PSE9e ISM chapter16 final tủ tài liệu training

16
Wave Motion
CHAPTER OUTLINE
16.1

Propagation of a Disturbance

16.2

Analysis Model: Traveling Wave

16.3

The Speed of Transverse Waves on Strings

16.4

Reflection and Transmission

16.5

Rate of Energy Transfer by Sinusoidal Waves on Strings

16.6

The Linear Wave Equation

* An asterisk indicates a question or problem new to this edition.

ANSWERS TO OBJECTIVE QUESTIONS
OQ16.1

(i)

Answer (a). As the wave passes from the massive string to the
less massive string, the wave speed will increase according to
T
v=
.
µ

(ii)

Answer (c). The frequency will remain unchanged. The rate at
which crests come up to the boundary is the same rate at which
they leave the boundary.

(iii) Answer (a). Since v = f λ , the wavelength must increase.
OQ16.2

OQ16.3

(i)

Answer (a). Higher tension makes wave speed higher.

(ii)

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

(i)

The ranking is (c) = (d) > (e) > (b) > (a). Look at the coefficients
of the sine and cosine functions: (a) 4, (b) 6, (c) 8, (d) 8, (e) 7.

(ii)

The ranking is (c) > (a) = (b) > (d) > (e). Look at the coefficients
of x. Each is the wave number, 2π/λ, so the smallest k goes with
854

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Chapter 16

855

the largest wavelength.
(iii) The ranking is (e) > (d) > (a) = (b) = (c). Look at the coefficients
of t. The absolute value of each is the angular frequency ω = 2π f.
(iv) The ranking is (a) = (b) = (c) > (d) > (e). Period is the reciprocal
of frequency, so the ranking is the reverse of that in part (iii).
(v)

OQ16.4

The ranking is (c) > (a) = (b) = (d) > (e). 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: (a) 5, (b) 5, (c) 7.5, (d) 5, (e) 4.

Answer (b). From v =

T
, we must increase the tension by a factor
µ

of 4 to make v double.
OQ16.5

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

OQ16.6

Answer (b). Not all waves are sinusoidal. A sinusoidal wave is a
wave of a single frequency. In general, a wave can be a superposition
of many sinusoidal waves.

OQ16.7

(a) through (d): Yes to all. The maximum element speed and the
wave speed are related by vy ,max = ω A = 2π fA = 2π vA / λ . Thus the
amplitude or the wavelength of the wave can be adjusted to make
either vy, max or v larger.

OQ16.8

Answer (c). The power carried by a wave is proportional to its
frequency, wave speed, and the square of its amplitude. If the

frequency does not change, the amplitude is increased by a factor of
2. The wave speed does not change.

OQ16.9

Answer (c). The distance between two successive peaks is the
wavelength: λ = 2 m, and the frequency is 4 Hz. The frequency,
wavelength, and speed of a wave are related by the equation f λ = v.

ANSWERS TO CONCEPTUAL QUESTIONS
CQ16.1

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

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856

Wave Motion

CQ16.2

The type of wave you generate depends upon the direction of the
disturbance (vibration) you generate and the direction of its travel
(propagation).
(a)

To use a spring (or slinky) to create a longitudinal wave, pull a
few coils back and release.

(b)

For a transverse wave, jostle the end coil side to side.

CQ16.3

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. A
wave inverts when it reflects off a medium in which the wave speed
is smaller.

CQ16.4

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

CQ16.5

Since the frequency is 3 cycles per second, the period is 1/3 second =
333 ms.

CQ16.6

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

CQ16.7

As the pulse moves down the string, the elements 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.

CQ16.8

No. The vertical speed of an element will be the same on any string
because it depends only on frequency and amplitude:

vy ,max = ω A = 2π fA
The elements of strings with different wave speeds will have the
same maximum vertical speed.
CQ16.9

(a)

Let Δt = ts − tp represent the difference in arrival times of the
two waves at a station at distance d = vsts = vptp from the focus.
−1

⎛ 1
1⎞
Then d = Δt ⎜ − ⎟ .
⎝ vs v p ⎠

(b)

Knowing the distance from the first station places the focus 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.

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Chapter 16

857

SOLUTIONS TO END-OF-CHAPTER PROBLEMS
Section 16.1
P16.1

Propagation of a Disturbance

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

or

t=

km s ) ( 17.3 s )

( 4.50 km s )(17.3 s ) = 23.6 s
(7.80 − 4.50)

km s

and the distance is d = ( 7.80 km s ) ( 23.6 s ) = 184 km
P16.2

(a)

ANS. FIG. P16.2(a) shows the sketch of y(x,t) at t = 0.

