CHAPTER 16 WAVE MOTION

ActivPhysics can help with these problems:Activities 10.1, 10.2, 10.7, 10.10

Section 16-2: Wave Properties

Problem1. Ocean waves with 18-m wavelength travel at

5.3 m/s. What is the time interval between wavecrests passing under a boat moored at a fixedlocation?

SolutionWave crests (adjacent wavefronts) take a time of oneperiod to pass a fixed point, traveling at the wavespeed (or phase velocity) for a distance of onewavelength. Thus T = λ/v = 18 m/(5.3 m/s) = 3.40 s.

Problem2. Ripples in a shallow puddle are propagating at

34 cm/s. If the wave frequency is 5.2 Hz, what are(a) the period and (b) the wavelength?

SolutionEquation 16-1 gives T = 1/f = 1/5.2 Hz = 0.192 s andλ = v/f = vT = (34 cm/s)/(5.2 Hz) = 6.54 cm.

Problem3. An 88.7-MHz FM radio wave propagates at the

speed of light. What is its wavelength?

SolutionFrom Equation 16-1, λ = v/f = (3×108 m/s)÷(88.7×106 Hz) = 3.38 m.

Problem4. One end of a rope is tied to a wall. You shake the

other end with a frequency of 2.2 Hz, producingwaves whose wavelength is 1.6 m. What is theirpropagation speed?

Solutionv = fλ = (2.2 Hz)(1.6 m) = 3.52 m/s (see Equa-tion 16-1).

Problem5. A 145-MHz radio signal propagates along a cable.

Measurement shows that the wave crests are spaced

1.25 m apart. What is the speed of the waves on thecable? Compare with the speed of light in vacuum.

SolutionThe distance between adjacent wave crests is onewavelength, so the wave speed in the cable (Equa-tion 16-1) is v = fλ = (145×106 Hz)(1.25 m) = 1.81×108 m/s = 0.604c, where c = 3×108 m/s is the wavespeed in vacuum.

Problem6. Calculate the wavelengths of (a) a 1.0-MHz AM

radio wave, (b) a channel 9 TV signal (190 MHz),(c) a police radar (10 GHz), (d) infrared radiationfrom a hot stove (4.0×1013 Hz), (e) green light(6.0×1014 Hz), and (f) 1.0×1018 Hz X rays. All areelectromagnetic waves that propagate at3.0×108 m/s.

SolutionWith Equation 16-1 in an equivalent form, λ = vT =v/f, we find: (a) λ = (3×108 m/s)/(106 Hz) =300 m, (b) λ = 1.58 m, (c) λ = 3 cm, (d) λ = 7.5 µm,(e) λ = 500 nm, (f) λ = 3A. (See Appendix C onunits.)

Problem7. Detecting objects by reflecting waves off them is

effective only for objects larger than about onewavelength. (a) What is the smallest object thatcan be seen with visible light (maximum frequency7.5×1014 Hz)? (b) What is the smallest object thatcan be detected with a medical ultrasound unitoperating at 5 MHz? The speed of ultrasoundwaves in body tissue is about 1500 m/s.

Solution(a) The wavelength of light corresponding to thismaximum frequency is λ = c/f = (3×108 m/s)÷(7.5×1014 Hz) = 400 nm, violet in hue (see Equa-tion 16-1). (b) The ultrasonic waves described havewavelength λ = v/f = (1500 m/s)/(5 MHz) = 0.3 mm.

Problem8. A seismograph located 1200 km from an earthquake

detects waves from the quake 5.0 min after the

2 CHAPTER 16

quake occurs. The seismograph oscillates in stepwith the waves, at a frequency of 3.1 Hz. Find thewavelength of the waves.

SolutionThe wave speed can be calculated from the distanceand the travel time, which, together with thefrequency and Equation 16-1, gives a wavelength ofλ = v/f = (d/t)/f = 1200 km/(5×60×3.1) = 1.29 km.

Problem9. In Fig. 16-28 two boats are anchored offshore and

are bobbing up and down on the waves at the rateof six complete cycles each minute. When one boatis up the other is down. If the waves propagate at2.2 m/s, what is the minimum distance between theboats?

SolutionThe boats are 180◦ = π rad out of phase, so theminimum distance separating them is half awavelength. (In general, they could be an odd numberof half-wavelengths apart.) The frequency is 6/60 s =0.1 Hz, so 1

2λ = 12v/f = 1

2 (2.2 m/s)/(0.1/s) = 11 m.(Fig. 16-28 shows the answer, not the question.)

figure 16-28 Problem 9 Solution.

Section 16-3: Mathematical Description ofWave Motion

Problem10. Ultrasound used in a particular medical imager

has frequency 4.8 MHz and wavelength 0.31 mm.Find (a) the angular frequency, (b) the wavenumber, and (c) the wave speed.

