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.
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
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.
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
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.
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
(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 =
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.
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].
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 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.
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.
(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?
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◦ ?
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.
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
(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.
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.
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
ignore the string mass in calculating the tension
but not in calculating the wave speed. Why?
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 =
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
CHAPTER 16 5
Problem pulley and tied to a wall as shown in Fig. 16-31.
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
d is µ = m/ = ρ 1 πd2 = (8.29 g/cm ) 1 π(1 mm)2 =
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
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.)
Solution 31. A steel wire can tolerate a maximum tension per
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
a density of 7.9 g/cm .
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),
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
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
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
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?
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) =
Section 16-5: Wave Power and Intensity
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?
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
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?
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
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) =
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) =
(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
noon, about 1 kW/m .
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.
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 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
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
beam, then I = P / 1 πd2 . (a) When the beam emerges,
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,
12.3 mW/m . (See Equation 16-9.) I = I(1/36)2 = 4.91 W/m .
40. The light intensity 3.3 m from a light bulb is 44. A large boulder drops from a cliﬀ into the ocean,
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 ×
(0.73 W/m ) = 99.9 W ≈ 100 W, typical for a Solution
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.
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
(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 )×
(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
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
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
(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
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
The derivative of a sum equals the sum of the 0 x
derivatives, i.e., ∂ 2 (f + g)/∂x2 = ∂ 2 f /∂x2 + ∂ 2 g/∂x2 ,
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.)
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
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.
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
Problem 48 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
one beat every six seconds. figure 16-34 Problem 52.
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
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 −
(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)×
cos 2 (α + β), we ﬁnd y1 + y2 = 2A cos 1 φ cos(kx − Problem
ωt + 1 φ) ≡ As cos(kx − ωt + φs ). This shows that the
53. The following equation arises in analyzing the
amplitude is As = 2A cos 1 φ, (and also φs = 2 φ).
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 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
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
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
unstretched length of the spring and (b) its spring
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
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) =
(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-
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) =
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)
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)
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.
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?
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
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.)
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?
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
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
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
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.
Utilizing the result of Problem 57, we have v1 =
k 1 ( 1 − 0 )/m and v2 = 3v1 = k2 1 (2 1 − 0 )/m.
Therefore 9 1 ( 1 − 0 ) = 2 1 (2 1 − 0 ) or 0 = 7 1 .
67. A 1-megaton nuclear explosion produces a shock
wave whose amplitude, measured as excess air
pressure above normal atmospheric pressure, is
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
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 ]
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