SM PDF Chapter13-1

8/6/2019 SM PDF Chapter13-1

1/20

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.


2/20

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.


3/20

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


4/20


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 .


5/20

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


6/20


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


7/20

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


8/20


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


9/20

Chapter 13 369

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. .


10/20


*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


11/20

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


12/20


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


13/20

Chapter 13 373

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


14/20


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 .


15/20

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..

. .


16/20


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.


17/20

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


18/20


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 .


19/20

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


20/20


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


P13.14 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


P13.50 130 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

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