# CHAPTER 16 WAVE MOTION

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```					CHAPTER 16                      WAVE MOTION

ActivPhysics can help with these problems:                1.25 m apart. What is the speed of the waves on the
Activities 10.1, 10.2, 10.7, 10.10                        cable? Compare with the speed of light in vacuum.
Section 16-2: Wave Properties                           Solution
Problem                                                 The distance between adjacent wave crests is one
1. Ocean waves with 18-m wavelength travel at           wavelength, so the wave speed in the cable (Equa-
5.3 m/s. What is the time interval between wave      tion 16-1) is v = f λ = (145×106 Hz)(1.25 m) = 1.81×
crests passing under a boat moored at a ﬁxed         108 m/s = 0.604c, where c = 3×108 m/s is the wave
location?                                            speed in vacuum.

Solution                                                Problem
Wave crests (adjacent wavefronts) take a time of one    6. Calculate the wavelengths of (a) a 1.0-MHz AM
period to pass a ﬁxed point, traveling at the wave         radio wave, (b) a channel 9 TV signal (190 MHz),
speed (or phase velocity) for a distance of one            (c) a police radar (10 GHz), (d) infrared radiation
wavelength. Thus T = λ/v = 18 m/(5.3 m/s) = 3.40 s.        from a hot stove (4.0×1013 Hz), (e) green light
(6.0×1014 Hz), and (f) 1.0×1018 Hz X rays. All are
Problem                                                    electromagnetic waves that propagate at
2. Ripples in a shallow puddle are propagating at          3.0×108 m/s.
34 cm/s. If the wave frequency is 5.2 Hz, what are
(a) the period and (b) the wavelength?               Solution
With Equation 16-1 in an equivalent form, λ = vT =
Solution                                                v/f, we ﬁnd: (a) λ = (3×108 m/s)/(106 Hz) =
Equation 16-1 gives T = 1/f = 1/5.2 Hz = 0.192 s and    300 m, (b) λ = 1.58 m, (c) λ = 3 cm, (d) λ = 7.5 µm,
λ = v/f = vT = (34 cm/s)/(5.2 Hz) = 6.54 cm.            (e) λ = 500 nm, (f) λ = 3˚. (See Appendix C on
A
units.)
Problem
3. An 88.7-MHz FM radio wave propagates at the          Problem
speed of light. What is its wavelength?              7. Detecting objects by reﬂecting waves oﬀ them is
eﬀective only for objects larger than about one
Solution                                                   wavelength. (a) What is the smallest object that
From Equation 16-1, λ = v/f = (3×108 m/s)÷                 can be seen with visible light (maximum frequency
(88.7×106 Hz) = 3.38 m.                                    7.5×1014 Hz)? (b) What is the smallest object that
can be detected with a medical ultrasound unit
Problem                                                    operating at 5 MHz? The speed of ultrasound
4. One end of a rope is tied to a wall. You shake the      waves in body tissue is about 1500 m/s.
other end with a frequency of 2.2 Hz, producing
waves whose wavelength is 1.6 m. What is their       Solution
propagation speed?                                   (a) The wavelength of light corresponding to this
maximum frequency is λ = c/f = (3×108 m/s)÷
Solution                                                (7.5×1014 Hz) = 400 nm, violet in hue (see Equa-
v = f λ = (2.2 Hz)(1.6 m) = 3.52 m/s (see Equa-         tion 16-1). (b) The ultrasonic waves described have
tion 16-1).                                             wavelength λ = v/f = (1500 m/s)/(5 MHz) = 0.3 mm.

Problem                                                 Problem
5. A 145-MHz radio signal propagates along a cable.     8. A seismograph located 1200 km from an earthquake
Measurement shows that the wave crests are spaced       detects waves from the quake 5.0 min after the
2    CHAPTER 16

quake occurs. The seismograph oscillates in step     Problem
with the waves, at a frequency of 3.1 Hz. Find the   11. An ocean wave has period 4.1 s and wavelength
wavelength of the waves.                                 10.8 m. Find (a) its wave number and (b) its
angular frequency.
Solution
The wave speed can be calculated from the distance       Solution
and the travel time, which, together with the            From Equations 16-3 and 4, (a) k = 2π/10.8 m =
frequency and Equation 16-1, gives a wavelength of       0.582 m−1 , and (b) ω = 2π/(4.1 s) = 1.53 s−1 .
λ = v/f = (d/t)/f = 1200 km/(5×60×3.1) = 1.29 km.
Problem
Problem
12. Find (a) the amplitude, (b) the wavelength,
9. In Fig. 16-28 two boats are anchored oﬀshore and          (c) the period, and (d) the speed of a wave whose
are bobbing up and down on the waves at the rate          displacement is given by y = 1.3 cos(0.69x + 31t),
of six complete cycles each minute. When one boat         where x and y are in cm and t is in seconds. (e) In
is up the other is down. If the waves propagate at        which direction is the wave propagating?
2.2 m/s, what is the minimum distance between the
boats?                                                Solution
(a) A = ymax = 1.3 cm, (b) λ = 2π/k = 2π÷
Solution                                                 0.69 cm−1 = 9.11 cm, (c) T = 2π/ω = 2π/31 s−1 =
The boats are 180◦ = π rad out of phase, so the          0.203 s−1 , and (d) v = ω/k = 44.9 cm/s. (e) A phase
minimum distance separating them is half a               of the form kx + ωt describes a wave propagating in
wavelength. (In general, they could be an odd number     the negative x direction.
of half-wavelengths apart.) The frequency is 6/60 s =
1
0.1 Hz, so 2 λ = 1 v/f = 1 (2.2 m/s)/(0.1/s) = 11 m.
2       2                               Problem
(Fig. 16-28 shows the answer, not the question.)
13. A simple harmonic wave of wavelength 16 cm and
amplitude 2.5 cm is propagating along a string in
the negative x direction at 35 cm/s. Find (a) the
angular frequency and (b) the wave number.
(c) Write a mathematical expression describing
the displacement y of this wave (in centimeters) as
a function of position and time. Assume the
displacement at x = 0 is a maximum when t = 0.

