SM chapter 16 - Solucionario capitulo 16 Serway 7ma edición PDF

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Solucionario capitulo 16 Serway 7ma edición ...

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16 Wave Motion ANSWERS TO QUESTIONS

CHAPTER OUTLINE 16.1 16.2 16.3 16.5 16.6

Propagation of a Disturbance The Traveling Wave Model The Speed of Waves on Strings Rate of Energy Transfer by Sinusoidal Waves on Strings The Linear Wave Equation

*Q16.3 (i)

(ii)

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.

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. Look at the coefficients of x. Each is the wave number, 2π Ⲑλ , 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 frequencyω = 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 = fλ = ω Ⲑk, 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 , we must increase the tension by a factor of 4 to make v double. Answer (b). µ

*Q16.5 Answer (b). Wave speed is inversely proportional to the square root of linear density. *Q16.6 (i) (ii)

Answer (a). Higher tension makes wave speed higher. 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 by vy,max = ω A = 2π fA = Here v is the speed of the wave.

2π v A . λ

428

Chapter 16

*Q16.9 (a) through (d): Yes to all. The maximum particle speed and the wave speed are related by 2π v A . Thus the amplitude or the wavelength of the wave can be adjusted vy,max = ω A = 2π fA = λ to make either vy,max or v larger. 1 Q16.10 Since the frequency is 3 cycles per second, the period is second = 333 ms. 3 Q16.11

Each element of the rope must support the weight of the rope below it. The tension increases with T increases height. (It increases linearly, if the rope does not stretch.) Then the wave speed v = µ 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)

(ii)

Answer a. As the wave passes from the massive string to the less massive string, the wave T . speed will increase according to v = µ 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 temporarily 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 = ts − tp represent the difference in arrival times of the two waves at a station at distance −1 ⎛ 1 1⎞ d = vs t s = v pt p from the hypocenter. Then d = ∆t ⎜ − ⎟ . Knowing the distance from the ⎝ vs v p ⎠ first station places the hypocenter on a sphere around it. A measurement from a second station 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 P16.1

Propagation of a Disturbance

Replace x by

x − vt = x − 4.5t to get y =

6 2 ⎡⎣( x − 4.5t ) + 3⎦⎤

Wave Motion

429

*P16.2

(a)

(b)

FIG. P16.2

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 R sin 30 .0 ° = 2 ( 6 .37 × 106 m) sin 30 .0 ° = 6 .37 × 106 m

at a speed of

7 800 m /s

Therefore, it takes

6 .37 × 106 m = 817 s 7 800 m s

The wave that travels along the Earth’s surface must travel π ⎞ s = Rθ = R ⎛ rad = 6.67 × 106 m a distance of ⎠ ⎝3 at a speed of

4 500 m /s

Therefore, it takes

6. 67 × 10 6 = 1 482 s 4 500

The time difference is P16.4

665 s = 11.1 min

The distance the waves have traveled is d = ( 7.80 km s ) t = ( 4.50 km s ) ( t + 17.3 s ) where t is the travel time for the faster wave. Then,

(7.80 − 4.50 ) ( km s )t = ( 4.50 km s )( 17.3 s )

or

t=

and the distance is

d = ( 7.80 km s ) (23.6 s) = 184 km .

( 4. 50 km s )( 17. 3 s ) = ( 7.80 − 4.50 ) km s

23.6 s

430

P16.5

Chapter 16

(a)

(b) (c) (d)

Section 16.2 *P16.6

P16.7

P16.8

Let u = 10π t − 3π x +

du dx = 10 π − 3 π = 0 at a point of constant phase dt dt dx 10 = = 3.33 m s 3 dt The velocity is in the positive x-direction .

