m s
The velocity is in the positive -directionx .
(b) y 0 100 0 0 350 0 3004
0 054 8 5 48. , . sin . . .b g a f= +FHGIKJ= = m m cm
(c) k= =2
3
: = 0 667. m = =2 10f : f= 5 00. Hz
(d) vy
tt xy =
= +FHG
IKJ0 350 10 10 3 4. cosa fa f
vy , . .max m s= =10 0 350 11 0a fa f
P13.6 y x t= 0 020 0 2 11 3 62. sin . .mb g a f in SI units A = 2 00. cm
k= 2 11. rad m
= =2
2 98k
. m
= 3 62. rad s f= =
20 576. Hz
v fk
= = = =
2
2 3 62
2 111 72
.
.. m s
P13.7 (a) = = =2 2 5 31 41f s rad se j .
(b) = = =
v
f
204 00
1
m s
5 sm.
k= = =2 2
41 57
mrad m.
(c) In y A kx t= +sin b g we take = 12 cm. At x = 0 and t = 0 we have y = 12 cma fsin. Tomake this fit y = 0, we take = 0. Then y x t= 12 0 1 57 31 4. sin . .cm rad m rad sa f b g b gd i
(d) The transverse velocity is
yt
A kx t= cosb g . Its maximum magnitude is
A = =12 3 77cm 31.4 rad s m sb g .
(e) av
t t A kx t A kx ty
y= = =
cos sinb gd i b g2
The maximum value is A2 12
0 12 31 4 118= =. .m s m s2a fe j .
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Chapter 13 365
*P13.8 At time t, the phase of y x t= 15 0 0 157 50 3. cos . .cma f a f at coordinate x is
= 0 157 50 3. .rad cm rad sb g b gx t . Since 60 03
. =
rad , the requirement for point B is that
B A= 3
rad , or (since xA = 0 ),
0 157 50 3 0 50 33
. . .rad cm rad s rad s radb g b g b gx t tB =
.
This reduces to xB =
= rad
rad cmcm
3 0 1 576 67
..
b g.
P13.9 (a) A y= = =max . .8 00 0 080 0cm m k= = =2 2
0 8007 85 1
..
mm
a f
= = =2 2 3 00 6 00f . .a f rad sTherefore, y A kx t= +sin b g
Or (where y t0 0,b g = at t = 0 ) y x t= +0 080 0 7 85 6. sin .b g b g m
(b) In general, y x t= + +0 080 0 7 85 6. sin . b g
Assuming y x, 0 0b g = at x = 0100. m
then we require that 0 0 080 0 0 785= +. sin . b g
or = 0 785.
Therefore, y x t= + 0 080 0 7 85 6 0 785. sin . .b g m
P13.10 y x t= +FHG
IKJ0120 8 4. sinma f
(a) vdy
dty = : v x ty = +
FHG
IKJ0 120 4 8 4. cosa fa f
vy 0 200 1 51. .s, 1.60 m m sa f =
adv
dty = : a x ty = +
FHG
IKJ0 120 4 8 4
2. sinma fa f
ay 0 200 0. s, 1.60 ma f =
(b) k= =
8
2: = 16 0. m
= =42
T: T= 0 500. s
vT
= = = 16 0
32 0.
.m
0.500 sm s
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P13.11 (a) Let us write the wave function as y x t A kx t, sinb g b g= + +
y A0 0 0 020 0, sin .b g = = m
dy
dtA
0 0
2 00,
cos .= = m s
Also,
= = =2 2
0025080 0
T ..
ss
A xv
ii2 2
22
2
002002 00
80 0= +
FHG
IKJ = +
FHG
IKJ
..
.m
m s
sb g
A = 0 021. 5 m
(b)A
A
sin
cos
.. tan
.
= = =
002002 51
280 0
Your calculators answer tan . .
