solucion al capitulo 16 del serway

8/18/2019 solucion al capitulo 16 del serway

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16Wave Motion

CHAPTER OUTLINE

16.1 Propagation of a Disturbance

16.2 The Traveling Wave Model

16.3 The Speed of Waves on Strings

16.5 Rate of Energy Transfer by Sinusoidal

Waves on Strings

16.6 The Linear Wave Equation

ANSWERS TO QUESTIONS

Q16.1 As the pulse moves down the string, the particles of the

string itself move side to side. Since the medium—here,

the string—moves perpendicular to the direction of wave

propagation, the wave is transverse by definition.

Q16.2 To use a slinky to create a longitudinal wave, pull a few

coils back and release. For a transverse wave, jostle the

end coil side to side.

*Q16.3 (i) Look at the coefficients of the sine and cosine functions: 2, 4, 6, 8, 8, 7. The ranking is d =

e > f > c > b > a.

(ii) Look at the coefficients of x . Each is the wave number, 2π λ , so the smallest k goes with the

largest wavelength. The ranking is d > a = b = c > e > f.

(iii) Look at the coefficients of t . The absolute value of each is the angular frequency ω = 2π f . The

ranking is f > e > a = b = c = d.

(iv) Period is the reciprocal of frequency, so the ranking is the reverse of that in part iii: d = c =

b = a > e > f.

(v) From v = f λ = ω k , we compute the absolute value of the ratio of the coefficient of t to the

coefficient of x in each case. From a to f respectively the numerical speeds are 5, 5, 5, 7.5,

5, 4. The ranking is d > a = b = c = e > f.

*Q16.4 From v =T

µ , we must increase the tension by a factor of 4 to make v double. Answer (b).

*Q16.5 Answer (b). Wave speed is inversely proportional to the square root of linear density.

*Q16.6 (i) Answer (a). Higher tension makes wave speed higher.

(ii) Answer (b). Greater linear density makes the wave move more slowly.

Q16.7 It depends on from what the wave reflects. If reflecting from a less dense string, the reflected

part of the wave will be right side up.

Q16.8 Yes, among other things it depends on. The particle speed is described byv y,max

v

= = =

ω π

π

λ A fA

A

2

2

.Here v is the speed of the wave.

427

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428 Chapter 16

*Q16.9 (a) through (d): Yes to all. The maximum particle speed and the wave speed are related by

v y,max

v

= = =ω π π

λ A fA

A2

2. Thus the amplitude or the wavelength of the wave can be adjusted

to make either v y,max

or v larger.

Q16.10 Since the frequency is 3 cycles per second, the period is 13

second = 333 ms.

Q16.11 Each element of the rope must support the weight of the rope below it. The tension increases with

height. (It increases linearly, if the rope does not stretch.) Then the wave speed v = T

µ increases

with height.

*Q16.12 Answer (c). If the frequency does not change, the amplitude is increased by a factor of 2.

The wave speed does not change.

*Q16.13 (i) Answer a. As the wave passes from the massive string to the less massive string, the wave

speed will increase according to

v = T

µ .

(ii) Answer c. The frequency will remain unchanged. However often crests come up to the

boundary they leave the boundary.

(iii) Answer a. Since v = f λ , the wavelength must increase.

Q16.14 Longitudinal waves depend on the compressibility of the fluid for their propagation. Transverse

waves require a restoring force in response to shear strain. Fluids do not have the underlying

structure to supply such a force. A fluid cannot support static shear. A viscous fluid can tempo-

rarily be put under shear, but the higher its viscosity the more quickly it converts input work

into internal energy. A local vibration imposed on it is strongly damped, and not a source of

wave propagation.

Q16.15 Let ∆t t t s p

= − represent the difference in arrival times of the two waves at a station at distance

d t t s s p p= =v v from the hypocenter. Then

d t s p

= −

−

∆1 1

1

v v

. Knowing the distance from the

first station places the hypocenter on a sphere around it. A measurement from a second sta-tion limits it to another sphere, which intersects with the first in a circle. Data from a third

non-collinear station will generally limit the possibilities to a point.

Q16.16 The speed of a wave on a “massless” string would be infinite!

SOLUTIONS TO PROBLEMS

Section 16.1 Propagation of a Disturbance

P16.1 Replace x by x t x t − = −v 4 5.

to get y x t

=−( ) +

6

4 5 32

.

428 Chapter

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Wave Motion 429

*P16.2

(a)

(b)

The graph (b) has the same amplitude and wavelength as graph (a). it differs just by being

shifted toward larger x by 2.40 m. The wave has traveled 2.40 m to the right.

P16.3 (a) The longitudinal P 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 6 Rsin . . sin . .° °= ×( ) = ×m m

at a speed of 7 800 m /s

Therefore, it takes6 37 10

8176

. ×=

m

7 800 m s

s

The wave that travels along the Earth’s surface must travel

a distance of s R R= = = ×θ

π

36 67 106rad m.

at a speed of 4 500 m /s

Therefore, it takes6 67 10

4 5001482

6. ×= s

The time difference is 665 11 1s min= .