Solution(a) Equation 16-3 gives ω = 2π(4.8 MHz) = 3.02×107 s−1. (b) Equation 16-4 gives k = 2π/0.31 mm =2.03×104 m−1. (c) Together, these give v = fλ =ω/k = 1.49 km/s (see Equation 16-1 or 6).

Problem11. An ocean wave has period 4.1 s and wavelength

10.8 m. Find (a) its wave number and (b) itsangular frequency.

SolutionFrom Equations 16-3 and 4, (a) k = 2π/10.8 m =0.582 m−1, and (b) ω = 2π/(4.1 s) = 1.53 s−1.

Problem12. Find (a) the amplitude, (b) the wavelength,

(c) the period, and (d) the speed of a wave whosedisplacement is given by y = 1.3 cos(0.69x + 31t),where x and y are in cm and t is in seconds. (e) Inwhich direction is the wave propagating?

Solution(a) A = ymax = 1.3 cm, (b) λ = 2π/k = 2π÷0.69 cm−1 = 9.11 cm, (c) T = 2π/ω = 2π/31 s−1 =0.203 s−1, and (d) v = ω/k = 44.9 cm/s. (e) A phaseof the form kx + ωt describes a wave propagating inthe negative x direction.

Problem13. A simple harmonic wave of wavelength 16 cm and

amplitude 2.5 cm is propagating along a string inthe negative x direction at 35 cm/s. Find (a) theangular frequency and (b) the wave number.(c) Write a mathematical expression describingthe displacement y of this wave (in centimeters) asa function of position and time. Assume thedisplacement at x = 0 is a maximum when t = 0.

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 wavemoving in the negative x direction, becomesy(x, t) = (2.5 cm) cos[(0.393 cm−1)x + (13.7 s−1)t].

Problem14. Figure 16-29 shows a simple harmonic wave at

time t = 0 and later at t = 2.6 s. Write amathematical description of this wave.

SolutionInspection of Fig. 16-29 shows that the wavelength is8 cm, the amplitude is 1.5 cm, and the velocity isv = ∆x/∆t = 2 cm/2.6 s = 0.769 cm/s. The phaseconstant is zero (since y = A at t = 0 and x = 0) andthe 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 −(0.604 s−1)t].

figure 16-29 Problem 14 Solution.

Problem15. What are (a) the amplitude, (b) the frequency in

hertz, (c) the wavelength, and (d) the speed of awater wave whose displacement is y =0.25 sin(0.52x− 2.3t), where x and y are in metersand t in seconds?

SolutionComparison 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 =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 presenceof a phase constant of φ = −π/2 in the expression fory(x, t) = A sin(kx− ωt) = A cos(kx− ωt + φ) does notaffect any of the quantities queried in this problem.)

Problem16. A sound wave with frequency 256 Hz (the musical

note middle C) is propagating in air at 343 m/s.How far apart are two points on the wave thatdiffer in phase by π/2 or 90◦?

SolutionTwo points in space separated by one wavelengthdiffer in phase by 2π (or one cycle). Therefore, a phasedifference of π/2 (one quarter of a cycle) correspondsto a separation of λ/4 = (v/f)/4 = (343 m/s)÷(4×256 Hz) = 33.5 cm.

Problem17. At time t = 0, the displacement in a transverse

wave pulse is described by y = 2(x4 + 1)−1, withboth x and y in cm. Write an expression for thepulse as a function of position x and time t if it ispropagating in the positive x direction at 3 cm/s.

SolutionFrom the shape of the pulse at t = 0, y(x, 0) = f(x), apulse with the same waveform, traveling in the

positive x direction with speed v, can be obtained byreplacing x by x− vt, y(x, t) = f(x− vt). For the givenf(x) and v, y(x, t) = 2[(x− 3t)4 + 1]−1, with x and y incm and t in s.

Problem18. Plot the answer to the previous problem as a

function of position x for the two cases t = 0 andt = 4 s, and verify that your plots are consistentwith the pulse speed of 3 cm/s.

Solution(Between 0 and 4 s, the pulse has moved(3 cm/s)(4 s) = 12 cm to the right.)

Problem 18 Solution.

Problem19. Figure 16-30a shows a wave plotted as a function

of position at time t = 0, while Fig. 16-30b showsthe same wave plotted as a function of time atposition x = 0. Find (a) the wavelength, (b) theperiod, (c) the wave speed, and (d) the directionof propagation.

figure 16-30 Problem 19.

4 CHAPTER 16

Solution(a) The wavelength is the distance between successivemaxima at the same time (say t = 0), so Fig. 16-30agives λ = 3 m. (b) The period is the time intervalbetween successive maxima (or some other specificphases differing by 2π) at the same point (say x = 0),so Fig. 16-30b gives T = 1.5 s. (c) v = ω/k = λ/T =2 m/s. (d) Fig. 16-30b shows that as t increases from0, the displacement at x = 0 first becomes negative.The waveform in Fig. 16-30a must therefore move tothe right, in the positive x direction. [For thesinusoidal wave pictured, y(x, 0) = A sin kx andy(0, t) = −A sin ωt = A sin(−ωt), so y(x, t) =A sin(kx− ωt).]