π y (0.100, 0 ) = ( 0.350 m ) sin ⎛ −0.300π + ⎞ = −0.054 8 m = −5.48 cm ⎝ 4⎠ π 2 λ = 0.667 m ω = 2π f = 10π : k= = 3π : f = 5.00 Hz λ π ∂y vy = v y, max = (10π ) (0.350) = 11.0 m s = ( 0.350) ( 10π ) cos⎛ 10π t − 3π x + ⎞ ⎝ 4⎠ ∂t

The Traveling Wave Model

(a)

a wave

(b)

later by TⲐ4

(c)

A is 1.5 times larger

(d)

λ is 1.5 times larger

(e)

λ is 2Ⲑ3 as large

40 .0 vibrations 4 425 cm = Hz v= = 42. 5 cm s 3 10.0 s 30.0 s v 42.5 cm s λ= = 4 = 31.9 cm = 0.319 m f 3 Hz f=

Using data from the observations, we have Therefore,

P16.9

π 4

v = λ f = (1. 20 m )

λ = 1.20 m and

⎛ 8. 00 ⎞ = . 0 800 m s ⎝12.0 s ⎠

y = ( 0.020 0 m )sin (2.11 x − 3.62 t ) in SI units

A = 2.00 cm

k = 2.11 rad m

λ=

2π = 2 .98 m k ω f= = 0.576 Hz 2π

ω = 3.62 rad s v = fλ =

ω 2π

f=

=

3.62

= 1 .72 m s

8 .00 12.0 s

Wave Motion

P16.10

v = f λ = ( 4.00 Hz) ( 60.0 cm) = 240 cm s = 2.40 m s

*P16.11 (a) (b)

(c)

ω = 2 π f = 2 π (5 s −1 ) = 31.4 rad s λ=

v 20 m s = = 4 .00 m 5 s−1 f

k=

2π 2π = = 1. 57 rad m λ 4m

In y = A sin ( kx − ω t + φ) we take A = 12 cm. At x = 0 and t = 0 we have y =(12 cm ) sin φ . To make this fit y = 0, we take φ= 0. Then y = ( 12.0 cm) sin (( 1.57 rad m) x − ( 31.4 rad s) t )

(d)

(e)

The transverse velocity is

∂y = − A ω cos ( kx − ωt ) ∂t

Its maximum magnitude is

Aω = 12 cm ( 31.4 rad s) = 3.77 m s

ay =

∂vy ∂t

=

∂ ( −Aω cos( kx −ω t ) ) = − Aω 2 sin( kx −ω t ) ∂t

The maximum value is P16.12

Aω2 = ( 0.12 m )(31.4 s−1 ) = 118 m s 2 2

At time t, the phase of y = (15.0 cm ) cos( 0.157 x − 50.3t ) at coordinate x is π φ = ( 0.157 rad cm ) x − (50.3 rad s) t. Since 60 .0 ° = rad, the requirement for point B is that 3 π φB = φ A ± rad, or (since x A = 0), 3

( 0.157 This reduces to x B = P16.13

431

rad cm ) x B − (50.3 rad s)t = 0 − (50.3 rad s)t ±

± π rad = ±6 .67 cm . 3 (0 .157 rad cm)

y = 0.250 sin (0.300 x − 40.0 t ) m Compare this with the general expression y = Asin ( kx −ω t) (a)

A = 0.250 m

(b)

ω = 40.0 rad s

(c)

k = 0.300 rad m

(d)

λ=

(e)

ω⎞ 40.0 rad s ⎞ λ=⎛ v=f λ=⎛ ( 20. 9 m ) = 133 m s ⎝ ⎝ 2π ⎠ ⎠ 2π

(f)

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

2π 2π = = 20.9 m k 0.300 rad m

π rad 3

432

Chapter 16

*P16.14 (a) (b)

y (cm)

See figure at right. 2π 2π = = 0. 125 s is the time from one peak .3 50 ω to the next one.

T=

10 0

t (s) 0.1

This agrees with the period found in the example in the text.