P16.4 The distance the waves have traveled is d t t = ( ) = ( ) +( )7 80 4 50 17 3. . .km s km s s where t is the travel time for the faster wave.

Then, 7 80 4 50 4 50 17 3. . . .−( )( ) = ( )( )km s km s st

or t = ( )( )

−( ) =

4 50 17 3

7 80 4 5023 6

. .

. ..

km s s

km ss

and the distance is d = ( )( ) =7 80 23 6 184. .km s s km .

FIG. P16.2

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430 Chapter 16

P16.5 (a) Let u t x = − +10 34

π π π

du

dt

dx

dt = − =10 3 0π π at a point of constant phase

dx

dt = =

10

33 33. m s

The velocity is in the positive -direction x .

(b) y 0 100 0 0 350 0 3004

0 05. , . sin . .( ) = ( ) − + = −m π

π 44 8 5 48m cm= − .

(c) k = =2

3π

λ π : λ = 0 667. m ω π π = =2 10 f : f = 5 00. Hz

(d) v y y

t t x =

∂∂

= ( )( ) − + 0 350 10 10 3 4

. cosπ π π π

v y, . .max m s= ( )( ) =10 0 350 11 0π

Section 16.2 The Traveling Wave Model

*P16.6 (a) a wave

(b) later by T 4

(c) A is 1.5 times larger

(d) λ is 1.5 times larger

(e) λ is 2 3 as large

P16.7 f = =40 0 4

3

. vibrations

30.0 sHz v = =

42542 5

cm

10.0 scm s.

λ = = = =v

f

42 531 9 0 319

43

.. .

cm s

Hzcm m

P16.8 Using data from the observations, we have λ = 1 20. m and f =8 00

12 0

.

. s

Therefore, v = = ( ) =λ f 1 20

8 00

12 00 800.

.

..m

sm s

P16.9 y x t = ( ) −( )0 020 0 2 11 3 62. sin . .m in SI units A = 2 00. cm

k = 2 11. rad m λ π = =2 2 98k

. m

ω = 3 62. rad s f = =ω

π 20 576. Hz

v = = = = f k

λ ω

π

π

2

2 3 62

2 111 72

.

.. m s

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Wave Motion 431

P16.10 v = = ( )( ) = = f λ 4 00 60 0 240 2 40. . .Hz cm cm s m s

*P16.11 (a) ω π π = = ( ) =−2 2 5 31 41 f s rad s.

(b) λ = = =−v

f 20 4 00

1m s

5 sm.

k = = =2 2

41 57

π

λ

π

mrad m.

(c) In y A kx t = − +( )sin ω φ we take A = 12 cm. At x = 0 and t = 0 we have y = ( )12 cm sinφ . To make this fit

y = 0, we take φ = 0. Then

y x t = ( ) ( ) − ( )( )12 0 1 57 31 4. sin . .cm rad m rad s

(d) The transverse velocity is∂∂

= − −( ) y

t A kx t ω ω cos

Its maximum magnitude is Aω = ( ) =12 31 4 3 77cm rad s m s. .

(e) at t

A kx t A kx t y

y= ∂∂

= ∂∂

− −( )( ) = − −( )v

ω ω ω ω cos sin2

The maximum value is Aω 2 12

0 12 31 4 118= ( )( ) =−. .m s m s2

P16.12 At time t , the phase of y x t = ( ) −( )15 0 0 157 50 3. cos . .cm at coordinate x is

φ = ( ) − ( )0 157 50 3. .rad cm rad s x t . Since 60 03

. ° = π

rad, the requirement for point B is that

φ φ π

B A= ±

3rad, or (since x A = 0),

0 157 50 3 0 50 3. . .rad cm rad s rad s( ) − ( ) = − ( ) x t t B ±± π

3rad

This reduces to x B = ±

( )

= ±π rad

rad cm

cm

3 0 157

6 67

.

. .

P16.13 y x t = −( )0 250 0 300 40 0. sin . . m

Compare this with the general expression y A kx t = −( )sin ω

(a) A = 0 250. m

(b) ω = 40 0. rad s

(c) k = 0 300. rad m

(d) λ π π

= = =2 2

0 30020 9

k ..

rad mm

(e) v = = = ( ) = f λ ω

π λ π 2

40 0

2 20 9 133.

.rad s

m m ss

(f) The wave moves to the right, in direction+ x .

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432 Chapter 16

*P16.14 (a) See figure at right.

(b) T = = =2 2

50 30 125

π

ω

π

.. s is the time from one peak

to the next one.

This agrees with the period found in the example

in the text.

P16.15 (a) A y= = =max . .8 00 0 080 0cm m k = =( )

= −2 2

0 8007 85

1π

λ

π

..

mm

ω π π π = = ( ) =2 2 3 00 6 00 f . . rad s

Therefore, y A kx t = +( )sin ω

Or (where y t 0 0,( ) = at = 0) y x t = ( ) +( )0 080 0 7 85 6. sin . π m

(b) In general, y x t = + +( )0 080 0 7 85 6. sin . π φ

Assuming y x , 0 0( ) = at x = 0 100. m

then we require that 0 0 080 0 0 785= +( ). sin . φ

or φ = −0 785.