ANS. FIG. P16.2(a)
(b)

ANS. FIG. P16.2(b) shows the sketch of y(x,t) at t = 2.00 s.

ANS. FIG. P16.2(b)
(c)

The graph in ANS. FIG. P16.2(b) has the same amplitude
and wavelength as the graph in ANS. FIG. P16.2(a). It

differs just by being shifted toward larger x by 2.40 m.

(d)

The wave has travelled d = vt = 2.40 m to the right.

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858
P16.3

Wave Motion
We obtain a function of the same shape by writing

y ( x,t ) =

6

⎡( x − x 0 )2 + 3 ⎤
⎣
⎦

where the center of the pulse is at x0 = 4.50t. Thus, we have
y=

6
⎡⎣( x − 4.50t )2 + 3 ⎤⎦

Note that for y to stay constant as t increases, x must increase by 4.50t,

as it should to describe the wave moving at 4.50 m/s.
P16.4

(a)

The longitudinal P wave travels a shorter distance and is
moving faster, so it will arrive at point B first.

(b)

The P wave that travels through the Earth must travel

(

)

a distance of 2R 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 ΔtP =

6.37 × 106 m
 817 s.
7 800 m/s

The Rayleigh wave that travels along the Earth’s surface must
travel a distance of

⎛π
⎞
s = Rθ = R ⎜ rad⎟ = 6.67 × 106 m

⎝3
⎠
at a speed of 4 500 m/s.
Therefore, it takes ΔtS =

6.67 × 106 m
 1 482 s.
4 500 m/s

The time difference is ΔT = ΔtS − ΔtP = 666 s = 11.1 min.

Section 16.2
P16.5

Analysis Model: Traveling Wave

Compare the specific equation to the general form:
y = (0.020 0 m) sin (2.11x − 3.62t) = y = A sin (kx − ω t + φ)
(a)

A = 2.00 cm

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Chapter 16

P16.6

(b)

k = 2.11 rad m → λ =

2π
= 2.98 m
k

(c)

ω = 3.62 rad s → f =

ω
= 0.576 Hz
2π

(d)

v = fλ =

(a)

ANS. FIG. P16.6(a) shows the snapshot of a wave on a string.

859

ω 2π 3.62
=
= 1.72 m s
2π k
2.11

ANS. FIG. P16.6(a)
(b)

ANS. FIG. P16.6(b) shows the wave from part (a) one-quarter
period later

ANS. FIG. P16.6(b)
(c)

ANS. FIG. P16.6(c) shows a wave with an amplitude 1.5 times
larger than the wave in part (a).

ANS. FIG. P16.6(c)
(d) ANS. FIG. P16.6(d) shows a wave with wavelength 1.5 times
larger than the wave in part (a).

ANS. FIG. P16.6 (d)

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860

Wave Motion
(e)

ANS. FIG. P16.6(e) shows a wave with frequency 1.5 times larger
than the wave in part (a): The wave appears the same as in ANS.
FIG. P16.6(a) because this is a snapshot of a given moment.

ANS. FIG. P16.6(e)
P16.7

The frequency of the wave is
f =

40.0 vibrations 4
= Hz
30.0 s
3

as the wave travels 425 cm in 10.0 s, its speed is
v=

425 cm
= 42.5 cm/s
10.0 s

and its wavelength is therefore

λ=
P16.8

v 42.5 cm s
=
= 31.9 cm = 0.319 m
f
1.33 Hz

Using data from the observations, we have λ = 1.20 m and
f =

8.00 crests 8.00 cycles 8.00
=
=
Hz
12.0 s
12.0 s
12.0

⎛ 8.00
⎞
Therefore, v = λ f = ( 1.20 m ) ⎜
Hz ⎟ = 0.800 m/s .
⎝ 12.0
⎠
P16.9

(a)

We note that sin θ = − sin ( −θ ) = sin ( −θ + π ) , so the given wave
function can be written as

y ( x,t ) = ( 0.350 ) sin ( −10π t + 3π x + π − π / 4 )
Comparing, 10π t − 3π x + π /4 = kx − ω t + φ . For constant phase, x
must increase as t increases, so the wave travels in the positive x
direction. Comparing the specific form to the general form, we
find that

ω 10π
v =   = 
 = 3.33 m/s.
k
3π

(

)

Therefore, the velocity is 3.33ˆi m/s .