=
1
2 51 1 19a f rad has a negative sine and positive cosine,just the reverse of what is required. You must look beyond your calculator to find = =1 19 1 95. .rad rad
(c) v Ay , . . .max m s m s= = = 0 021 5 80 0 5 41b g
(d) = = =v Tx 30 0 0 025 0 750. . .m s 0 s mb ga f
k= = =2 2
0 750
. m8.38 m = 80 0. s
y x t x t, . sin . . .b g b g b g= + +0 021 5 8 38 80 0 1 95m rad m rad s rad
P13.12 The linear wave equation is
=
2
2 2
2
2
1y
x v
y
t
If y eb x vt= a f
then
=
y
tbveb x vta f and
=
y
xbeb x vta f
=
2
22 2y
tb v eb x vta f and
=
2
22y
xb eb x vta f
Therefore,
=
2
22
2
2yt
v yx
, demonstrating that eb x vta f is a solution
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Chapter 13 367
Section 13.4 The Speed of Transverse Waves on Strings
P13.13 The down and back distance is 4 00 4 00 8 00. . .m m m+ = .
The speed is then vd
t
T= = = =
total m
sm s
4 8 00
0 80040 0
.
..
a f
Now, = = 0 200
5 00 10 2.
.kg
4.00 mkg m
So T v= = = 2 22
5 00 10 40 0 80 0. . .kg m m s Ne jb g
P13.14 T Mg= is the tension; vT Mg MgL
m
L
tmL= = = =
is the wave speed.
Then,MgL
m
L
t=
2
2
and gLm
Mt= =
=
2
3
2 2
1 60 4 00 10
3 00 101 64
. .
..
m kg
kg 3.61 sm s2
e j
e j
*P13.15 Since
is constant, = =T
v
T
v2
22
1
12 and
Tv
vT2
2
1
2
1
230 0
20 06 00 13 5=
FHG
IKJ
=FHG
IKJ
=.
.. .
m s
m sN Na f .
P13.16 From the free-body diagram mg T= 2 sin
T mg=2sin
The angle is found from cos= =38
2
3
4
L
L
= 41 4.
r
Tr
T
mr
g
FIG. P13.16
(a) vT
=
vmg
m=
=
F
HGG
I
KJJ2 41 4
9 80
2 8 00 10 41 43sin .
.
. sin .
m s
kg m
2
e j
or v m=F
HG
I
KJ30 4.
m s
kg
(b) v m= =60 0 30 4. . and m = 3 89. kg
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P13.17 The total time is the sum of the two times.
In each wire tL
vL
T= =
LetA represent the cross-sectional area of one wire. The mass of one wire can be written both asm V L= = and also as m L= .
Then we have
= =Ad 2
4
Thus, t Ld
T=
FHG
IKJ
21 2
4
For copper, t =
L
N
MMM
O
Q
PPP
=
20 08 920 1 00 10
4 1500 137
3 21 2
..
.a fa fb ge j
a fa f
s
For steel, t =
L
N
MMM
O
Q
PPP =
30 0
7 860 1 00 10
4 150 0 192
3 21 2
.
.
.a f
a fb ge ja fa f
s
The total time is 0137 0192 0 329. . .+ = s
Section 13.5 Reflection and Transmission of Waves
P13.18 (a) If the end is fixed, there is inversion of the pulse upon reflection. Thus, when they meet,
they cancel and the amplitude is zero .
(b) If the end is free, there is no inversion on reflection. When they meet, the amplitude is2 2 0 150 0 300A = =. .m ma f .
Section 13.6 Rate of Energy Transfer by Sinusoidal Waves on Strings
P13.19 fv
= = =
30 0
0 50060 0
.
.. Hz = =2 120f rad s
P= =FHG
IKJ =
1
2
1
2
0 180
3 60120 0 100 30 0 1 072 2
2 2 A v
.
.. . .a f a f a f kW
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P13.20 = = 30 0 30 0 10 3. .g m kg m
=
= = =
= =
1 50
50 0 2 314
2 0 150 7 50 10
1
2
.