Solution
figure 16-28 Problem 9 Solution.              (b) Equation 16-4 gives k = 2π/16 cm = 0.393 cm−1 ,
and (a) Equation 16-6 gives ω = kv = (0.393 cm−1 )×
(35 cm/s) = 13.7 s−1 . (c) Equation 16-5, for a wave
Section 16-3: Mathematical Description of                moving in the negative x direction, becomes
Wave Motion                                              y(x, t) = (2.5 cm) cos[(0.393 cm−1 )x + (13.7 s−1 )t].

Problem                                                  Problem
10. Ultrasound used in a particular medical imager       14. Figure 16-29 shows a simple harmonic wave at
has frequency 4.8 MHz and wavelength 0.31 mm.            time t = 0 and later at t = 2.6 s. Write a
Find (a) the angular frequency, (b) the wave             mathematical description of this wave.
number, and (c) the wave speed.
Solution
Solution                                                 Inspection of Fig. 16-29 shows that the wavelength is
(a) Equation 16-3 gives ω = 2π(4.8 MHz) = 3.02×          8 cm, the amplitude is 1.5 cm, and the velocity is
107 s−1 . (b) Equation 16-4 gives k = 2π/0.31 mm =       v = ∆x/∆t = 2 cm/2.6 s = 0.769 cm/s. The phase
2.03×104 m−1 . (c) Together, these give v = f λ =        constant is zero (since y = A at t = 0 and x = 0) and
ω/k = 1.49 km/s (see Equation 16-1 or 6).                the wave is traveling in the positive x direction. Thus,
k = 2π/λ = 0.785 cm−1 , ω = kv = 0.604 s−1 , and the
CHAPTER 16       3

waveform is y(x, t) = (1.5 cm) cos[(0.785 cm−1 )x −         positive x direction with speed v, can be obtained by
(0.604 s−1 )t].                                             replacing x by x − vt, y(x, t) = f (x − vt). For the given
f (x) and v, y(x, t) = 2[(x − 3t)4 + 1]−1 , with x and y in
cm and t in s.

Problem
18. Plot the answer to the previous problem as a
function of position x for the two cases t = 0 and
t = 4 s, and verify that your plots are consistent
with the pulse speed of 3 cm/s.
figure 16-29 Problem 14 Solution.
Solution
(Between 0 and 4 s, the pulse has moved
Problem                                                     (3 cm/s)(4 s) = 12 cm to the right.)
15. What are (a) the amplitude, (b) the frequency in
hertz, (c) the wavelength, and (d) the speed of a
water wave whose displacement is y =
0.25 sin(0.52x − 2.3t), where x and y are in meters
and t in seconds?

Solution
Comparison of the given displacement with Equa-
tion 16-5 reveals that (a) A = 0.25 m, (b) f =
ω/2π = (2.3 s−1 )/2π = 0.366 Hz, (c) λ = 2π/k =                               Problem 18 Solution.
2π/(0.52 m−1 ) = 12.1 m, and v = ω/k =
(2.3 s−1 )/0.52 m−1 = 4.42 m/s. (Note: The presence
of a phase constant of φ = −π/2 in the expression for       Problem
y(x, t) = A sin(kx − ωt) = A cos(kx − ωt + φ) does not      19. Figure 16-30a shows a wave plotted as a function
aﬀect any of the quantities queried in this problem.)           of position at time t = 0, while Fig. 16-30b shows
the same wave plotted as a function of time at
Problem                                                         position x = 0. Find (a) the wavelength, (b) the
16. A sound wave with frequency 256 Hz (the musical             period, (c) the wave speed, and (d) the direction
note middle C) is propagating in air at 343 m/s.            of propagation.
How far apart are two points on the wave that
diﬀer in phase by π/2 or 90◦ ?

Solution
Two points in space separated by one wavelength
diﬀer in phase by 2π (or one cycle). Therefore, a phase
diﬀerence of π/2 (one quarter of a cycle) corresponds
to a separation of λ/4 = (v/f )/4 = (343 m/s)÷
(4×256 Hz) = 33.5 cm.

Problem
17. At time t = 0, the displacement in a transverse
wave pulse is described by y = 2(x4 + 1)−1 , with
both x and y in cm. Write an expression for the
pulse as a function of position x and time t if it is
propagating in the positive x direction at 3 cm/s.

Solution                                                                   figure 16-30 Problem 19.
From the shape of the pulse at t = 0, y(x, 0) = f (x), a
pulse with the same waveform, traveling in the
4   CHAPTER 16

Solution                                                  Problem
(a) The wavelength is the distance between successive     23. A transverse wave with 3.0-cm amplitude and
maxima at the same time (say t = 0), so Fig. 16-30a           75-cm wavelength is propagating on a stretched
gives λ = 3 m. (b) The period is the time interval            spring whose mass per unit length is 170 g/m. If
between successive maxima (or some other speciﬁc              the wave speed is 6.7 m/s, ﬁnd (a) the spring
phases diﬀering by 2π) at the same point (say x = 0),         tension and (b) the maximum speed of any point
so Fig. 16-30b gives T = 1.5 s. (c) v = ω/k = λ/T =           on the spring.
2 m/s. (d) Fig. 16-30b shows that as t increases from
0, the displacement at x = 0 ﬁrst becomes negative.       Solution
The waveform in Fig. 16-30a must therefore move to        (a) Equation 16-7 gives F = µv 2 = (0.17 kg/m)×
the right, in the positive x direction. [For the          (6.7 m/s)2 = 7.63 N. (b) The unnumbered equation for
sinusoidal wave pictured, y(x, 0) = A sin kx and          the vertical velocity of the medium in Section 16.5
y(0, t) = −A sin ωt = A sin(−ωt), so y(x, t) =            gives umax = (dy/dt)max = ωA = (2πv/λ)A =
A sin(kx − ωt).]                                          2π(6.7 m/s)(3 cm)/(75 cm) = 1.68 m/s.