–10

FIG. P16.14

P16.15

(a)

k=

A = ymax = 8.00 cm = 0.080 0 m

2π 2π = = 7.85 m −1 λ ( 0.800 m )

ω = 2π f = 2π (3.00 ) = 6.00π rad s y = A sin ( kx +ω t)

Therefore, Or (where y ( 0, t) = 0 at t= 0) (b)

y = (0.080 0 ) sin ( 7.85 x + 6π t ) m

In general,

y = 0.080 0 sin ( 7.85 x +6π t +φ )

Assuming

y( x, 0) = 0 at x = 0.100 m

then we require that

0 = 0. 080 0sin ( 0. 785 + φ)

or

φ = −0.785 y = 0.080 0 sin ( 7.85 x + 6π t − 0.785 ) m

Therefore, P16.16

(a)

y (m) 0.2 0.1 0.0 –0.1 –0.2

t=0 0.2 0.4

x (m)

FIG. P16.16(a)

(b)

2π 2π = = 18. 0 rad m 0. 350 m λ 1 1 = 0.083 3 s T= = f 12.0 s k=

ω = 2 π f = 2 π 12.0 s = 75.4 rad s v = f λ = (12.0 s )(0.350 m ) = 4.20 m s (c)

y = A sin (kx + ω t + φ ) specializes to y = 0 .200 m sin( 18 .0 x m + 75 .4t s + φ ) at x = 0, t = 0 we require −3.00 × 10 2 m = 0.200 m sin ( + φ ) −

φ = −8 .63 ° = − 0 .151 rad so

y ( x , t ) = ( 0.200 m ) sin( 18.0 x m + 75.4t s − 0 .151 rad )

0.2

Wave Motion

P16.17

433

π y = ( 0.120 m) sin ⎛ x + 4π t ⎞ ⎝8 ⎠ (a)

v=

dy : dt

v = (0.120 ) (4 π ) cos

⎛π x +4 π t ⎞ ⎠ ⎝8

v ( 0.200 s, 1.60 m ) = −1.51 m s a=

dv : dt

2 a = (− 0. 120 m ) ( 4π ) sin

⎛π ⎞ x + 4π t ⎠ ⎝ 8

a ( 0.200 s, 1.60 m ) = 0 (b)

π 2π : = λ 8 2π : ω = 4π = T

k=

λ = 16.0 m T = 0.500 s v=

P16.18

(a)

λ 16.0 m = = 32.0 m s T 0.500 s y ( x , t ) = A sin( kx + ω t + φ)

Let us write the wave function as

y ( 0, 0) = A sin φ = 0.020 0 m dy = A ωcos φ = − 2. 00 m s dt 0, 0

ω=

Also,

2π 2π = = 80.0 π s T 0.025 0 s ⎛ 2.00 m s ⎞ 2 ⎛ vi ⎞ = ( 0.020 0 m) + ⎜ ⎝ ω⎠ ⎝ 80.0 π s ⎟⎠ 2

A 2 = xi2 +

2

A = 0.021 5 m (b)

0. 020 0 A sin φ = = − 2.51 = tanφ A cos φ − 2 / 80.0π 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 rad = 1.95 rad (c)

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

(d)

λ = v xT = (30.0 m s ) (0.025 0 s) = 0.750 m k=

2π 2π = = 8.38 m λ 0 .750 m

ω = 80.0π s

y ( x, t ) = ( 0.021 5 m ) sin (8.38 x rad m + 80 .0π t rad s + 1.95 rad )

434

P16.19

Chapter 16

(a)

f=

v (1 .00 m s) = 0. 500 Hz = λ 2. 00 m

ω = 2π f = 2π ( 0.500 s) = 3.14 rad s 2π 2π = = 3 .14 rad m λ 2 .00 m

(b)

k=

(c)

y = A sin (kx − ω t + φ ) becomes y = ( 0.100 m) sin ( 3.14x m − 3.14t s + 0 )

(d)

For x = 0 the wave function requires y = ( 0.100 m) sin ( − 3.14t s )

(e) (f)

y = ( 0.100 m) sin ( 4.71 rad− 3.14 t s ) ∂y = 0.100 m ( − 3.14 s ) cos ( 3. 14 x m − 3.14 t s ) ∂t The cosine varies between +1 and –1, so vy =

v y ≤ 0.314 m s P16.20

(a)

At x = 2.00 m , y = ( 0.100 m ) sin (1.00 rad − 20 .0 t ) Because this disturbance varies sinusoidally in time, it describes simple harmonic motion.