. :
. : .
m
Hz s
m m
f f
A
(a) y A x t= FHG
IKJsin
2
y x t= 7 50 10 4 19 3142. sin .e j a f FIG. P13.20
(b) P= = FHG
IKJ
1
2
1
230 0 10 314 7 50 10
314
4 192 2 3 2 2 2 A v . .
.e ja f e j W P= 625 W
P13.21 A = 5 00 10 2. m = 4 00 10 2. kg m P= 300 W T= 100 N
Therefore, vT
= =
50 0. m s
P =1
22 2 A v :
22 2 2 2
2 2 300
4 00 10 5 00 10 50 0= =
P
A v
a f
e je j a f. . .
=
= =
346
255 1
rad s
Hzf .
*P13.22 Originally,
P
P
P
02 2
02 2
02 2
1
21
2
1
2
=
=
=
A v
AT
A T
The doubled string will have doubled mass-per-length. Presuming that we hold tension constant, it
can carry power larger by 2 times.
21
220
2 2P = A T
Section 13.7 Sound Waves
P13.23 Since v vlight sound>> : d =343 16 2 5 56m s s kmb ga f. .
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*P13.24 The sound pulse must travel 150 m before reflection and 150 m after reflection. We have d vt=
td
v= = =
3000196
m
1 533 m ss.
P13.25 (a) = = =
v
f
343
1 480 0 2321
m s
s m.
(b) =
= =
v
f
343
13970 246
1
m s
sm. , = = 13 8. mm
P13.26 = =
=
v
f
340
60 0 105 67
3 1
m s
smm
..
*P13.27 The sound speed is v = + =331 0 6 26 347m s m s C C m s. a f
(a) Let t represent the time for the echo to return. Then
d vt= = =1
2
1
2347 24 10 4 163m s s m. .
(b) Let t represent the duration of the pulse:
tv f f
= = = =
=10 10 10 10
22 100 455
6
1 ss. .
(c) Lv
f= = =
=10
10 10 347
22 100158
6
m s
1 smm
b g.
*P13.28 (a) Iff= 2 4. MHz , = =
=v
f
1500
2 4 100 625
6
m s
smm
..
(b) Iff= 1 MHz, = = =v
f
1500
101 50
6
m s
smm.
Iff= 20 MHz, =
=1500
2 1075 0
7
m s
sm.
P13.29 P v s vv
smax max max= =FHG
IKJ
2
= =
=
2 2 1 20 343 5 50 10
0 8405 81
2 2 6v s
Pmax
max
. .
..
a fa f e jm
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Chapter 13 371
P13.30 (a) A = 2 00. m
= = =2
15 70 400 40 0
.. .m cm
vk
= = = 858
15 754 6
.. m s
(b) s = = 2 00 15 7 0 050 0 858 3 00 10 0 4333. cos . . . .a fb g a fe j m
(c) v Amax . .= = = 2 00 858 1 721m s mm sb ge j
P13.31 k= = = 2 2
0 10062 8 1
..
mm
a f
= = =
2 2 343
01002 16 104 1
v m s
ms
b ga f.
.
Therefore, P x t= 0 200 62 8 2 16 104
. sin . .N m m s2
e j .
P13.32 P vsmax max . .= = 1 20 2 2 000 343 2 00 101 8kg m s m s m3e j e j b ge j
Pmax .= 0 103 Pa
Section 13.8 The Doppler Effect
P13.33 (a) =+
f
f v v
v v
o
s
b gb g
=+
=f 2 500
343 25 0
343 40 03 04
.
..
a fa f
kHz
(b) =+
FHG
IKJ
=f 2 500343 25 0
343 40 02
.
( . )
a f.08 kHz
(c) =+
FHG
IKJ
=f 2 500343 25 0
343 40 02
.
.
a f.62 kHz while police car overtakes
=+
F
HG
I
KJ=f 2 500
343 25 0
343 40 0
2.
.a f
.40 kHz after police car passes
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P13.34 (a) = =FHG
IKJ
=2 2115
60 012 0f
min
s minrad s
..
v Amax . . .= = = 12 0 1 80 10 0 021 73rad s m m sb ge j
(b) The heart wall is a moving observer.