Problem                                                   Problem
20. Write a mathematical description of the wave in       24. A rope is stretched between supports 12 m apart;
the preceding problem.                                    its tension is 35 N. If one end of the rope is
tweaked, the resulting disturbance reaches the
Solution                                                      other end 0.45 s later. What is the total mass of
From Fig. 16-30, the amplitude is probably 2 m, so            the rope?
y(x, t) = (2 m)sin[(2π/3 m)x − (2π/1.5 s)t] =
(2 m)sin[(2.09 m−1 )x − (4.19 s−1 )t]. (See solution to   Solution
preceding problem.)                                       The wave speed is v = /t = F/(m/ ), where
= 12 m is the length and t = 0.45 s is the travel
Section 16-4: Waves on a String                           time. Therefore, the mass is m = F t2 / = (35 N)×
(0.45 s)2 /(12 m) = 591 g.
Problem
21. The main cables supporting New York’s George          Problem
Washington Bridge have a mass per unit length of
25. A 3.1-kg mass hangs from a 2.7-m-long string
4100 kg/m and are under tension of 250 MN. At
whose total mass is 0.62 g. What is the speed of
what speed would a transverse wave propagate on
transverse waves on the string? Hint: You can
these cables?
ignore the string mass in calculating the tension
but not in calculating the wave speed. Why?
Solution
v = F/µ = (2.5×108 N)/(4100 kg/m) = 247 m/s               Solution
(from Equation 16-7).
The tension in the string is approximately equal to the
weight of the 3.1 kg mass (since the weight of the
Problem                                                   string is only 2% of this). Thus, v = F/µ =
22. A transverse wave 1.2 cm in amplitude is                                2
propagating on a string; the wave frequency is          (3.1 kg)(9.8 m/s )(2.7 m)/(0.62 g) = 364 m/s.
44 Hz. The string is under 21 N tension and has       (0.62 g is small compared to 3.1 kg, but not small
mass per unit length of 15 g/m. Determine (a) the     compared to zero!)
wave speed and (b) the maximum speed of a point
on the string.                                        Problem
26. Transverse waves propagate at 18 m/s on a string
Solution                                                      whose tension is 14 N. What will be the wave
(a) The wave speed is v = F/µ =                               speed if the tension is increased to 40 N?
(21 N)/(0.015 kg/m) = 37.4 m/s. (b) From the
second equation in Section 16.5, umax = ωA =
Solution
2
2π(44 Hz)(1.2 cm) = 3.32 m/s.                             Since µ = F1 /v1 , v2 = F2 /µ = F2 /F1 v1 =
40/14(18 m/s) = 30.4 m/s is the speed at the
increased tension.
CHAPTER 16       5

Problem                                                        pulley and tied to a wall as shown in Fig. 16-31.
3
27. The density of copper is 8.29 g/cm . What is the           The speed of transverse waves on the horizontal
tension in a 1.0-mm-diameter copper wire that              section of wire is observed to be 20 m/s. If a
propagates transverse waves at 120 m/s?                    second mass m2 is added to the ﬁrst, the wave
speed increases to 45 m/s. Find the second mass.
Solution                                                       Assume the string does not stretch appreciably.
The linear mass density of copper wire with diameter
3
d is µ = m/ = ρ 1 πd2 = (8.29 g/cm ) 1 π(1 mm)2 =
Solution
4                    4
6.51×10−3 kg/m, so F = µv 2 = (6.51×10−3 kg/m)×            Since the wire is at rest, the tension in the horizontal
(120 m/s)2 = 93.8 N.                                       section equals the weight attached (provided the
pulley is frictionless). Then the wave speeds are
Problem                                                    v1 = m1 g/µ and v2 = (m1 + m2 )g/µ, from which
2   2
one ﬁnds that m2 = µ(v2 − v1 )/g = (5.6 g/m)×
28. A 100-m-long wire has a mass of 130 g. A sample                  2              2         2
of the wire is tested and found to break at a          [(45 m/s) − (20 m/s) ]/(9.8 m/s ) = 929g. (If the
tension of 150 N. What is the maximum                  wire doesn’t stretch, its diameter stays the same and µ
propagation speed for transverse waves on this         is constant.)
wire?
Problem
Solution                                                   31. A steel wire can tolerate a maximum tension per
2
unit cross-sectional area of 2.7 GN/m before it
undergoes permanent distortion. What is the
vmax = Fmax /µ =        150 N/(0.13 kg/100 m)              maximum possible speed for transverse waves in a
= 340 m/s.                                            steel wire if it is to remain undistorted? Steel has
3
a density of 7.9 g/cm .
Problem
29. A 25-m-long piece of 1.0-mm-diameter wire is put       Solution
under 85 N tension. If a transverse wave takes         The linear density is the (volume) density times the
0.21 s to travel the length of the wire, what is the   cross-sectional area (see solution to Problem 27),
2
density of the material comprising the wire?           whereas the maximum tension is 2.7 GN/m times
the same cross-sectional area. Therefore vmax =
Solution                                                                2             3
(2.7 GN/m )/(7.9 g/cm ) = 585 m/s. (Recall that
From the length of wire, travel time, and Equa-                                                        3
the preﬁx giga equals 109 and that 1 g/cm =
tion 16-7, v = 25 m/0.21 s = 85 N/µ, so µ = 6.00×                   3
103 kg/m .)
10−3 kg/m. But for a uniform wire of length and
1
diameter d, ρ = µ/ 4 πd2 = (6.00×10−3 kg/m)÷
1          2             3                                 Problem
4 π(1 mm) = 7.64 g/cm (see solution to Problem 27).
32. A uniform cable hangs vertically under its own
weight. Show that the speed of waves on the cable
√
Problem                                                        is given by v = yg, where y is the distance from
the bottom of the cable.
30. A mass m1 is attached to a wire of linear density
5.6 g/m, and the other end of the wire run over a
Solution
The tension in the cable can be found by integrating
Newton’s second law, applied to a small element at
rest. With quantities deﬁned in the sketch,
0 = T + dT − T − g dm, or dT = g dm. For a uniform
cable, dm = µ dy where the linear density µ is a
constant, so T = µgy (the constant of integration is
m1                  zero for y measured from the bottom of the cable). It
√
follows from Equation 16-7 that v = T /µ = gy.