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Chapter 16
(b)

861

Substituting t = 0 and x = 0.100 m, we have

π⎞
⎛
y ( 0.100 0 ) = ( 0.350 m ) sin ⎜ −0.300π + ⎟ = −0.054 8 m
⎝
4⎠
= −5.48 cm

2π

= 3π : λ = 0.667 m
λ

(c)

k=

(d)

vy =

ω = 2π f = 10π : f = 5.00 Hz

∂y
π⎞
⎛
= ( 0.350 ) ( 10π ) cos ⎜ 10π t − 3π x + ⎟
⎝
∂t
4⎠

vy , max = ( 10π ) ( 0.350 ) = 11.0 m/s
P16.10

The speed of waves along this wire is

v = f λ = ( 4.00 Hz ) ( 60.0 cm ) = 240 cm s = 2.40 m s
P16.11

(a)

ω = 2π f = 2π ( 5.00 s −1 ) = 31.4 rad s

(b)

λ=

v 20.0 m/s
=
= 4.00 m
f
5.00 s −1

k=

2π
2π
=
= 1.57 rad/m
λ
4.00 m

(c)

In y = A sin ( kx − ω t + φ ) we take A = 12.0 cm. At x = 0 and t = 0

we have y = ( 12.0 cm ) sin φ . To make this fit y = 0, we take φ = 0.
Then

y = 0.120 sin (1.57x − 31.4t), where x and y are in meters and t is

in seconds
(d) The transverse velocity is

∂y
= −Aω cos ( kx − ω t ) .
∂t

Its maximum magnitude is

Aω = ( 12.0 cm ) ( 31.4 rad s ) = 3.77 m s
(e)

ay =

∂vy
∂t

=

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

The maximum value is Aω 2 = ( 0.120 m ) ( 31.4 s −1 ) = 118 m/s 2
2

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862

P16.12

Wave Motion
At time t, the motion at point A, where x = 0, is

y A = ( 1.50 cm ) cos ( −50.3t )
At point B, the motion is

π⎞
⎛
y B = ( 15.0 cm ) cos ( 15.7xB − 50.3t ) = ( 15.0 cm ) cos ⎜ −50.3t ± ⎟
⎝
3⎠
which implies
15.7xB = ( 15.7 m −1 ) xB = ±

xB = −0.066 7 m = ±6.67 cm

or
P16.13

π
3

v ( 1.00 m s )
=
= 0.500 Hz
λ
2.00 m

(a)

f =

(b)

ω = 2π f = 2π ( 0.500 s ) = π s = 3.14 rad s

(c)

k=

(d)

y = A sin ( kx − ω t + φ ) becomes

2π
2π
=
= π m = 3.14 rad m
λ 2.00 m

y = 0.100 sin (π x − π t )
(e)

For x = 0 the wave function requires

y = 0.100 sin (π t )
(f)
(g)

y = 0.100 sin ( 4.71 − π t )
vy =

∂y
= 0.100 m ( − 3.14 s ) cos ( 3.14x m − 3.14t s )
∂t

The cosine varies between +1 and −1, so maximum
vy = 0.314 m s .
P16.14

(a)

ANS. FIG. P16.14 shows the y vs. t plot of the given wave.

ANS. FIG. P16.14

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Chapter 16
(b)

The time from one peak to the next one is
T=

(c)
P16.15

863

2π
2π
=
= 0.125 s
ω 50.3 s −1

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

The wave function is given as

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

We differentiate the wave function with respect to time to obtain
the velocity:

v=

∂y
⎛π
⎞
: v = ( 0.120 ) ( 4π ) cos ⎜ x + 4π t ⎟
⎝8
⎠

∂t

v ( 0.200 s, 1.60 m ) = −1.51 m/s
(b)

Differentiating the velocity function gives the acceleration:

a=

∂v
2
⎛π
⎞
: a = ( −0.120 m ) ( 4π ) sin ⎜ x + 4π t ⎟
⎝8
⎠
∂t

a ( 0.200 s, 1.60 m ) = 0

P16.16

π 2π
=
: λ = 16.0 m
8
λ

(c)

k=

(d)

ω = 4π =

(e)

v=

(a)

At x = 2.00 m, y = 0.100 sin ( 1.00 − 20.0t ) . Because this

2π
: T = 0.500 s
T

λ 16.0 m
=
= 32.0 m s
T 0.500 s

disturbance varies sinusoidally in time, it describes simple
harmonic motion.
(b)

At x = 2.00 m, compare y = 0.100 sin ( 1.00 − 20.0t ) to A cos (ω t + φ ) :

y = 0.100sin ( 1.00 − 20.0t ) = −0.100sin ( 20.0t − 1.00 )

= 0.100cos(20.0t − 1.00 + π )
= 0.100cos ( 20.0t + 2.14 )
so

ω = 20.0 rad s and f =

ω
= 3.18 Hz
2π

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864
P16.17

Wave Motion
The wave function is: y = 0.25 sin (0.30x − 40t) m
Compare this with the general expression y = A sin (kx − ω t):

P16.18

(a)

A = 0.250 m

(b)

ω = 40.0 rad s

(c)

k = 0.300 rad m

(d)

λ=

(e)

⎛ 40.0 rad s ⎞
⎛ω ⎞
v = fλ = ⎜ ⎟ λ = ⎜
⎟⎠ ( 20.9 m ) = 133 m s
⎝ 2π ⎠
⎝
2π

(f)

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

(a)

ANS. FIG. P16.18(a) shows a sketch of the wave at t = 0.