Therefore, y x t = + −( )0 080 0 7 85 6 0 785. sin . .π m

P16.16 (a)

(b) k = = =2 2

0 35018 0

π

λ

π

..

mrad m

T f

= = =1 1

12 00 0833

..

ss

ω π π = = =2 2 12 0 75 4 f . .s rad s

v = = ( )( ) = f λ 12 0 0 350 4 20. . .s m m s

(c) y A kx t = + +( )sin ω φ specializes to

y x t = + +( )0 200 18 0 75 4. sin . .m m s φ

at x = 0, t = 0 we require

− × = +( )

= − = −

−3 00 10 0 200

8 63 0 151

2. . sin

. .

m m φ

φ ° rrad

so y x t x t , . sin . . .( ) = ( ) + −(0 200 18 0 75 4 0 151m m s rad))

0.0 –0.1 –0.2

0.10.2

y (m)

x (m)

t = 0

0.2

0.4

FIG. P16.16(a)

y (cm)

t (s)

10

0

0.1 0.2–10

FIG. P16.14

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Wave Motion 433

P16.17 y x t = ( ) + 0 120 8

4. sinm π

π

(a) v = dy

dt : v = ( )( ) +

0 120 4 8

4. cosπ π

π x t

v 0 200 1 51. .s, 1.60 m m s( ) = −

a d

dt =

v: a x t = −( )( ) +

0 120 4 8

42

. sinm π π

π

a 0 200 0. s, 1.60 m( ) =

(b) k = =π π

λ 8

2: λ = 16 0. m

ω π π

= =42

T : T = 0 500. s

v = = =λ

T

16 032 0

..

m

0.500 sm s

P16.18 (a) Let us write the wave function as y x t A kx t , sin( ) = + +( )ω φ

y A0 0 0 020 0, sin .( ) = =φ m

dy

dt A

0 0

2 00,

cos .= = −ω φ m s

Also, ω π π

π = = =2 2

0 025080 0

T ..

ss

A x i

i2 2

22

0 020 02 00

80 0= +

= ( ) +

v

ω π .

.

.m

m s

s

2

A = 0 021. 5 m

(b) A

A

sin

cos

.

/ .

. tanφ

φ π

φ =

−

= − =0 020 0

2 80 0

2 51

Your calculator’s answer tan . .− −( ) = −1 2 51 1 19 rad has a negative sine and positive cosine, just the reverse of what is required. You must look beyond your calculator to find

φ π = − =1 19 1 95. .rad rad

(c) v y A, . . .max m s m s= = ( ) =ω π 0 021 5 80 0 5 41

(d) λ = = ( )( ) =v x T 30 0 0 025 0 750. . .m s 0 s m

k = = =2 2

0 750

π

λ

π

. m8.38 m ω π = 80 0. s

y x t x t , . sin . .( ) = ( ) + +0 021 5 8 38 80 0m rad m rad sπ 11 95. rad( )

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Wave Motion 435

P16.23 v = = ⋅

× =−

T

µ

1350

5 00 10520

3

kg m s

kg mm s

2

.

P16.24 v = T

µ

T A r

T

= = =

= ( )( ) × −µ ρ ρπ

π

v v v2 2 2 2

48 920 7 50 10kg m3 . mm m s

N

( ) ( )

=

2 2200

631T

P16.25 T Mg= is the tension; v = = = =T Mg

m L

MgL

m

L

t µ is the wave speed.

Then, MgL

m

L

t =

2

2

and g Lm

Mt = =

×( )×

−

−2

31 60 4 00 10

3 00 10

. .

.

m kg

kg 3.61 222 1 64

sm s

2

( ) = .

P16.26 The period of the pendulum is T L

g= 2π

Let F represent the tension in the string (to avoid confusion with the period) when the pendulum

is vertical and stationary. The speed of waves in the string is then:

v = = =F Mg

m L

MgL

mµ

Since it might be difficult to measure L precisely, we eliminate L T g=

2π

so

v = = Mg

m

T g Tg M

m2 2π π

P16.27 Since µ

is constant, µ = =T T

2

2

2

1

1

2v v

and

T T 2

2

1

2

1

2

30 0

20 06 00=

=

v

v

.

..

m s

m sNN N( ) = 13 5.

P16.28 From the free-body diagram mg T = 2 sinθ

T mg

=2sinθ

The angle θ is found from cosθ = =3 8

2

3

4

L

L

∴ =θ 41 4. °

(a) v = T µ

v = =×( )−

mg

2 41 4

9 80

2 8 00 10 43µ sin .

.

. sin°

m s

kg m

2

11 4. °

m

or v =

30 4.

m s

kgm

(b) v = =60 0 30 4. . m and m = 3 89. kg

FIG. P16.28

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436 Chapter 16

P16.29 If the tension in the wire is T , the tensile stress is

Stress = = ( )T

AT Aso stress

The speed of transverse waves in the wire is

v = = ( ) = =T Am L m AL mµ Stress Stress Stress

Volume == Stress

ρ

where ρ is the density. The maximum velocity occurs when the stress is a maximum:

vmax.