figure 16-31 Problem 30.
6   CHAPTER 16

Problem
35. A 600-g Slinky is stretched to a length of 10 m.
You shake one end at the frequency of 1.8 Hz,
applying a time-average power of 1.1 W. The
resulting waves propagate along the Slinky at
2.3 m/s. What is the wave amplitude?

Solution
We assume that the elastic properties of a
stretched string are shared by the Slinky, so
Equation 16-8 applies. Then A =
Problem 32 Solution.                       2(1.1 W)/(0.06 kg/m)(2.3 m/s)/(2π×1.8 Hz) =
35.3 cm.
Section 16-5: Wave Power and Intensity
Problem
Problem
36. A simple harmonic wave of amplitude 5.0 cm,
33. A rope with 280 g of mass per meter is under             wavelength 70 cm, and frequency 14 Hz is
550 N tension. A wave with frequency 3.3 Hz and          propagating on a wire with linear density 40 g/m.
amplitude 6.1 cm is propagating on the rope.             Find the wave energy per unit length of the wire.
What is the average power carried by the wave?
Solution
Solution
Using the expression found in the solution to the
The average power transmitted by transverse traveling                            ¯
next problem, we ﬁnd dE/dx = 1 (0.04 kg/m)×
¯
waves in a string is given by Equation 16-8, P =                                         2
1    2 2      1                        2         2
(2π×14 Hz)2 (0.05 m)2 = 0.387 J/m.
2 µω A v = 2 (0.28 kg/m)(2π×3.3 Hz) (0.061 m) ×
550 N/(0.28 kg/m) = 9.93 W. (We used Equa-            Problem
tion 16-7 for v.)
37. Figure 16-32 shows a wave train consisting of two
cycles of a sine wave propagating along a string.
Problem                                                      Obtain an expression for the total energy in this
34. A motor drives a mechanism that produces simple          wave train, in terms of the string tension F, the
harmonic motion at one end of a stretched cable.         wave amplitude A, and the wavelength λ.
The frequency of the motion is 30 Hz, and the
motor can supply energy at an average rate of                                 λ
350 W. If the cable has linear density 450 g/m and
A
is under 1.7 kN tension, (a) what is the maximum
wave amplitude that can be driven down the
cable? (b) If the motor were replaced by a larger
one capable of supplying 700 W, how would the                         figure 16-32 Problem 37.
maximum amplitude change?

Solution                                                 Solution
¯
The average wave energy, dE, in a small element of
(a) With all the other quantities in Equation 16-8
string of length dx, is transmitted in time, dt, at the
ﬁxed, the amplitude is proportional to the square root
same speed as the waves, v = dx/dt. From Equa-
of the average power transmitted. If there are no                        ¯     ¯                    1
tion 16-8, dE = P dt = 1 µω 2 A2 v dt = 2 µω 2 A2 dx, so
losses, the power transmitted equals the power                                       2
¯        1
the average linear energy density is dE/dx = 2 µω 2 A2 .
supplied by the motor, so the maximum wave
¯              ¯ √
amplitude is 2P /µv/ω = 2P / F µ/ω =
The total average energy in a wave train of length
¯      ¯
= 2λ is E = (dE/dx) = 1 µω 2 A2 (2λ). In terms of
[2(350 W)/ (0.45 kg/m)(1.7 kN)]1/2 /(2π×30 Hz) =
√
2
the quantities speciﬁed in this problem (see Equa-
2.67 cm. (Note: µv = F µ from Equation 16-7.)                                 ¯
tions 16-1 and 7) E = 1 (F/v 2 )(2πv/λ)2 A2 (2λ) =
2
(b) If the motor’s power is doubled, the maximum
√                    4π 2 F A2 /λ. (Note: The relation derived can be written
amplitude increases by a factor of 2 to 3.77 cm.             ¯       ¯                                      ¯
as P = (dE/dx)v. For a one-dimensional wave, P is
the intensity, so the average intensity equals the
CHAPTER 16       7

average energy density times the speed of wave energy         Problem
propagation. This is a general wave property, e.g., see       42. A 9-W laser produces a beam 2 mm in diameter.
the ﬁrst unnumbered equation for S in Section 34-10.)             Compare its light intensity with that of sunlight at
2
Problem
38. A steel wire with linear density 5.0 g/m is under         Solution
450 N tension. What is the maximum power that             For a beam of constant cross-sectional area, the
can be carried by transverse waves on this wire if        intensity of the laser beam is 9 W/π(1 mm)2 =
the wave amplitude is not to exceed 10% of the                         2
2.86 MW/m , which is 2.86×103 times the given
wavelength?                                               intensity of sunlight at the ground.
Solution
Problem
Equation 16-8, written in terms of the tension and
¯
wavelength, is P = 1 (F/v 2 )(2πv/λ)2 A2 F/µ =                43. Light emerges from a 5.0-mW laser in a beam
2
2 3/2 −1/2
2π F µ        (A/λ)2 . If A/λ < 0.1, then                         1.0 mm in diameter. The beam shines on a wall,
¯
P < 2π 2 (450 N)3/2 (0.005 kg/m)−1/2 (0.1)2 = 26.6 kW.            producing a spot 3.6 cm in diameter. What are
the beam intensities (a) at the laser and (b) at the
Problem                                                           wall?
39. A loudspeaker emits energy at the rate of 50 W,
spread in all directions. What is the intensity of        Solution
sound 18 m from the speaker?                              If we assume that the power output of the laser is
spread uniformly over the cross-sectional area of its
Solution                                                                      ¯
beam, then I = P / 1 πd2 . (a) When the beam emerges,
4
The wave power is spread out over a sphere of area                        1                         2
I = 5 mW/ 4 π(1 mm)2 = 6.37 kW/m , while (b) after
4πr2 , so the intensity is 50 W/4π(18 m)2 =                   its diameter has expanded by 36 times, at the wall,
2                                                                          2
12.3 mW/m . (See Equation 16-9.)                              I = I(1/36)2 = 4.91 W/m .