2π
2π
=
= 20.9 m

k
0.300 rad m

ANS FIG. P16.18(a)
(b)

k=

2π
2π
=
= 18.0 rad m
λ
0.350 m

(c)

T=

1
1
=
= 0.083 3 s
f 12.0/s

(d)

ω = 2π f = 2π 12.0 s = 75.4 rad s

(e)

v = f λ = ( 12.0 s ) ( 0.350 m ) = 4.20 m s

(f)

y = A sin ( kx + ω t + φ ) specializes to
y = ( 0.200 m ) sin ( 18.0 x m + 75.4t s + φ )

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Chapter 16
(g)

865

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 sin ( 18.0x + 75.4t − 0.151) , where x and y are in
meters and t is in seconds.

P16.19

Using the traveling wave model, we can put constants with the right
values into y = A sin ( kx + ω t + φ ) to have the mathematical
representation of the wave. We have the same (positive) signs for both

kx and ωt so that a point of constant phase will be at a decreasing value
of x as t increases—that is, so that the wave will move to the left.
The amplitude is

A = ymax = 8.00 cm = 0.080 0 m

The wave number is

k=

2π
2π
=
= 2.50π m −1
λ 0.800 m

ω = 2π f = 2π ( 3.00 s −1 ) = 6.00π rad/s

The angular frequency is
(a)

In y = A sin ( kx + ω t + φ ) , choosing φ = 0 will make it true that
y(0, 0) = 0. Then the wave function becomes upon substitution of
the constant values for this wave

y = ( 0.080 0 ) sin ( 2.50π x + 6.00π t )
(b)

In general, y = ( 0.080 0 ) sin ( 2.50π x + 6.00π t + φ )
If y(x, 0) = 0 at x = 0.100 m, we require

0 = ( 0.080 0 ) sin ( 2.50π + φ )
so we must have the phase constant be φ = −0.250π rad.
Therefore, the wave function for all values of x and t is
y = 0.080 0 sin ( 2.50π x + 6.00π t − 0.250π ) , where x and y are in
meters and t is in seconds.

P16.20

(a)

Let us write the wave function as y ( x,t ) = A sin ( kx + ω t + φ ) .
We have y i = y ( 0, 0 ) = A sin φ = 0.020 0 m
and vi = v ( 0, 0 ) =

∂y
∂t

= Aω cos φ = −2.00 m/s.
0, 0

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866

Wave Motion
Also, ω =

2π

2π
=
= 80.0π s −1 .
T
0.025 0 s

Use the identity sin 2 φ + cos 2 φ = 1 and the expressions for yi and
vi:

( A sin φ )2 + ( Aω cos φ )2 = 1
A 2ω 2

A2

( A sin φ )

2

2
Aω cos φ )
(
+
= A2

ω2

2

2
⎛v ⎞

⎛ −2.00 m/s ⎞
A = y + ⎜ i ⎟ = ( 0.020 0 m ) + ⎜
⎝ω⎠
⎝ 80.0π s −1 ⎟⎠
2

2

2
i

A = 0.021 5 m
(b)

ω y i ω ( A sin φ )
80.0π ( 0.020 0 )
=
= tan φ → tan φ =
= −2.51
vi
ω A cos φ
−2.00
Your calculator’s answer φ = tan−1 (−2.51) = −1.19 rad is an angle
in the fourth quadrant with a negative sine and positive cosine,
just the reverse of what is required. Recall on the unit circle, an
angle with a negative tangent can be in either the second or
fourth quadrant. The sine is positive and the cosine is negative in
the second quadrant. The angle in the second quadrant is

φ = π − 1.19 rad = 1.95 rad

= Aω = ( 0.021 5 m ) ( 80.0π s ) = 5.41 m/s

(c)

vy ,

(d)

λ = vxT = ( 30.0 m s ) ( 0.025 0 s ) = 0.750 m
k=

max

2π
2π
=
= 8.38 m −1 ,
λ
0.750 m

ω = 80.0π s −1

y ( x, t ) = ( 0.021 5 ) sin ( 8.38x + 80.0π t + 1.95 )