= ×

=2 70 10

18 Pa

7 860 kg m85 m s

3

*P16.30 (a) f has units Hz s= 1 , so T f

=1

has units of seconds, s . For the other T we have T = µ v2,

with unitskg

m

m

s

kg m

sN

2

2 2=

⋅= .

(b) The first T is period of time; the second is force of tension.

P16.31 The total time is the sum of the two times.

In each wire t L

LT

= =v

µ

Let A represent the cross-sectional area of one wire. The mass of one wire can be written both as

m V AL= =ρ ρ and also as m L= µ .

Then we have µ ρ πρ

= = A d

2

4

Thus, t L d

T =

πρ 21 2

4

For copper, t = ( ) ( )( ) ×( )

( )( )

−

20 08 920 1 00 10

4 150

3 2

..π

=

1 2

0 137. s

For steel, t = ( ) ( )( ) ×( )( )( )

−

30 0 7 860 1 00 104 150

3 2

. .π

=

1 2

0 192. s

The total time is 0 137 0 192 0 329. . .+ = s

Section 16.5 Rate of Energy Transfer by Sinusoidal Waves on Strings

P16.32 f = = =v

λ

30 0

0 50060 0

.

.. Hz ω π π = =2 120 f rad s

P = = ( ) ( )

1

2

1

2

0 180

3 60120 0 100 302 2

2 2µω π A v

.

.. .. .0 1 07( ) = kW

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Wave Motion 437

P16.33 Suppose that no energy is absorbed or carried down into the water. Then a fixed amount of power

is spread thinner farther away from the source. It is spread over the circumference 2π r of anexpanding circle. The power-per-width across the wave front

P

2π r is proportional to amplitude squared so amplitude is proportional to

P

2π r

P16.34 T = constant; v = T

µ ; P =

1

2

2 2µω A v

(a) If L is doubled, v remains constant and P is constant .

(b) If A is doubled and ω is halved, P ω 2 2 A remains constant .

(c) If λ and A are doubled, the product ω λ

2 22

2 A

A remains constant, so

P remains constant .

(d) If L and λ are halved, then ω λ

2

2

1 is quadrupled, so P is quadrupled .

(Changing L doesn’t affect P .)

P16.35 A = × −5 00 10 2. m µ = × −4 00 10 2. kg m P = 300 W T = 100 N

Therefore, v = =T

µ 50 0. m s

P =1

2

2 2µω A v : ω 2 = 2P _____ µ A2v

2 2 2

2 300

4 00 10 5 00 10 50 0=

( )

×( ) ×( )− −. . .(( )

ω

ω

π

=

= =

346

255 1

rad s

Hz f .

P16.36 µ = = × −30 0 30 0 10 3. .g m kg m

λ

ω π

=

= = =

=

−

1 50

50 0 2 314

2 0 150

1

.

. :

. :

m

Hz s

m

f f

A A == × −7 50 10 2. m

(a) y A x t = − sin

2π

λ ω

y x t = ×( ) −( )−7 50 10 4 19 3142. sin .

(b) P = = ×( )( ) ×( )− −1

2

1

230 0 10 314 7 50 10

32 2 3 2 2 2µω A v . .114

4 19.

W P = 625 W

FIG. P16.36

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438 Chapter 16

P16.37 (a) v = = = = = f k k

λ ω

π

π ω

2

2 50 0

0 80062 5

.

..m s m s

(b) λ π π

= = =2 2

0 8007 85

k ..m m

(c) f = =50 0

27 96

..

π Hz

(d) P = = ×( )( ) ( )−1

2

1

212 0 10 50 0 0 150 62 52 2 3

2 2µω A v . . . .(( ) =W W21 1.

P16.38 Comparing y t x = − + 0 35 10 3 4

. sin π π π

with y A kx t A t kx = − +( ) = − − +( )sin sinω φ ω φ π

we have k

m=

3π , ω π = 10 s, A = 0 35. m. Then v = = = = = f f

k λ π

λ

π

ω π

π 2

2

10

33 33

s

mm s. .

(a) The rate of energy transport is

P = = ×( )( ) ( )−1

2

1

275 10 10 0 35 32 2 3

2 2µω π A v kg m s m. .. .33 15 1m s W=

(b) The energy per cycle is

E λ = P T Aµω λ π = = ×( )( )−

1

2

1

275 10 10 0 352 2 3

2kg m s m.(( ) =2

23 02

π

π

m

3J.

P16.39 Originally,

P

P

P

0

2 2

0

2 2

0

2 2

1

2

1

2

1

2

=

=

=

µω

µω µ

ω µ

A

A T

A T

v

The doubled string will have doubled mass-per-length. Presuming that we hold tension constant,

it can carry power larger by 2 times.

