CHAPTER 16 WAVE MOTION

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CHAPTER 16 WAVE MOTION Powered By Docstoc
					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
                                                                  noon, about 1 kW/m .
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