Problem                                                       Problem
40. The light intensity 3.3 m from a light bulb is            44. A large boulder drops from a cliﬀ into the ocean,
2
0.73 W/m . What is the power output of the bulb,              producing circular waves. A small boat 18 m from
assuming it radiates equally in all directions?               the impact point measures the wave amplitude at
Solution                                                          130 cm. At what distance will the amplitude be
¯                                             50 cm?
From Equation 16-9, P = 4πr2 I = 4π(3.3 m)2 ×
2
(0.73 W/m ) = 99.9 W ≈ 100 W, typical for a                   Solution
lightbulb.
The intensity of a surface wave decreases inversely
Problem                                                       with the distance from the source (see diagram), and
41. Use data from Appendix E to determine the                 is proportional to the square of the amplitude. Then
intensity of sunlight at (a) Mercury and (b) Pluto.       A2 ∼ 1/r or, at two distances from the source,
(A/A )2 = r /r. Thus, r = (130/50)2 (18 m) = 122 m
for the wave in this problem.
Solution
Equation 16-9 gives the ratio of intensities at two
distances from an isotropic source of spherical waves
as I2 /I1 = (r1 /r2 )2 . If we use the average intensity of
sunlight given in Table 16-1 and mean orbital
distances to the sun from Appendix E, we obtain
2
(a) IMerc = IE (rE /rMerc )2 = (1368 W/m )(150÷
2                  2                              2
57.9) = 9.18 kW/m , and (b) IPluto = (1368 W/m )×
2
(150/5.91×103 )2 = 0.881 W/m . (Alternatively, the                             Problem 44 Solution.
¯
luminosity of the sun, P = 3.85×1026 W, from
Appendix E, could be used directly in Equation 16-9,
with only slightly diﬀerent numerical results.)
8   CHAPTER 16

Problem                                                           Solution
45. Use Table 16-1 to determine how close to a rock               (a) The absolute value of the maximum displacement
band you should stand for it to sound as loud as a            for each pulse is 2 cm, a value attained when the
jet plane at 200 m. Treat the band and the plane              denominators are minimal (x − t = 0 for the ﬁrst pulse
as point sources. Is this assumption reasonable?              and x − 5 + t = 0 for the second). (b) At t = 0, the
peak of the ﬁrst pulse is at x = 0 moving in the
Solution                                                          positive x direction. (x − t = 0 represents the peak, so
To have the same loudness, the soundwave intensities              if t increases so does x. This is why a wave traveling in
should be equal, i.e., Iband (r) = Ijet (200 m). Regarded         the positive x direction is represented by a function of
as isotropic point sources, use of Equation 16-9 gives            x − vt.) For the second pulse, the peak is at x = 5,
¯           ¯
Pband /r2 = Pjet /(200 m)2 . The average power of each            moving in the negative x direction, when t = 0
source can be found from Table 16-1 and a second                  (x − 5 + t = 0 implies x = 5 − t and dx/dt = −1 < 0).
¯
application of Equation 16-9, Pband =                             (c) y1 (x, t) + y2 (x, t) = 0 for all values of x implies
2         2
4π(4 m) (1 W/m ) and P     ¯jet = 4π(50 m)2 (10 W/m2 ).           (x − t)2 = (x − 5 + t)2 . This is true for all x, only if
¯        ¯
Then r2 = (Pband /Pjet )(200 m)2 = (200 m)2 (4 m)2 ×              (x − t) = +(x − 5 + t) or at t = 5 = 2.5 s. (The other
2
2          2          2
(1 W/m )/(50 m) (10 W/m ), or r = 5.06 m. The size                root, (x − t) = −(x − 5 + t), shows that x = 2.5 cm is
of a rock band is several meters, nearly equal to this            always a node, i.e., the net displacement there is zero
distance, so a point source is not a good                         at all times.)
approximation. Besides, the acoustical output of a
rock band usually emanates from an array of speakers,             Problem
which is not point-like. Moreover, the size of a jet              48. The triangular wave of Fig. 16-33 can be described
plane is also not very small compared to 50 m.                        by the following sum of simple harmonic terms:
8    sin x sin 3x sin 5x
Section 16-6: The Superposition Principle and                             y(x) =                −      +       − ··· .
π2    12     32     52
Wave Interference
Plot the sum of the ﬁrst three terms in this series
Problem                                                               for x ranging from 0 to 2π, and compare with the
46. Consider two functions f (x ± vt) and g(x ± vt)                   ﬁrst cycle shown in Fig. 16-33. (See also
that both satisfy the wave equation (Equa-                        ActivPhysics Activity 10.7.)
tion 16-12). Show that their sum also satisﬁes the
wave equation.                                                                y