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Chapter 16

Section 16.3

P16.21

867

The Speed of Transverse Waves on Strings

If the tension in the wire is T, the tensile stress is
stress =

T
A

so

T = A ( stress )

The speed of transverse waves in the wire is
v=

A ( Stress )
=
m/L

T
=
µ

Stress
=
m / AL

Stress
=
m / Volume

Stress
ρ

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

vmax =
P16.22

2.70 × 108 Pa
= 185 m s
7860 kg m 3

The speed is given by

1 350 kg ⋅ m s 2
= 520 m s
5.00 × 10−3 kg m

T
v=
=
µ
P16.23

The two wave speeds can be written as

v1 = T1 µ
Since µ is constant, µ =

and

v2 = T2 µ

T2 T1
= , and
v22 v12

2

2

⎛ 30.0 m s ⎞
⎛v ⎞
T2 = ⎜ 2 ⎟ T1 = ⎜
(6.00 N ) = 13.5 N
⎝ v1 ⎠
⎝ 20.0 m s ⎟⎠
P16.24

(a)

For the first equation,

1

1
1
1
f  =      →    T  =      →    [T ] = 
 =  −1  = T 
T
f
[f] T
units are seconds
2

T
M ⎛ L⎞
ML
v = 
    →    T  =  µ v 2  →    [T ] =  ⎡⎣ µ v 2 ⎤⎦  =  ⎜ ⎟  =  2
⎝
⎠
µ
L T
T
units are newtons

(b)

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

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868
P16.25

Wave Motion
The down and back distance is 4.00 m + 4.00 m = 8.00 m.
The speed is then v =

P16.26

dtotal 4 ( 8.00 m )
T
=
= 40.0 m s =
.
t
0.800 s
µ

0.200 kg
= 5.00 × 10−2 kg/m.
4.00 m

Now,

µ=

So

T = µ v 2 = ( 5.00 × 10−2 kg/m ) ( 40.0 m/s ) = 80.0 N .

(a)

2

To write the equation, we determine the angular frequency and
wave number:

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

ω 3 140
=
= 16.0 m −1
v
196

y = ( 2.00 × 10−4 ) sin ( 16.0x − 3 140t ) , where y and x are in meters
and t is in seconds.
(b)
P16.27

v = 196 m s =

T
→ T = 158 N
4.10 × 10−3 kg m

The total time interval is the sum of the two time intervals.
In each wire

Δt =

L
µ
=L
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.
Then we have µ = ρ A =
Thus,

⎛ πρ d 2 ⎞
Δt = L ⎜
⎝ 4T ⎟⎠

πρd 2
.
4
12

For copper,

⎡ (π ) ( 8 920 kg/m 3 ) ( 1.00 × 10−3 m )2 ⎤
⎥
Δt = ( 20.0 m ) ⎢
4 ) ( 150 N )
(
⎢

⎥
⎣
⎦

12

= 0.137 s

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 16

869

For steel,

⎡ (π ) ( 7 860 kg/m 3 ) ( 1.00 × 10−3 m )2 ⎤
⎥
Δt = ( 30.0 m ) ⎢
4 ) ( 150 N )
(
⎢
⎥
⎣
⎦

12

= 0.192 s

The total time interval is 0.137 + 0.192 = 0.329 s
P16.28

The tension in the string is T = mg, where g is the acceleration of
gravity on the Moon, about one-sixth that of Earth. From the data
given, what is the acceleration of gravity on the Moon?
The wave speed is

v=

T
Mg
=
=
µ
m/L

MgL L
MgL L2
mL
= →
= 2 →g=
m
t
m
t
Mt 2

−3

mL ( 4.00 × 10 kg ) ( 1.60 m )
g=
=
= 3.13 m/s 2
Mt 2 ( 3.00 kg ) ( 26.1 × 10−3 s )2

The calculated gravitational acceleration of the Moon is almost twice
that of the accepted value.
P16.29

(a)

The tension in the string is

F = mg = ( 3.00 kg ) ( 9.80 m s 2 ) = 29.4 N
Then, from v =

µ=
(b)

F
, the mass per unit length is
µ

F
29.4 N
=
= 0.0510 kg m
2
v

( 24.0 m s )2

When m = 2.00 kg, the tension is

F = mg = ( 2.00 kg ) ( 9.80 m s 2 ) = 19.6 N
and the speed of transverse waves in the string is
v=