=+F
HGIKJ=
+FHG
IKJ
=f fv v
vO 2 000 000
1 500 0 021 7
15002 000 028 9Hz Hzb g . .
(c) Now the heart wall is a moving source.
=
FHG
IKJ
=
FHG
IKJ
=f fv
v vs2 000 029
1500
1 500 0 021 72 000 057 8Hz Hzb g
..
P13.35 Approaching ambulance: =
ff
v vS1b g
Departing ambulance: =
ff
v vS1 b gd i
Since =f 560 Hz and = 480 Hz 560 1 480 1FHG
IKJ= +
FHG
IKJ
v
v
v
vS S
1 040 80 0
80 0 343
104026 4
v
v
v
S
S
=
= =
.
..
a fm s m s
P13.36 The maximum speed of the speaker is described by
12
12
20 0
5 000 500 1 00
2 2mv kA
vk
mA
max
max
.
.. .
=
= = =N m
kgm m sa f
The frequencies heard by the stationary observer range from
=+
FHG
IKJ
f fv
v vmin
max
to =
FHG
IKJ
f fv
v vmax
max
where v is the speed of sound.
=+
FHG
IKJ
=
=
FHG
IKJ
=
f
f
min
max
.
.
440343 1 00
439
440343 1 00
441
Hz343 m s
m s m sHz
Hz343 m s
m s m sHz
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P13.37 =
FHG
IKJ
f fv
v vs485 512
340
340 9 80=
FHG
IKJ. tfallb g
485 340 485 9 80 512 340
512 485
485
340
9 801 93
a f a fd i a fa f+ =
=F
HG I
KJ =
.
..
t
t
f
f s
d gtf121
218 3= = . m : treturn s= =
18 3
34000538
..
The fork continues to fall while the sound returns.
t t t
d gt
ftotal fall return
total total fall2
s s s
m
= + = + =
= =
1 93 0 053 8 1 985
1
219 3
. . .
.
P13.38 (a) v = +
=331 0 6 10 325m sm
s CC m sb g a f.
(b) Approaching the bell, the athlete hears a frequency of = +FHG IKJf fv v
vO
After passing the bell, she hears a lower frequency of =+ F
HGIKJ
f fv v
v
Ob g
The ratio is
=
+=
f
f
v v
v vO
O
5
6
which gives 6 6 5 5v v v vo o = + or vv
O = = =11
325
1129 5
m sm s.
P13.39 (a) Sound moves upwind with speed 343 15
a f m s . Crests pass a stationary upwind point atfrequency 900 Hz.
Then = = =v
f
328
9000 364
m s
sm.
(b) By similar logic, = =+
=v
f
343 15
9000 398
a f m ss
m.
(c) The source is moving through the air at 15 m/s toward the observer. The observer isstationary relative to the air.
=+
F
HG
I
KJ=
+
F
HG
I
KJ=f f
v v
v v
o
s
900343 0
343 15941Hz Hz
(d) The source is moving through the air at 15 m/s away from the downwind firefighter. Herspeed relative to the air is 30 m/s toward the source.
=+
FHG
IKJ
=+
FHG
IKJ
=FHG
IKJ=f f
v v
v vo
s
900343 30
343 15900
373
358938Hz Hz Hz
a f
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Section 13.9 Context ConnectionSeismic Waves
P13.40 (a) The longitudinal wave travels a shorter distance and is moving faster, so it will arrive at
point B first.
(b) The wave that travels through the Earth must travel
a distance of 2 30 0 2 6 37 10 30 0 6 37 106 6Rsin . . sin . . = = m me j
at a speed of 7 800 m/s
Therefore, it takes6 37 10
8176.
=m
7 800 m ss
The wave that travels along the Earths surface must travel
a distance of s R R= =FHG
IKJ=
36 67 106rad m.
at a speed of 4 500 m/s
Therefore, it takes6 67 10
45001 482
6. = s
The time difference is 665 111s min= .