1
Solution
The derivative of a sum equals the sum of the                                 0                                  x
2π      4π
derivatives, i.e., ∂ 2 (f + g)/∂x2 = ∂ 2 f /∂x2 + ∂ 2 g/∂x2 ,
–1
etc., so if f and g satisfy Equation 16-12, so does f + g.
(The wave equation is a linear diﬀerential equation,
i.e., it does not involve products or powers of the                                   figure 16-33 Problem 48.
function and its derivatives, so any linear combination
of solutions af ± bg, is itself a solution.)
Solution
Problem                                                           The amplitudes of the ﬁrst three harmonic
47. Two wave pulses are described by                              components are 8/π 2 = 0.81057, 8/9π 2 = 0.09006, and
8/25π 2 = 0.03243, and their wavelengths are
2                           −2             λ1 = 3λ3 = 5λ5 . The phases alternate by 180◦ . A
y1 (x, t) =                , y2 (x, t) =                  ,
(x − t)2 + 1               (x − 5 + t)2 + 1     sketch of the components, their superposition, and the
where x and y are in cm and t in seconds.                     ﬁrst cycle of the triangular wave is shown.
(a) What is the amplitude of each pulse?
(b) At t = 0, where is the peak of each pulse, and            Problem
in what direction is it moving? (c) At what time              49. You’re in an airplane whose two engines are
will the two pulses exactly cancel?                               running at 560 rpm and 570 rpm. How often do
you hear the sound intensity increase as a result of
wave interference?
CHAPTER 16      9

multiple of π = 180◦ in general, insures complete
destructive interference.) From Example 16-5,
2 ∆r = 16.0 m, so λ2 = 2 ∆r/3 = 5.34 m.

Problem
52. The two loudspeakers shown in Fig. 16-34 emit
identical 500-Hz sound waves. Point P is on the
ﬁrst nodal line of the interference pattern. Use the
numbers shown to calculate the speed of the sound
waves.

Problem 48 Solution.

Solution
As mentioned in the text, pilots of twin-engine
airplanes use the beat frequency to synchronize the
rpm’s of their engines. The beat frequency is simply
the diﬀerence of the two interfering frequencies,
fbeat = (570 − 560)/60 s = 1 s−1 , so you would hear
6
one beat every six seconds.                                            figure 16-34 Problem 52.

Problem                                                  Solution
50. Two waves have the same angular frequency ω,         The path diﬀerence between the two loudspeakers
wave number k, and amplitude A, but they             and a point on the ﬁrst nodal line is one half-
diﬀer in phase: y1 = A cos(kx − ωt) and y2 =         wavelength, ∆r = 1 λ = 1 v/f, or v = 2f ∆r, where
2    2
A cos(kx − ωt + φ). Show that their superposition    f is the frequency. From Fig. 16-34 and use of the
is also a simple harmonic wave, and determine its    Pythagorean theorem,
amplitude As as a function of the phase              v = 2(500 Hz)( (3.5 m)2 + (0.75 m + 0.83 m)2 −
diﬀerence φ.
(3.5 m)2 + (0.83 m − 0.75 m)2 ) = 339 m/s.
Solution                                                 Section 16-7: The Wave Equation
Using the identity cos α + cos β = 2 cos 1 (α − β)×
1
2
cos 2 (α + β), we ﬁnd y1 + y2 = 2A cos 1 φ cos(kx −      Problem
2
ωt + 1 φ) ≡ As cos(kx − ωt + φs ). This shows that the
2
53. The following equation arises in analyzing the
1
amplitude is As = 2A cos 1 φ, (and also φs = 2 φ).
2
behavior of shallow water:
∂2y    1 ∂2y
Problem                                                                          −        = 0,
dx2   gh dt2
51. What is the wavelength of the ocean waves in             where h is the equilibrium depth and y the
Example 16-5 if the calm water you encounter at          displacement from equilibrium. Give an expression
33 m is the second calm region on your voyage            for the speed of waves in shallow water. (Here
from the center line?                                    shallow means the water depth is much less than
the wavelength.)
Solution
The second node occurs when the path diﬀerence is        Solution
three half-wavelengths, or AP − BP ≡ ∆r = 3 λ2 . (A
2           The equation given is in the standard form for the one-
phase diﬀerence of k2 ∆r = (2π/λ2 )∆r = 3π, or an odd    dimensional linear wave equation (Equation 16-12), so
10   CHAPTER 16