21

22

0

2 2P = ω µ A T

*P16.40 As for a string wave, the rate of energy transfer is proportional to the square of the amplitude

and to the speed. The rate of energy transfer stays constant because each wavefront carries con-

stant energy and the frequency stays constant. As the speed drops the amplitude must increase.

We write P = F Av 2 where F is some constant. With no absorption of energy,

F A F Av v

v

bedrock bedrock

2

mudfill mudfill

2

bedro

=

cck

mudfill

mudfill

bedrock

mudfill

mudv

v

v

= = A

A

25

f fill

= 5

The amplitude increases by 5.00 times.

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Wave Motion 439

Section 16.6 The Linear Wave Equation

P16.41 (a) A = +( )7 00 3 00 4 00. . . yields A = 40 0.

(b) In order for two vectors to be equal, they must have the same magnitude and the samedirection in three-dimensional space. All of their components must be equal.

Thus, 7 00 0 3 00. ˆ ˆ . ˆ ˆ ˆ ˆi j k i j k+ + = + + A B C requires A B C = = =7 00 0 3 00. , , .and .

(c) In order for two functions to be identically equal, they must be equal for every value

of every variable. They must have the same graphs.

In

A B Cx Dt E x t + + +( ) = + + +cos . cos . . .0 7 00 3 00 4 00 2 0mm 00( )

the equality of average values requires that A = 0 . The equality of maximum values requires

B = 7 00. mm . The equality for the wavelength or periodicity as a function of x requires

C = 3 00. rad m . The equality of period requires D = 4 00. rad s , and the equality of

zero-crossings requires E = 2 00. rad .

P16.42 The linear wave equation is∂

∂=

∂

∂

2

2 2

2

2

1 y

x

y

t v

If y eb x t

= −( )v

then∂

∂= −

−( ) y

t b e

b x t v

v

and∂

∂=

−( ) y

x be

b x t v

∂

∂=

−( )2

2

2 2 y

t b e

b x t v

v

and∂

∂=

−( )2

2

2 y

x b e

b x t v

Therefore, ∂∂

= ∂

∂

2

2

2 2

2

y

t

y

x v , demonstrating that eb x t

−( )v is a solution.

P16.43 The linear wave equation is1

2

2

2

2

2v

∂

∂=

∂

∂

y

t

y

x

To show that y b x t = −( )[ ]ln v is a solution, we find its first and second derivatives with respect

to x and t and substitute into the equation.

∂

∂=

−( ) −( )

y

t b x t b

1

v

v ∂

∂=

− −( )

−( ) = −

−( )

2

2

2

2 2

2

2

1 y

t

b

b x t x t

v

v

v

v

∂

∂= −( )[ ]

− y

x b x t bv

1

∂

∂= − − = −

−

−

2

2

2

2

1 y

x

b

b x t

x t

( )

( )

v

v

Then 1 1 12

2

2 2

2

2 2

2

2v v

v

v v

∂

∂=

−( )−( )

= −−( )

= ∂

∂

y

t x t x t

y

x so the given wave function is a solution.

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14/22

440 Chapter 16

P16.44 (a) From y x t = +2 2 2v ,

evaluate∂∂

= y

x x 2

∂∂

=2

22

y

x

∂

∂ =

y

t t v

2

2

∂

∂ =

2

2

2

2

y

t v

Does∂∂

= ∂

∂

2

2 2

2

2

1 y

t

y

t v?

By substitution, we must test1

22

2=v

v and this is true, so the wave function does satisfy

the wave equation.

(b) Note

1

2

1

2

2 2 x t x t +( ) + −( )v v = + + + − +

1

2

1

2

1

2

1

2

2 2 2 2 2 2 x x t t x x t t v v v v

= + x t 2 2 2v as required.

So

f x t x t +( ) = +( )v v122 and g x t x t −( ) = −( )v v12

2

(c) y x t = sin cosv makes

∂∂

= y

x x t cos cosv

∂∂

= −2

2

y

x x t sin cosv

∂∂

= − y

t x t v vsin sin

∂∂

= −2

2

2 y

t x t v vsin cos

Then

∂∂

= ∂

∂

2

2 2

2

2

1 y

x

y

t v

becomes − = −

sin cos sin cos x t x t vv

v v

1

2

2 which is true as required.

Note sin sin cos cos sin x t x t x t +( ) = +v v v

sin sin cos cos sin x t x t x t −( ) = −v v v

So sin cos x t f x t g x t v v v= +( )+ −( ) with

f x t x t +( ) = +( )v v1

2sin and g x t x t −( ) = −( )v v

1

2sin

Additional Problems

P16.45 Assume a typical distance between adjacent people ~1 m.

Then the wave speed is v = ∆∆ x

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

T r

= ( )

=2 2 10

1063 1

2π π

v

~ ~m s

s min

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15/22

Wave Motion 441

*P16.46 (a) From y = 0.150 m sin(0.8 x – 50t )

we compute dy dt = 0.150 m (–50) cos(0.8 x – 50t )

and a = d 2 y dt 2 = –0.150 m (–50 s)2 sin(0.8 x – 50t )

Then amax = 375 m/s2

(b) For the 1-cm segment with maximum force acting on it, ΣF = ma = [12 g (100 cm)]1 cm 375 m /s2 = 0.045 0 N

We find the tension in the string from v = fλ = ω k = (50 s) (0.8 m) = 62.5 m /s = (T µ )1 2

T = v2µ = (62.5 m /s)2(0.012 kg m) = 46.9 N.