P13.41 The distance the waves have traveled is d t t= = +7 80 4 50 17 3. . .km s km s sb g b ga f
where t is the travel time for the faster wave.
Then, 7 80 4 50 4 50 17 3. . . . =a fb g b ga fkm s km s st
or t =
=4 50 17 37 80 4 50
23 6. .. .
.km s skm s
sb ga fa f
and the distance is d = =7 80 23 6 184. .km s s kmb ga f .
Additional Problems
P13.42 Assume a typical distance between adjacent people ~ 1 m .
Then the wave speed is vx
t=
~ ~
110
m
0.1 sm s
Model the stadium as a circle with a radius of order 100 m. Then, the time for one circuit around the
stadium is
Tr
v= =
2 2 10
1063 1
2
~ ~e jm s
s min .
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Chapter 13 375
P13.43 Assuming the incline to be frictionless and taking the positive x-direction to be up the incline:
F T Mg x = =sin 0 or the tension in the string is T Mg= sin
The speed of transverse waves in the string is then vT Mg
m L
MgL
m= = =
sin sin
The time interval for a pulse to travel the strings length is tL
vL
m
MgL
mL
Mg= = =
sin sin
P13.44 Mgx kx=1
22
(a) T kx Mg = = 2
(b) L L x LMg
k= + = +0 0
2
(c) vT TL
m
Mg
mL
Mg
k= = = +
FHG
IKJ
2 20
*P13.45 Let M = mass of block, m= mass of string. For the block, F ma = implies Tmv
rm rb= =
22 . The
speed of a wave on the string is then
vT M r
rM
m
t rv mM
tm
M
mr
= = =
= =
= = = =
2
1
0003200843
..
kg
0.450 kgrad
P13.46 (a) Assume the spring is originally stationary throughout, extended to have a length L muchgreater than its equilibrium length. We start moving one end forward with the speed v atwhich a wave propagates on the spring. In this way we create a single pulse of compressionthat moves down the length of the spring. For an increment of spring with length dx and
mass dm, just as the pulse swallows it up, F ma = becomes kdx adm= ork
admdx
= . But
dm
dx = so ak
= . Also, adv
dt
v
t= = when vi = 0. But L vt= , so av
L=
2
. Equating the two
expressions for a, we havek v
L=
2
or vkL
=
.
(b) Using the expression from part (a) vkL kL
m= = = =
2 2100 2 00
0 40031 6
N m m
kgm s
b ga f..
. .
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376 Mechanical Waves
P13.47 vT
=
where T xg= , the weight of a length x, of rope.
Therefore, v gx=
But vdx
dt= , so that dt
dx
gx=
and tdx
gx g
x L
g
L L
= = =z0
12 0
12
P13.48 At distance x from the bottom, the tension is Tmxg
LMg=
FHG
IKJ+ , so the wave speed is:
vT TL
mxg
MgL
m
dx
dt= = = +
FHG
IKJ= .
(a) Then t dt xg MgL
mdx
t L
= = +FHG
IKJ
LNM
OQPz z
0
1 2
0
tg
xg MgL m
x
x L
=+
=
=
11 2
12
0
b g
tg
LgMgL
m
MgL
m= +
FHG
IKJ
FHG
IKJ
L
NMM
O
QPP
21 2 1 2
tL
g
m M M
m=
+ FHG
IKJ
2
(b) When M = 0 , as in the previous problem, tL
g
m
m
L
g=
FHG
IKJ
=20
2
(c) As m 0 we expand m M Mm
MM
m
M
m
M+ = +
FHG
IKJ = + +
FHG
IKJ
1 11
2
1
8
1 2 2
2K
to obtain tL
g
M m M m M M
m
=+ + F
H
G
G
I
K
J
J
2
12
18
2 3 2e j e j K
tL
g
m
M
mL
Mg
FHG
IKJ
=21
2
P13.49 (a) P x A v A ek k
A ebx bxa f = = FHGIKJ=
1
2
1
2 22 2 2
02 2
3
02 2
(b) P 02
3
02a f =
kA
(c) PP
x e bxa fa f02
=
P13.50 v = = =4 450
468 130km
9.50 hkm h m s
dv
g= = =
2 2130
9 801730
m s
m sm
2
b ge j.