the wave speed is the reciprocal of the square root of
√             Problem
the quantity multiplying ∂ 2 y/∂t2 . Thus v = gh.          57. A spring of mass m and spring constant k has an
unstretched length 0 . Find an expression for the
Problem                                                        speed of transverse waves on this spring when it
54. Use the chain rule for diﬀerentiation to show              has been stretched to a length .
explicitly that any function of the form f (x ± vt)
satisﬁes the wave equation (Equation 16-12).           Solution
The spring may be regarded as a stretched string
Solution
with tension, F = k( − 0 ), and linear mass density
Let primes denote diﬀerentiation with respect to the       µ = m/ . Equation 16-7 gives the speed of transverse
whole argument φ = (x ± vt). Then the chain rule           waves as v = k ( − 0 )/m.
gives ∂f /∂x = (df /dφ)(∂φ/∂x) = f , and ∂ 2 f ÷
∂x2 = f . Similarly, ∂f /∂t = f (±v), and ∂ 2 f /∂t2 =     Problem
v 2 f . Therefore, any function of φ satisﬁes the wave
58. When a 340-g spring is stretched to a total length
equation ∂ 2 f /∂x2 − (1/v 2 )∂ 2 f /∂t2 = f − (1/v 2 )×
of 40 cm, it supports transverse waves propagating
v 2 f = 0.
at 4.5 m/s. When it’s stretched to 60 cm, the
waves propagate at 12 m/s. Find (a) the
Paired Problems
unstretched length of the spring and (b) its spring
Problem                                                        constant.
55. A wave on a taut wire is described by the equation
y = 1.5 sin(0.10x − 560t), where x and y are in cm     Solution
and t is in seconds. If the wire tension is 28 N,      From the solution to the previous problem, mv 2 =
what are (a) the amplitude, (b) the wavelength,        k ( − 0 ). (a) With v1 and v2 given for 1 and 2 , k
(c) the period, (d) the wave speed, and (e) the        may be eliminated by division, before solving for
power carried by the wave?                                           2
0 : (v2 /v1 ) = 2 ( 2 − 0 )/ 1 ( 1 − 0 ) or
2          2
Solution                                                                         − 2
1 (v2 /v1 ) 2
0   =              2
The wave has the form of Equation 16-5, with a phase                             − 2
1 (v2 /v1 )
constant of − π = −90◦ , y(x, t) = A sin(kx − ωt) =
2
2
(40) (12/4.5)2 − (60)2
=                        cm = 34.7 cm.
A cos(kx− ωt − π ). Comparison reveals that
2                                                    (40)(12/4.5)2 − (60)
k = 0.1 cm−1 , ω = 560 s−1 , and (a) A = 1.5 cm
(b) From either pair of values of wave speed and
(b) λ = 2π/k = 2π/(0.1 cm−1 ) = 62.8 cm (Equa-
length,
tion 16-4). (c) T = 2π/ω = 2π/(560 s−1 ) = 11.2 ms
(Equation 16-3). (d) v = ω/k = 56 m/s (Equa-                                          (0.34 kg)(4.5 m/s)2
¯                                            k = mv 2 / ( −       0)   =
1
tion 16-6). (e) P = 2 µω 2 A2 v = 1 (ωA)2 (F/v) =
2                                                (0.4 m)(0.4 m − 0.347 m)
1        −1
2 (560 s    ×0.015 m)2 (28 N)/(56 m/s) = 17.6 W                       (0.34 kg)(12 m/s)2
(Equation 16-8, and Equation 16-7 to eliminate µ).              =                            = 322 N/m.
(0.6 m)(0.6 m − 0.347 m)
Problem                                                    Problem
56. A wave given by y = 23 cos(0.025x − 350t), with x      59. At a point 15 m from a source of spherical
and y in mm and t in seconds, is propagating on a          sound waves, you measure a sound intensity of
cable with mass per unit length 410 g/m. Find                         2
750 mW/m . How far do you need to walk,
(a) the amplitude, (b) the wavelength, (c) the             directly away from the source, until the intensity
frequency in Hz, (d) the wave speed, and (e) the                         2
is 270 mW/m ?
power carried by the wave.
Solution
Solution
The intensity of spherical waves from a point source
Inspection of the given function and use of Equa-
is given by Equation 16-9. At a distance r1 , I1 =
tions 16-3 through 8 gives (a) A = 23 mm; (b) λ =           ¯     2
P /4πr1 , while after increasing the radial distance by
2π/k = 2π/(0.025 mm−1 ) = 25.1 cm; (c) f = ω/2π =                   ¯
d, I2 = P /4π(r1 + d)2 . Dividing and solving for d, one
350 s−1 /2π = 55.7 Hz; (d) v = ω/k = (350 s−1 ) ÷
¯   1                        ﬁnds d = r1 ( I1 /I2 − 1) = (15 m)( (750/270) − 1) =
(0.025 mm−1 ) = 14 m/s; (e) P = 2 µω 2 A2 v =
1                             2                            10.0 m.
2 (0.41 kg/m)(350×23 mm/s) (14 m/s) = 186 W.
CHAPTER 16      11

Problem                                                     500 Hz, so the percent diﬀerence is (100)∆f /f =
60. Figure 16-35 shows two observers 20 m apart, on a       (100)(500 Hz/50 MHz) = 10−3 %.
line that connects them and a spherical light
source. If the observer nearest the source measures     Supplementary Problems
a light intensity 50% greater than the other            Problem
observer, how far is the nearest observer from the
63. For a transverse wave on a stretched string, the
source?                                                     requirement that the string be nearly horizontal
is met if the amplitude is much less than the
wavelength. (a) Show this by drawing an
appropriate sketch. (b) Show that, under this
approximation that A      λ, the maximum speed u
20 m                     x=?                         of the string must be considerably less than the
wave speed v. (c) If the amplitude is not to exceed
figure 16-35 Problem 60.
1% of the wavelength, how large can the string
speed u be in relation to the wave speed v?
Solution
Solution
If we assume the source emits spherical waves, the
ratio of the intensities for the two observers is           (a) The relative “ﬂatness” or “peakedness” of a
I1 /I2 = (x2 /x1 )2 (Equation 16-9), where the closer       sinusoidal waveform is determined by its maximum
observer is at x1 . Then I1 = 1.5I2 and x2 = x1 + 20 m,     slope, |dy/dx|max = |∂/∂x[A cos(kx − ωt)]|max =
so this equation becomes (x1 + 20 m)2 = 1.5x2 . The         kA = 2π(A/λ). If A      λ (or kA     1), the slope is
1
positive solution of this quadratic (when both              nearly horizontal. (b) In terms of the speeds, kA =
observers are on the same side of the source) is            ωA/v = umax /v, so the string is nearly ﬂat if
x1 = 2(20 m)(1 + 3/2) = 89.0 m. (The negative root,         umax     v. (c) If A/λ < 1%, then umax /v = 2π(A/λ) <
2π(1%) = 6.3%.
2(20 m)(1 − 3/2) = −8.99 m, corresponds to
observers on opposite sides of the source, i.e., with the
lamp in Fig. 16-35 between the two observers.)

Problem
61. Two motors in a factory produce sound waves
with the same frequency as their rotation rates. If                      Problem 63 Solution.
one motor is running at 3600 rpm and the other at
3602 rpm, how often will workers hear a peak in
the sound intensity?                                    Problem
64. A 64-g spring has unstretched length 25 cm. With
Solution                                                        a 940-g mass attached, the spring undergoes
The beat frequency equals the diﬀerence in the                  simple harmonic motion with angular frequency
motors’ rpm’s, so the period of the beats is Tbeat =            6.1 s−1 . What will be the speed of transverse
1/fbeat = 1/(3602− 3600) min−1 = 30 s. (See also                waves on this spring when it’s stretched to a total
Problem 49.)                                                    length of 40 cm?