F
=
µ

19.6 N
= 19.6 m s
0.0510 kg m

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


870
P16.30

Wave Motion
From the free-body diagram mg = 2T sin θ

T=

mg
2 sin θ

The angle θ is found from
cos θ =

3L/8 3
=
L/2 4

ANS. FIG. P16.30

∴ θ = 41.4°

(a)

v=
=
or

(b)
P16.31

T
mg
=
µ
2 µ sin θ
⎛
⎞
mg
9.80 m/s 2
=⎜

⎟ m
2 µ sin 41.4° ⎝ 2 ( 8.00 × 10−3 kg/m ) sin 41.4° ⎠
v = ( 30.4 ) m, where v is in meters per second and
m is in kilograms.

v = 60.0 = 30.4 m and

We use v =

m = 3.89 kg

T
to solve for the tension:
µ

T = µ v 2 = ρ Av 2 = ρπ r 2 v 2
T = ( 8920 kg m 3 ) (π ) ( 7.50 × 10−4 m ) ( 200 m s )
2

2

T = 631 N

Section 16.5
P16.32

(a)

Rate of Energy Transfer by Sinusoidal
Waves on Strings

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.

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Chapter 16
(b)

871

We write P = FvA2, where F is some constant. With no absorption
of energy,
2
2
Fvgranite Agranite
= Fvmudfill Amudfill

vgranite
vgranite
25.0vgranite
Amudfill
=
=
=
= 5.00
vgranite

Agranite
vmudfill
vgranite
25.0
The amplitude increases by 5.00 times.
P16.33

We are given T = constant; we use the equation for the speed of a wave
T
, and the power supplied to a vibrating string,
on a string, v =
µ
1
P = µω 2 A 2 v.
2
(a)

If L is doubled, µ is still the same, so v remains constant: therefore
P is constant: 1 .

(b)

If A is doubled and ω is halved, P ∝ ω 2 A 2 remains constant: 1 .

(c)

If λ and A are doubled, the product ω 2 A 2 ∝

A2
remains constant,

λ2

so 1 .
(d) If L and λ are halved, µ is still the same, and ω 2 ∝

1
is
λ2

quadrupled, so P is increased by a factor of 4 .
P16.34

We will use the expression for power carried by a wave on a string.
T
100 N
=
= 50.0 m/s
The wave speed is v =
µ
4.00 × 10−2 kg/m
From P =

1
µω 2 A 2 v, we have
2

ω2 =

2P
2(300 N ⋅ m/s)

=
2
–2
µ A v ( 4.00 × 10 kg m ) ( 5.00 × 10 –2  m )2 (50.0 m s)

Computing,

ω = 346 rad/s

and

f=

ω
= 55.1 Hz
2π

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


872
P16.35

P16.36

Wave Motion
Comparing the given wave function, y = (0.15) sin (0.80x − 50t), with
the general wave function, y = A sin (kx − ωt), we have k = 0.80 rad/m,
ω = 50 rad/s, and A = 0.15 m.

ω 2 π ω 50.0
= =
m s = 62.5 m s
2π k
k 0.800

(a)

v = fλ =

(b)

λ=

2π
2π
=
m = 7.85 m
k
0.800

(c)

f=

50.0
= 7.96 Hz
2π

(d)

1
1
2
2
P = µω 2 A 2 v = (12.0 × 10−3 ) ( 50.0) ( 0.150) (62.5) W = 21.1 W
2
2

The frequency and angular frequency of the wave are
f=

v 30.0 m/s
=
= 60.0 Hz and ω = 2π f = 120π rad s
λ
0.500 s

The power that is required is then

1
µω 2 A 2 v
2
1 ⎛ 0.180 kg ⎞
2
2
= ⎜
⎟⎠ ( 120π rad/s ) ( 0.100 m ) ( 30.0 m/s )
⎝
2 3.60 m

P=

= 1.07 kW
P16.37

We are given µ = 30.0 g m = 30.0 × 10−3 kg m , with

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
⎞
From y = A sin ⎜
x − ω t⎟ ,
⎝ λ
⎠

(b)

2 ⎛ 314 ⎞
1
1
2
P = µω 2 A 2 v = ( 30.0 × 10−3 ) ( 314) (7.50 × 10−2 ) ⎜

⎟W
⎝ 4.19 ⎠
2
2

y = ( 0.075 ) sin ( 4.19x − 314t )

P = 625 W

ANS. FIG. P16.37
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 16
P16.38

873

Originally,

1
P0 = µω 2 A 2 v
2
1
T
P0 = µω 2 A 2
2
µ
1
P0 = ω 2 A 2 T µ

2
The doubled string will have doubled mass per length. Presuming that
we hold tension constant, it can carry power larger by 2 times:

1
⎛1
⎞
P =  ω 2 A 2 T ( 2 µ )  =  2 ⎜ ω 2 A 2 T µ ⎟  =  2P0
⎝2
⎠
2
P16.39

Comparing

π⎞
⎛
y = 0.350sin ⎜ 10π t − 3π x + ⎟
⎝
4⎠
with
y = A sin ( kx − ω t + φ ) = A sin (ω t − kx − φ + π )

we have

k = 3π m −1 , ω = 10π s −1 , and A = 0.350 m
Then,
−1
⎛ λ ⎞ ω 10π s
v = f λ = ( 2π f ) ⎜

=
=
= 3.33 m/s
⎝ 2π ⎟⎠ k 3π m −1

(a)

The rate of energy transport is

1
µω 2 A 2 v
2
2
1
2
= ( 75 × 10−3 kg/m ) ( 10π s −1 ) ( 0.350 m ) ( 3.33 m/s )
2

P=

= 15.1 W
(b)

Recall that vT = λ. The energy per cycle is

Eλ = P T =
=

1
µω 2 A 2 λ

2

2
1
2π ⎞
2⎛
75.0 × 10−3 kg m ) ( 10π s −1 ) ( 0.350 m ) ⎜
(
⎝ 3π m −1 ⎟⎠
2

= 3.02 J
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


874
P16.40

Wave Motion
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
2π r

Section 16.6
P16.41

The Linear Wave Equation

The important thing to remember with partial derivatives is that you
treat all variables as constants, except the single variable of interest.
Keeping this in mind, we must apply two standard rules of differentiation
to the function y = ln[b(x − vt)]:
∂
1 ∂[ f (x)]
ln f (x)] =
[
∂x
f (x) ∂x

∂ ⎡ 1 ⎤ ∂
−1
−2 ∂[ f (x)]
⎢
⎥ = [ f (x)] = (−1)[ f (x)]
∂x ⎣ f (x) ⎦ ∂x
∂x
1 ∂[ f (x)]
=−
2
[ f (x)] ∂x

[1]

[2]

Applying [1],

∂y ⎛
1 ⎞ ∂(bx – bvt) = ⎛
1 ⎞
1
=⎜
⎟⎠
⎜⎝ b(x– vt) ⎟⎠ ( b ) = x – vt
b(x–
vt)
∂x
⎝
∂x
Applying [2],

∂2 y
1
2 = –
∂x
(x – vt)2
In a similar way,

∂y
−v
=
∂t x − vt

and

∂2 y
v2
=
∂t 2 ( x − vt )2

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 16

875

From the second-order partial derivatives, we see that it is true that

∂2 y 1 ∂2 y
=
∂x 2 v 2 ∂t 2
so the proposed function is one solution to the wave equation.
P16.42

(a)

A = (7.00 + 3.00)( 4.00) yields A = 40.0

A = 7.00, B = 0, and C = 3.00

(b)

In order for two vectors to be equal, they must have the same
magnitude and the same direction in three-dimensional space.
All of their components must be equal, so all coefficients of the
unit vectors must be equal.

(c)

A=0

(d)

B = 7.00 in meters, C = 3.00 in m−1, D = 4.00 in s−1,

E = 2.00 in rad.
Identify corresponding parts. In order for two functions to be
identically equal, corresponding parts must be identical. The
argument of the sine function must have no units, or be equivalent to units of radians.

(e)

P16.43

∂2 y 1 ∂2 y
The linear wave equation is 2 = 2 2 .
∂x
v ∂t
If

y = eb(x−vt)

Then

∂y
b x−vt
= −bve ( )
∂t

and

∂y
b x−vt
= be ( )
∂x

∂2 y
b x−vt
= b2 v2e ( )
2
∂t

and

∂2 y
b x−vt
= b2e ( )
2
∂x

Therefore,
P16.44

(a)

2
∂2 y
b(x−vt)
2 ∂ y
is a solution.
=
v
, demonstrating that e
2
2
∂t
∂x

From y = x 2 + v 2t 2 ,
evaluate

∂y
= 2x
∂x

and

∂2 y
=2
∂x 2

Also,

∂y
= v 2 2t
∂t

and

∂2 y
= 2v 2
∂t 2

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876

Wave Motion
∂2 y 1 ∂2 y
?
=
∂t 2 v 2 ∂t 2

Does

1
2v 2 ) and this is true, so the
2 (
v
wave function does satisfy the wave equation.