The maximum transverse force is very small compared to the tension, more than a

thousand times smaller.

P16.47 The equation v = λ f is a special case of

speed = (cycle length)(repetition rate)

Thus,

v = ×( )( ) =−

19 0 10 24 0 0 4563

. . .m frame frames s m s

P16.48 (a) 0 175 0 350 99 6. . sin .m m rad s= ( ) ( ) t

∴ ( ) =sin . .99 6 0 5rad s t

The smallest two angles for which the sine function is 0.5 are 30° and 150°, i.e., 0.523 6 rad

and 2.618 rad.

99 6 0 523 61. .rad s rad( ) =t , thus t 1 5 26= . ms

99 6 2 6182. .rad s rad( ) =t , thus t 2 26 3= . ms

t t t 2 1 26 3 5 26 21 0− = − =. . .ms ms ms

(b) Distance traveled by the wave = =

× −ω

k t ∆

99 6

1 2521 0 10 3

.

..

rad s

rad ms m( ) = 1 68. .

P16.49 Energy is conserved as the block moves down distance x :

K U U E K U U

Mgx

g s g s+ +( ) + = + +( )

+ + + = +

top bottom∆

0 0 0 0 001

2

2

2+

=

kx

x Mg

k

(a) T kx Mg= = = ( )( ) =2 2 2 00 9 80 39 2. . .kg m s N2

(b) L L x L Mg

k = + = +0 0

2

L = + =0 50039 2

0 892..

.mN

100 N m

m

(c) v = =T TL

mµ

v

v

= ×

×

=

−

39 2 0 892

10

83 6

3

. .

.

N m

5.0 kg

m s

13794_16_ch1 6_p427 448.indd 44113794_16_ch16_p427-448.indd 441 12/9/06 12:47:0112/9/06 12:47:01


16/22

442 Chapter 16

P16.50 Mgx kx =1

2

2

(a) T kx Mg= = 2

(b) L L x L Mg

k = + = +

0 0

2

(c) v = = = +

T TL

m

Mg

m L

Mg

k µ

2 20

*P16.51 (a) The energy a wave crest carries is constant in the absence of absorption. Then the rate at

which energy passes a stationary point, which is the power of the wave, is constant. The

power is proportional to the square of the amplitude and to the wave speed. The speed

decreases as the wave moves into shallower water near shore, so the amplitude must

increase.

(b) For the wave described, with a single direction of energy transport, the intensity is the same at

the deep-water location 1 and at the place 2 with depth 9 m. To express the constant inten-sity we write

A A A gd

A

1

2

1 2

2

2 2

2

2

2

2

21 8 9 8

v v= =

( ) =. .m 200 m s m s22 m

m s

m s

m s

( )

=

=

9

9 39

1 8200

9 39

2

2

2

A

A

.

..

11 2

8 31= . m

(c) As the water depth goes to zero, our model would predict zero speed and infinite amplitude.

In fact the amplitude must be finite as the wave comes ashore. As the speed decreases the

wavelength also decreases. When it becomes comparable to the water depth, or smaller, ourformula gd for wave speed no longer applies.

P16.52 Assuming the incline to be frictionless and taking the positive x -direction to be up the incline:

F T Mg x ∑ = − =sinθ 0 or the tension in the string is T Mg= sinθ

The speed of transverse waves in the string is then v = = =T Mg

m L

MgL

mµ

θ θ sin sin

The time interval for a pulse to travel the string’s length is ∆t L

L m

MgL

mL

Mg= = =

v sin sinθ θ

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17/22

Wave Motion 443

*P16.53 (a) In P = 12

2 2µω A v where v is the wave speed, the quantity ω A is the maximum particle

speed v ymax

. We have µ = 0.5 × 10–3 kg m and v = (T µ )1 2 = (20 N 0.5 × 10–3 kg m)1 2 = 200 m /s

Then P =1

2 (0.5 × 10–3

kg m) v2

ymax (200 m s) = (0.050 0 kg/s)v2

y,max

(b) The power is proportional to the square of tthe maximum particle speed.

(c) In time t = (3 m) v = (3 m) (200 m /s) = 1.5 × 10–2 s, all the energy in a 3-m length of stringgoes past a point. Therefore the amount of this energy is

E = P t = (0.05 kg s) v2 y, max

(0.015 s) = 7.5 × 10−4 kg v y,max2

The mass of this section is m3 = (0.5 × 10–3 kg m)3 m = 1.5 × 10–3 kg so (1 2)m

3 = 7.5 × 10−4 kg

and E = (1/2) m3v

2 y,max

= K max

. The string also contains potential energy. We could write its

energy as U max

or as U avg

+ K avg

(d) E = P t = (0.05 kg s) v2 y,max

(6 s) = 0.300 kg v y,max

2

P16.54 v = T

µ and in this case T mg= ; therefore, m

g=

µ v2

Now = f λ implies v = ω

k so that

mg k

= =

−µ ω π

π

2 10 250

9 80

18

0 750

.