Problem                                                     Solution
62. Two radio waves with frequencies of                     When used as a nearly ideal mass-spring system
approximately 50 MHz interfere. The composite           (since the spring’s mass is much less than the attached
wave is detected and fed to a loudspeaker, which        mass), ω 2 = k/m, which allows the spring constant to
emits audible sound at 500 Hz. What is the              be determined, k = (0.94 kg)(6.1 s−1 )2 . When used in
percentage diﬀerence between the frequencies of         a diﬀerent way to support transverse waves (see
the two radio waves?                                    Problem 57), v = k ( − 0 )/ms = [(0.94 kg)×
(6.1 s−1 )2 (0.4 m)(0.4 m − 0.25 m)/(0.064 kg)]1/2 =
Solution                                                    5.73 m/s.
The diﬀerence in the frequencies (really its absolute
value) equals the beat frequency, ∆f = fbeat =
12   CHAPTER 16

Problem                                                      Problem
65. An ideal spring is compressed until its total length     68. Show that the time it takes a wave to propagate
is 1 , and the speed of transverse waves on the              up the cable in Problem 32 is t = 2 /g, where
spring is measured. When it’s compressed further             is the cable length.
to a total length 2 , waves propagate at the same
speed. Show that the uncompressed spring length          Solution
is just 1 + 2 .                                          The wave speed in the cable of Problem 32 was
√
v = dy/dt = gy, where y is the distance from the
Solution                                                     bottom of the cable. The time for a transverse wave
The tension in a compressed spring has magnitude             signal to propagate from the bottom to the top (y = 0
k( 0 − ) while its linear mass-density is µ = m/ .           to ) is
Therefore, the speed of transverse waves is v =
F/µ = k ( 0 − )/m (as in Problem 57 for a                                   dy         dy  1 √
t=    dt =          =       √ =√ 2 y            =2       .
stretched spring). If v1 = v2 for two diﬀerent                             0    v     0    gy  g                   g
0
compressed lengths, then 1 ( 0 − 1 ) = 2 ( 0 − 2 ) or
( 1 − 2 ) 0 = 2 − 2 = ( 1 − 2 )( 1 + 2 ). Since 1 = 2 ,
1    2                                        Problem
division by 1 − 2 gives 0 = 1 + 2 .                          69. In Example 16-5, how much farther would you
have to row to reach a region of maximum wave
Problem                                                          amplitude?
66. An ideal spring is stretched to a total length 1 .
When that length is doubled, the speed of
transverse waves on the spring triples. Find an
expression for the unstretched length of the spring.

Solution
Utilizing the result of Problem 57, we have v1 =
k 1 ( 1 − 0 )/m and v2 = 3v1 = k2 1 (2 1 − 0 )/m.
5
Therefore 9 1 ( 1 − 0 ) = 2 1 (2 1 − 0 ) or 0 = 7 1 .

Problem
67. A 1-megaton nuclear explosion produces a shock
wave whose amplitude, measured as excess air
pressure above normal atmospheric pressure, is
2
1.4×105 Pa (1 Pa = 1 N/m ) at a distance of
1.3 km from the explosion. An excess pressure of
3.5×104 Pa will destroy a typical woodframe
house. At what distance from the explosion will
such houses be destroyed? Assume the wavefront
is spherical.

Solution
The intensity of a spherical wavefront varies inversely
with the square of the distance from the central source                 figure 16-36 Problem 69 Solution.
(see Fig. 16-18b). In general, the intensity is
proportional to the amplitude squared, so A ∼ 1/r for
a spherical wave. (This can be proved rigorously by          Solution
solution of the spherical wave equation, a                   In general, the interference condition for waves in the
generalization of Equation 16-12.) Therefore A1 /A2 =        geometry of Example 16-5 is AP − BP = nλ/2, where
r2 /r1 , or the overpressure reaches the stated limit at a   n is an odd integer for destructive interference (a node)
distance r2 = (1.4×105 Pa/3.5×104 Pa)(1.3 km) =              and n is an even integer for constructive interference
5.2 km from the explosion.                                   (a maximum amplitude). (In Example 16-5, n = 1
gave the ﬁrst node and in Problem 51, n = 3 gave
CHAPTER 16         13

the second node.) If d = 20 m is the distance between     wavelengths of the ﬁrst and second nodes, for n = 1
the openings, = 75 m is the perpendicular distance        and 3, calculated in Example 16-5 and Problem 51,
from the breakwater, and x is the distance parallel       respectively.)
to the breakwater measured from the midpoint
of the openings, the interference condition is            Problem
2              1 2   2 + (x −1 2
+ (x +       2 d)
−             2 d) = nλ/2 (see          70. Suppose the wavelength of the ocean waves in
Fig. 16-36). In this problem, we wish to ﬁnd x for the        Example 16-5 were 8.4 m. How far would you
ﬁrst maximum, n = 2, and the wavelength calculated            have to row from the center line, staying 75 m
in Example 16-5, λ = 16.01 m. Solving for x, we ﬁnd:          from the breakwater, in order to ﬁnd (a) the ﬁrst
and (b) the second region of relative calm?
[ + ( 1 d)2 − ( 1 nλ)2 ]
2
x2 =          2         4
(2d/nλ)2 − 1                                 Solution
[(75)2 + (10)2 − (8.005)2 ] m2                   From the general solution for x in the previous
=                                = (100.5 m)2 .
(40/32.02)2 − 1                           problem, (a) x = 16.2 m to the ﬁrst node (n = 1),
This is 100.5 m − 33 m = 67.5 m farther than the ﬁrst     and (b) x = 61.2 m to the second node (n = 3). (Use
node in Example 16-5. (Note: We rounded oﬀ to three        = 75 m, d = 20 m, and λ = 8.4 m for the other
ﬁgures; if you round oﬀ to two ﬁgures, the answer is      values.)
67 m. Also, if x = 33 m is substituted into the general
interference condition, one can recapture the

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