By substitution, we must test 2 =

(b)

Note

1
1
1
1
1
1
( x + vt )2 + ( x − vt )2 = x 2 + xvt + v 2t 2 + x 2 − xvt + v 2t 2
2
2
2
2
2
2
2
2 2
=x +v t
as required. So
1
2
f ( x + vt ) = ( x + vt )
2

(c)

1
2
g ( x − vt ) = ( x − vt )
2

and

y = sin x cos vt makes
∂y
= cos x cos vt
∂x

∂2 y
= −sin x cos vt
∂x 2

∂y
= −v sin x sin vt
∂t

∂2 y
= −v 2 sin x cos vt
∂t 2

−1
∂2 y 1 ∂2 y
= 2 2 becomes −sin x cos vt = 2 v 2 sin x cos vt which
2
v
∂x

v ∂t
is true, as required.

Then

sin ( x + vt ) = sin x cos vt + cos x sin vt

Note

sin ( x − vt ) = sin x cos vt − cos x sin vt
sin x cos vt = f ( x + vt ) + g ( x − vt ) with

So

f ( x + vt ) =

1
sin ( x + vt )
2

and

g ( x − vt ) =

1
sin ( x − vt )
2

Additional Problems
P16.45

The equation v = λ f is a special case of
speed = (cycle length)(repetition rate)
Thus,

v = (19.0 × 10−3 m frame) ( 24.0 frames s) = 0.456 m s
© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.


Chapter 16
P16.46

877

Assume a typical distance between adjacent people ∼ 1 m.
Then the wave speed is v =

Δx 1 m
~
~ 10 m/s.
Δt 0.1 s

Model the stadium as a circle with a radius of order 100 m. Then, the
time for one circuit around the stadium is
2
2 π r 2 π (10 )
T=
~
= 63 s ~ 1 min
v

10 m s

P16.47

The speed of the wave on the rope is v =
therefore, m =

T
and in this case T = mg;
µ

µv2
.
g

Now v = fλ implies v =

ω
so that
k

2

2

0.250 kg m ⎡ 18π s −1 ⎤
µ⎛ω⎞
m= ⎜ ⎟ =
⎢
⎥ = 14.7 kg

g⎝ k⎠
9.80 m s 2 ⎣ 0.750π m −1 ⎦
*P16.48

v=

2d
gives
t

d=
P16.49

vt 1
= ( 6.50 × 103 m s ) ( 1.85 s ) = 6.01 km
2 2

The block-cord-Earth system is isolated, so energy is conserved as the
block moves down distance x:

ΔK + ΔU = 0 →

( K +U

g

+U s )top = ( K +U g +U s ) bottom

1
0 + Mgx + 0 + 0 = 0 + 0 + kx 2

2
2Mg
x=
k
(a)

T = kx = 2Mg = 2 ( 2.00 kg ) ( 9.80 m s 2 ) = 39.2 N

(b)

L = L0 + x = L0 +

L = 0.500 m +

2Mg
k

39.2 N
= 0.892 m
100 N m

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878

Wave Motion

(c)

v=

T
TL
=
µ
m

v=

39.2 N × 0.892 m
5.0 × 10−3 kg

v = 83.6 m/s

P16.50

The block-cord-Earth system is isolated, so energy is conserved as the
block moves down distance x:

ΔK + ΔU = 0 →

(K + U

g

+ Us

)

top

(

= K + U g + Us

0 + Mgx + 0 + 0 = 0 + 0 +
Mgx =

P16.51

)

bottom

1 2
kx
2

1 2
kx
2

(a)

T = kx = 2Mg

(b)

L = L0 + x = L0 +

(c)

v=

(a)

The wave function becomes

2Mg
k

2Mg ⎛
2Mg ⎞
L
+
0
m ⎜⎝
k ⎟⎠

T
TL
=
=
µ
m

0.175 m = ( 0.350 m) sin ⎡⎣( 99.6 rad s) t⎤⎦
or

sin ⎡⎣( 99.6 rad s ) t ⎤⎦ = 0.500

The smallest two angles for which the sine function is 0.500 are
30.0° and 150°, i.e., 0.523 6 rad and 2.618 rad.

( 99.6 rad s ) t1 = 0.523 6 rad, thus t1 = 5.26 ms

( 99.6 rad s ) t

2

= 2.618 rad, thus t2 = 26.3 ms

Δt ≡ t2 − t1 = 26.3 ms − 5.26 ms = 21.0 ms
(b)

Distance traveled by the wave
⎛ 99.6 rad s ⎞
⎛ω ⎞
= ⎜ ⎟ Δt = ⎜
21.0 × 10−3 s ) = 1.68 m
(
⎟
⎝ k⎠
⎝ 1.25 rad m ⎠

© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.