(b)

y = ( 0.100 m) sin ( 0.500x − 20.0t ) = A sin (kx − ωt ) so

Section 16.3 P16.21

ω = 20.0 rad s

f =

ω = 3.18 Hz 2π

The Speed of Waves on Strings

The down and back distance is 4.00 m + 4.00 m = 8 .00 m. The speed is then Now,

(a)

dtotal 4 (8. 00 m ) T = = 40.0 m s = t µ 0.800 s 0.200 kg µ= = 5.00 × 10 −2 kg m 4.00 m

v=

T = µv 2 = (5.00 ×10 −2 kg m ) (40.0 m s ) = 80.0 N 2

So P16.22

and

ω = 2π f = 2π ( 500) = 3 140 rad s, k =

ω 3 140 = = 16.0 rad m v 196

y = (2.00 × 10 −4 m )sin (16.0x − 3 140t ) (b)

v = 196 m s = T = 158 N

T − 4. 10 × 10 3 kg m

Wave Motion

P16.23

v=

1 350 kg ⋅ m s2 T = 520 m s = µ 5. 00 × 10− 3 kg m

P16.24

v=

T µ

435

T = µv2 = ρ Av2 = ρπ r2 v 2 T = (8 920 kg m 3 ) (π )( 7 .50 ×10 −4 m) ( 200 m s) 2

2

T = 631 N P16.25

P16.26

T Mg = = µ m ⲐL

T = Mg is the tension;

v=

Then,

MgL L2 = 2 m t

and

g=

MgL L = is the wave speed. m t

−3 Lm 1.60 m ( 4.00 × 10 kg ) 2 = 2 = 1 .64 m s Mt2 3 .00 kg (3.61 × 10 −2 s)

The period of the pendulum is T = 2 π

L g

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: F = µ

v=

Mg = mⲐ L

MgL m

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

L=

T g 2π

so Tg Mg T g = m 2π 2π

v=

P16.27

Since µ is constant, µ =

M m

T2 T1 = and v22 v12 2

2

⎛v ⎞ ⎛ 30.0 m s ⎞ ( 6.00 N ) = 13.5 N T2 = ⎜ 2 ⎟ T1 = ⎜ ⎝ v1 ⎠ ⎝ 20.0 m s ⎟⎠ P16.28

From the free-body diagram

mg = 2T sinθ mg T= 2 sin θ

The angle θ is found from

cosθ =

3LⲐ 8 3 = L Ⲑ2 4

∴ θ = 41.4° (a)

v=

or

T µ

v=

FIG. P16.28

⎛ ⎞ mg 9 .80 m s 2 =⎜ ⎟ m −3 2µ sin 41 .4° ⎝ 2 (8.00 ×10 kg m)sin 41.4° ⎠

⎛ m s⎞ v = ⎜ 30.4 m ⎝ k ⎟⎠

436

P16.29

Chapter 16

If the tension in the wire is T, the tensile stress is T so Stress = A The speed of transverse waves in the wire is v=

T = A( stress)

Stress A( Stress) Stress Stress = = = mⲐ L mⲐ AL mⲐVolume ρ

T = µ

where ρ is the density. The maximum velocity occurs when the stress is a maximum: vmax = f has units Hz = 1 s, so T =

*P16.30 (a)

with units (b) P16.31

1 has units of seconds, s . For the other T we have T = µ v 2, f

kg m2 kg ⋅ m = 2 = N . s m s2

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

The total time is the sum of the two times. µ L t= =L In each wire v T Let A represent the cross-sectional area of one wire. The mass of one wire can be written both as m = ρV = ρ AL and also as m = µL .