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Chapter 13 377
P13.51 (a) xa f is a linear function, so it is of the form x mx ba f = + To have 0 0a f = we require b = 0 . Then L mLLa f = = + 0
so mL
L=
0
Then
x xL
La f b g=
+0 0
(b) From vdx
dt= , the time required to move from x to x dx+ is
dx
v. The time required to move
from 0 to L is
tdx
v
dx
Tx dx
tT
x
L Ldx
L
tT
L x
L
tL
T
tL
T
L
T
L L
L L
L
L
L
L
L
LL
L L L
L L
= = =
=
+FHG
IKJ
FHG
IKJ
FHG
IKJ
=
FHG
IKJ
+
FHG
IKJ
=
= + +
z z z
z
0 0 0
00
1 2
0
00
0
00
3 2
32 0
0
3 203 2
0 0 0
0
1
1
1 1
2
3
2
3
a f
b g
b g
b g e j
e je je j +
=+ +
+
FHG
IKJ
0
0 0
0
2
3
e j
tL
T
L L
L
P13.52 Sound takes this time to reach the man:20 0 1 75
3435 32 10 2
. ..
m m
m ss
=
a f
so the warning should be shouted no later than 0 300 5 32 10 0 3532. . .s s s+ = before the pot strikes.
Since the whole time of fall is given by y gt=1
22 : 18 25
1
29 80 2. .m m s2= e jt
t = 1 93. s
the warning needs to come 1 93 0 353 158. . .s s s =
into the fall, when the pot has fallen1
2
9 80 1 58 12 22
. . .m s s m2e ja f =
to be above the ground by 20 0 12 2 7 82. . .m m m =
P13.53 Since cos sin2 2 1 + = , sin cos = 1 2 (each sign applying half the time)
P P kx t v s kx t= = max maxsin cos b g b g1 2
Therefore P v s s kx t v s s= = max max maxcos2 2 2 2 2b g
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378 Mechanical Waves
P13.54 The trucks form a train analogous to a wave train of crests with speed v = 19 7. m s
and unshifted frequency f= = 2
3 000 667 1
..
minmin .
(a) The cyclist as observer measures a lower Doppler-shifted frequency:
=+
FHG IKJ=+
FHG IKJ=f f v vv o 0 667 19 7 4 4719 7 0 5151. . .. .min mine j a f
(b) =+ F
HGIKJ=
+ FHG
IKJ
=f f
v v
vo 0 667
19 7 1 56
19 70 6141.
. .
..min mine j
a f
The cyclists speed has decreased very significantly, but there is only a modest increase inthe frequency of trucks passing him.
P13.55 vd
t=
2: d
vt= = =
2
1
26 50 10 1 85 6 013. . .m s s kme ja f
P13.56(a)
=
f
fv
v u =
f
v
v ua f =
+
F
HG
I
KJ f f fv v u v u1 1
ffv v u v u
v u
uvf
vf
uv
uv
uv
=+ +
=
=
a fe j
2 2 2
2
1
2
12
2
2
2
(b) 130 36 1km h m s= . =
=f2 36 1 400
340 185 9
36 1340
2
2
..
.
a fa fHz
*P13.57 (a) =
f fv
v vdiverb g
so 1 =
vv f
diver
=
FHG
IKJ
v vf
fdiver 1
with v = 343 m s , f= 1 800 Hz and =f 2 150 Hz
we find
vdiver m s= FHG
IKJ
=343 11800
215055 8. .
(b) If the waves are reflected, and the skydiver is moving into them, we have
= +
=
LNMM
OQPP
+f f
v v
vf f
v
v v
v v
v
diver
diver
diverb gb g
b g
so =+
=f 1800
343 55 8
343 55 82 500
.