. .

kg m

m s

s2 mm

kg−

=1

2

14 7.

P16.55 Let M = mass of block, m = mass of string. For the block, F ma∑ = implies T mr

m r b= =v

22ω

The speed of a wave on the string is then

v

v

= = =

= =

= = =

T M r

m r r

M

m

t r m

M

t m

M

µ

ω ω

ω

θ ω

2

1

0 0032

. kg

00.450 kgrad= 0 084 3.

P16.56 (a) µ ρ ρ = = =dm

dL A

dx

dx A

v = = =+( )[ ]

=+( )

− −

T T

A

T

ax b

T

x µ ρ ρ ρ 10 103 2 cm2

With all SI units,

v =+( ) − − −T

x ρ 10 10 103 2 4m s

(b) v x = − −

=( ) +( )( )

=0 2 4

24 0

2 700 0 10 1094 3

.. m s

v x = − − −= ( ) +( )( ) =

10 0 2 2 4

24 0

2 700 10 10 1066

.

..77 m s

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18/22

444 Chapter 16

P16.57 v = T

µ where T xg= µ , to support the weight of a length x , of rope.

Therefore, v = gx

But v = dx

dt , so that dt

dx

gx =

and t dx

gx g

x L

g

L L

= = =∫ 0

12 0

12

P16.58 At distance x from the bottom, the tension is T mxg

L Mg=

+ , so the wave speed is:

v = = = + =

T TL

m xg

MgL

m

dx

dt µ

(a) Then t dt xg MgL

mdx

t L

= = +

∫ ∫

−

0

1 2

0

t g

xg MgL m

x

x L

= + ( )

=

=

11 2

12

0

t g

Lg MgL

m

MgL

m= +

−

2 1 2 1 2 t

L

g

m M M

m= + −

2

(b) When M = 0, as in the previous problem, t L

g

m

m

L

g=

−

=2

02

(c) As m → 0 we expand m M M m

M M

m

M

m

M + = +

= + − +

1 11

2

1

8

1 2 2

2…

to obtain t L

g

M m M m M M

m=

+ ( ) − ( ) + −

2

12

18

2 3 2 …

t L

g

m

M

mL

Mg≈

=2

1

2

P16.59 (a) The speed in the lower half of a rope of length L is the same function of distance (from the

bottom end) as the speed along the entire length of a rope of length L

2

.

Thus, the time required = ′

2 L

g with ′ = L

L

2

and the time required = =

2

20 707 2

L

g

L

g.

It takes the pulse more that 70% of the total time to cover 50% of the distance.

(b) By the same reasoning applied in part (a), the distance climbed in τ is given by d g=

τ 2

4

For τ = =t L

g2 , we find the distance climbed = L

4 .

In half the total trip time, the pulse has climbed1

4 of the total length.

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19/22

Wave Motion 445

P16.60 (a) Consider a short section of chain at the top of the loop. A

free-body diagram is shown. Its length is s = R(2θ ) and its

mass is µ R2θ . In the frame of reference of the center of the

loop, Newton’s second law is

F ma y y∑ = 2 20

2

0

2

T m R

R R

sinθ µ θ down down= =v v

For a very short section, sinθ θ = and T = µ v02

(b) The wave speed is v v= =T

µ 0

(c) In the frame of reference of the center of the loop, each pulse moves with equal speed

clockwise and counterclockwise.

In the frame of reference of the ground, once pulse moves backward at speed v v v0 02+ =

and the other forward at v v0 0− = .

The one pulse makes two revolutions while the loop makes one revolution and the other

pulse does not move around the loop. If it is generated at the six-o’clock position, it will

stay at the six-o’clock position.

P16.61 Young’s modulus for the wire may be written as Y = T A

L L

∆, where T is the tension maintained in

the wire and ∆ L is the elongation produced by this tension. Also, the mass density of the wire

may be expressed as ρ µ =

A

The speed of transverse waves in the wire is then

v = = = ( )T T A

A

Y L L

µ µ ρ

∆

and the strain in the wire is∆ L

L=

ρ v2

Y

If the wire is aluminum and v = 100 m s, the strain is

∆ L

L=

×( )( )×

=2 70 10 100

7 00 103

3 2

10

.

.

kg m m s

N m

3

2..86 10 4× −

R 2θ

θ θ

T T

FIG. P16.60(a)

v0 v0 v0

v v

FIG. P16.60(c1)

v0 v0 v0

FIG. P16.60(c2)

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20/22

446 Chapter 16

P16.62 (a) Assume the spring is originally stationary throughout, extended to have a length L much

greater than its equilibrium length. We start moving one end forward with the speed v at

which a wave propagates on the spring. In this way we create a single pulse of compression

that 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

dm dx a

=

Butdm

dx = µ so a

k =

µ

Also, a d

dt t = =

v v

when vi = 0

But L = vt, so a L

=v

2

Equating the two expressions for a, we havek

Lµ =

v2

or v = kL

µ

(b) Using the expression from part (a) v = = = ( )( )

=kL kL

mµ

2 2100 2 00

0 400 31 6N m m

kg m.