πρ d 2 4

Then we have

µ = ρA =

Thus,

⎛ πρd 2 ⎞ t = L⎜ ⎝ 4T ⎟⎠

⎡( π ) ( 8 920) ( 1. 00 × 10− 3 )2 ⎤ t = ( 20. 0) ⎢ ⎥ ( 4 )(150) ⎢⎣ ⎦⎥

12

For copper,

⎡( π ) ( 7 860) ( 1. 00 × 10− 3 )2 ⎤ t = ( 30. 0) ⎢ ⎥ ( 4) (150) ⎢⎣ ⎦⎥

12

For steel, The total time is

0.137 + 0.192 = 0.329 s

Section 16.5 P16.32

2 .70 × 108 Pa = 185 m s 7 860 kg m 3

f=

12

= 0.137 s

= 0.192 s

Rate of Energy Transfer by Sinusoidal Waves on Strings 30.0 v = = 60.0 Hz λ 0.500

ω = 2π f = 120π rad s

P = 1 µω A v = 1 ⎛⎝ 0 .180 ⎞⎠ (120 π) 2

2

2

2

3.60

2

( 0.100 )2 ( 30.0 )= 1.07 kW

Wave Motion

P16.33

437

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 an expanding circle. The power-per-width across the wave front

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

P 2 πr P16.34

(a)

1 T 2 2 ; P = µω A v 2 µ If L is doubled, v remains constant and

(b)

If A is doubled and ω is halved, P ⬀ ω 2A 2 remains constant .

T = constant; v =

P is constant . 2

(c) (d)

P16.35

A remains constant, so P remains constant . λ2 1 If L and λ are halved, then ω2 ⬀ 2 is quadrupled, so P is quadrupled . λ (Changing L doesn’t affect P .) If λ and A are doubled, the product ω 2 A2 ⬀

A = 5. 00 × 10− 2 m Therefore,

P

1 = µω 2 A 2 v : 2

µ = 4.00 × 10− 2 kg m v=

P = 300 W

T = 100 N

T = 50.0 m s µ

2( 300 ) 2P = ω2 = _____ µA2v ( 4. 00 × 10− 2 ( 5. 00 × 10− 2 2 ( 50. 0) ) )

ω = 346 rad s ω = 55. 1 Hz f = 2π P16.36

µ = 30. 0 g m = 30. 0 × 10− 3 kg m λ = 1.50 m f = 50 .0 Hz :

ω = 2π f = 314 s −1

2A = 0. 150 m:

A = 7.50 × 10 −2 m

(a)

2π ⎞ y = Asin ⎛ x −ω t ⎝λ ⎠ 2 y = ( 7. 50 × 10 − )sin (4. 19 x − 314t )

(b)

P

FIG. P16.36

2 314 ⎞ 1 1 = µω 2A 2v = ( 30. 0 × 10 −3 ) ( 314) 2 ( 7. 50 × 10 − 2) ⎛ W ⎝ 4.19 ⎠ 2 2

P

= 625 W

438

Chapter 16

P16.37

(a) (b) (c) (d)

P16.38

ω 2π ω 50 .0 = = m s = 62.5 m s k 0.800 2π k 2π 2π = m = 7.85 m λ= k 0.800 v = fλ =

f=

50.0 = 7. 96 Hz 2π

P = 1 µω 2

2

A2 v =

π Comparing y = 0.35 sin ⎛10π t −3π x + ⎞ with y = Asin ( kx − ω t + φ) = Asin ( ω t − kx − φ + π ) ⎝ 4⎠ we have k = (a)

λ ω 10 π s 3π = = = 3. 33 m s . , ω = 10π s, A = 0.35 m. Then v = f λ = 2π f 2π k 3π m m

The rate of energy transport is

P = 1 µω 2

(b)

2

A2 v =

1 (75 × 10 − 3 kg m )(10 π s)2 (0 .35 m )2 3 .33 m s = 15.1 W 2

The energy per cyc...