.
a fa f
Hz .
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Chapter 13 379
*P13.58 (a)
FIG. P13.58(a)
(b) = = =
v
f
343
10000 343
1
m s
sm.
(c) =
=F
HGIKJ=
=
v
f
v
f
v v
vS 343 40 0
10000 303
1
..
a f m ss
m
(d) =
=+F
HGIKJ=
+=
v
f
v
f
v v
vS 343 40 0
10000 383
1
..
a f m ss
m
(e) =
FHG
IKJ
=
=f f
v v
v vO
S
1000343 30 0
343 40 01 03Hz
m s
m skHzb g a fa f
.
..
*P13.59 Use the Doppler formula, and remember that the bat is a moving source. If the velocity of the insectis vx ,
40 4 40 0340 5 00 340
340 5 00 340. .
.
.=
+
+
a fb ga fb g
v
v
x
x
.
Solving, vx = 3 31. m s . Therefore, the bat is gaining on its prey at 1.69 m s .
P13.60 (a) If the source and the observer are moving away from each other, we have: S = 0 180 ,
and since cos 180 1 = , we get Equation 13.30 with negative values for both vO and vS .
(b) If vO = 0 m s then =
fv
v vf
S Scos. Also, when the train is 40.0 m from the intersection,
and the car is 30.0 m from the intersection, cosS =4
5so
=
f343
343 0 800 25 0500
m s
m s m sHz
. .b ga f
or =f 531 Hz . Note that as the train approaches, passes, and departs from the
intersection, S varies from 0 to 180 and the frequency heard by the observer varies from:
=
=
=
=
=+
=
fv
v vf
fv
v vf
S
S
max
min
cos .
cos .
0
343
343 25 0500 539
180
343
343 25 0500 466
m s
m s m sHz Hz
m s
m s m sHz Hz
a f
a f
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380 Mechanical Waves
ANSWERS TO EVEN PROBLEMS
P13.2 see the solution
P13.4 0 800. m s
P13.6 2.00 cm, 2.98 m, 0.576 Hz, 1.72 m/s
P13.8 6 67. cm
P13.10 (a) 1.51 m/s, 0 m s2 ; (b) 16.0 m, 0.500 s,
32.0 m/s
P13.12 see the solution
P13.14 1 64. m s2
P13.16 (a) v m= FHG I
KJ30 4.
m s
kg; (b) 3.89 kg
P13.18 (a) zero; (b) 0.300 m
P13.20 (a) y x t= 7 50 10 4 19 3142. sin .e j a f ;(b) 625 W
P13.22 2 0P
P13.24 0.196 s
P13.26 5.67 mm
P13.28 (a) 0.625 mm; (b) 1.50 mm to 75 0. m
P13.30 (a) 2 00. m , 40.0 cm, 54.6 m/s;
(b) 0 433. m ; (c) 1.72 mm/s
P13.32 0.103 Pa
P13.34 (a) 0.021 7 m/s; (b) 2 000 028.9 Hz;(c) 2 000 057.8 Hz
P13.36 439 Hz and 441 Hz
P13.38 (a) 325 m/s; (b) 29.5 m/s
P13.40 (a) longitudinal ; (b) 665 s
P13.42 ~1 min
P13.44 (a) 2Mg; (b) LMg
k0
2+ ;
(c)2 2
0
Mg
mL
Mg
k+
FHG
IKJ
P13.46 (a) vkL
=
; (b) 31.6 m/s
P13.48 see the solution
P13.50 130 m/s, 1730 m
P13.52 7.82 m
P13.54 (a) 0 515. min; (b) 0 614. min
P13.56 (a)2
12
2
uv
uv
f
; (b) 85.9 Hz
P13.58 (a) see the solution; (b) 0.343 m;(c) 0.303 m; (d) 0.383 m; (e) 1.03 kHz
P13.60 (a) see the solution; (b) 531 Hz