. . ss

P16.63 (a) P x A A ek k

A ebx ( ) = = =

−1

2

1

2 2

2 2 2

0

2 2

3

0

2µω µω ω µω

v −−2bx

(b) P 02

3

0

2( ) = µω

k A

(c)P ( x )

_____

P (0) e bx = −2

P16.64 v = = =4 450

468 130km

9.50 h

km h m s

d g

= = ( )

( ) =

v2 2130

9 801 730

m s

m sm

2.

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21/22

Wave Motion 447

P16.65 (a) µ x ( ) is a linear function, so it is of the form µ x mx b( ) = +

To have µ µ 0 0( ) = we require b = µ 0. Then µ µ µ L mL L( ) = = + 0

so m L

L

= −µ µ 0

Then µ µ µ

µ x x

L

L( ) = −( )

+00

(b) From v = dx

dt , the time required to move from x to x + dx is dx

v

. The time required to move

from 0 to L is

∆

∆

t dx dx

T T x dx

t T

x

L L L

L

= = = ( )

= −( )

∫ ∫ ∫ v

0 0 0

0

1

1

µ µ

µ µ

L L Ldx

L L

L

L

+

−

−

∫ µ

µ µ

µ µ 0

1 2

0

00

∆

∆

t T

L x

L L

L

L

=−

−( ) +

1 1

0

0

0

3 2

32 0

µ µ µ µ µ

t t L

T

t L

L

L

L L L

=−( )

−( )

= −( ) +

2

3

2

0

3 2

0

3 2

0

µ µ µ µ

µ µ µ µ µ ∆ 00 0

0 0

0 0

3

2

3

+( )−( ) +( )

= + +

+

µ

µ µ µ µ

µ µ µ µ

µ

T

t L

T

L L

L L

L

∆µ µ 0

ANSWERS TO EVEN PROBLEMS

P16.2 See the solution. The graph (b) has the same amplitude and wavelength as graph (a). It differs justby being shifted toward larger x by 2.40 m. The wave has traveled 2.40 m to the right.

P16.4 184 km

P16.6 See the solution

P16.8 0 800. m s

P16.10 2 40. m s

P16.12 ±6 67. cm

P16.14 (a) see the solution (b) 0.125 s, in agreement with the example

P16.16 (a) see the solution (b) 18.0 m; 83 3. ms; 75 4. rad s; 4 20. m s

(c) 0 2 18 75 4 0 151. sin . .m( ) + −( ) x t

P16.18 (a) 0.021 5 m (b) 1.95 rad (c) 5 41. m s (d) y x t x t , . sin . . .( ) = ( ) + +( )0 021 5 8 38 80 0 1 95m π

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22/22

448 Chapter 16

P16.20 (a) see the solution (b) 3.18 Hz

P16.22 (a) y x t = ( ) −( )0 2 16 3140. sinmm (b) 158 N

P16.24 631 N

P16.26 v = Tg M

m2π

P16.28 (a) v =⋅

30 4.

m

s kgm (b) 3 89. kg

P16.30 (a) s and N (b) The first T is period of time; the second is force of tension.

P16.32 1.07 kW

P16.34 (a), (b), (c) P is a constant (d) P is quadrupled

P16.36 (a) y x t = ( ) −( )0 075 0 4 19 314. sin . (b) 625 W

P16.38 (a) 15.1 W (b) 3.02 J

P16.40 As for a string wave, the rate of energy transfer is proportional to the square of the amplitude and to the

speed. The rate of energy transfer stays constant because each wavefront carries constant energy and the

frequency stays constant. As the speed drops the amplitude must increase. It increases by 5.00 times.

P16.42 see the solution

P16.44 (a) see the solution (b)1

2

1

2

2 2 x t x t +( ) + −( )v v (c)

1

2

1

2sin sin x t x t +( )+ −( )v v

P16.46 (a) 375 m /s2 (b) 0.0450 N. This force is very small compared to the 46.9-N tension, more than a

thousand times smaller.

P16.48 (a) 21.0 ms (b) 1.68 m

P16.50 (a) 2 Mg (b) L Mgk

0 2+ (c) 2 20 Mgm

L Mgk

+

P16.52 ∆t mL

Mg=

sinθ

P16.54 14.7 Kg

P16.56 (a) v =+( )− −

T

x ρ 10 107 6 in SI units (b) 94 3. m s; 66 7. m s

P16.58 See the solution.

P16.60 (a) µ v02 (b) v0 (c) One travels 2 rev and the other does not move around the loop.

P16.62 (a) see the solution (b) 31.6 m

/s

P16.64 130 m s; 1.73 km

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solucion al capitulo 16 del serway