Problem20. Write a mathematical description of the wave in

the preceding problem.

SolutionFrom Fig. 16-30, the amplitude is probably 2 m, soy(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 topreceding problem.)

Section 16-4: Waves on a String

Problem21. The main cables supporting New York’s George

Washington Bridge have a mass per unit length of4100 kg/m and are under tension of 250 MN. Atwhat speed would a transverse wave propagate onthese cables?

Solutionv =

√F/µ =

√(2.5×108 N)/(4100 kg/m) = 247 m/s

(from Equation 16-7).

Problem22. A transverse wave 1.2 cm in amplitude is

propagating on a string; the wave frequency is44 Hz. The string is under 21 N tension and hasmass per unit length of 15 g/m. Determine (a) thewave speed and (b) the maximum speed of a pointon the string.

Solution(a) The wave speed is v =

√F/µ =√

(21 N)/(0.015 kg/m) = 37.4 m/s. (b) From thesecond equation in Section 16.5, umax = ωA =2π(44 Hz)(1.2 cm) = 3.32 m/s.

Problem23. A transverse wave with 3.0-cm amplitude and

75-cm wavelength is propagating on a stretchedspring whose mass per unit length is 170 g/m. Ifthe wave speed is 6.7 m/s, find (a) the springtension and (b) the maximum speed of any pointon the spring.

Solution(a) Equation 16-7 gives F = µv2 = (0.17 kg/m)×(6.7 m/s)2 = 7.63 N. (b) The unnumbered equation forthe vertical velocity of the medium in Section 16.5gives umax = (dy/dt)max = ωA = (2πv/λ)A =2π(6.7 m/s)(3 cm)/(75 cm) = 1.68 m/s.

Problem24. A rope is stretched between supports 12 m apart;

its tension is 35 N. If one end of the rope istweaked, the resulting disturbance reaches theother end 0.45 s later. What is the total mass ofthe rope?

SolutionThe wave speed is v = `/t =

√F/(m/`), where

` = 12 m is the length and t = 0.45 s is the traveltime. Therefore, the mass is m = Ft2/` = (35 N)×(0.45 s)2/(12 m) = 591 g.

Problem25. A 3.1-kg mass hangs from a 2.7-m-long string

whose total mass is 0.62 g. What is the speed oftransverse waves on the string? Hint: You canignore the string mass in calculating the tensionbut not in calculating the wave speed. Why?

SolutionThe tension in the string is approximately equal to theweight of the 3.1 kg mass (since the weight of thestring is only 2% of this). Thus, v =

√F/µ =√

(3.1 kg)(9.8 m/s2)(2.7 m)/(0.62 g) = 364 m/s.(0.62 g is small compared to 3.1 kg, but not smallcompared to zero!)

Problem26. Transverse waves propagate at 18 m/s on a string

whose tension is 14 N. What will be the wavespeed if the tension is increased to 40 N?

SolutionSince µ = F1/v2

1 , 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

Problem27. The density of copper is 8.29 g/cm3

. What is thetension in a 1.0-mm-diameter copper wire thatpropagates transverse waves at 120 m/s?

SolutionThe linear mass density of copper wire with diameterd is µ = m/` = ρ 1

4πd2 = (8.29 g/cm3) 14π(1 mm)2 =

6.51×10−3 kg/m, so F = µv2 = (6.51×10−3 kg/m)×(120 m/s)2 = 93.8 N.

Problem28. A 100-m-long wire has a mass of 130 g. A sample

of the wire is tested and found to break at atension of 150 N. What is the maximumpropagation speed for transverse waves on thiswire?

Solution

vmax =√

Fmax/µ =√

150 N/(0.13 kg/100 m)= 340 m/s.

Problem29. A 25-m-long piece of 1.0-mm-diameter wire is put

under 85 N tension. If a transverse wave takes0.21 s to travel the length of the wire, what is thedensity of the material comprising the wire?

SolutionFrom the length of wire, travel time, and Equa-tion 16-7, v = 25 m/0.21 s =

√85 N/µ, so µ = 6.00×

10−3 kg/m. But for a uniform wire of length ` anddiameter d, ρ = µ/1

4πd2 = (6.00×10−3 kg/m)÷14π(1 mm)2 = 7.64 g/cm3 (see solution to Problem 27).

Problem30. 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

m1

figure 16-31 Problem 30.

pulley and tied to a wall as shown in Fig. 16-31.The speed of transverse waves on the horizontalsection of wire is observed to be 20 m/s. If asecond mass m2 is added to the first, the wavespeed increases to 45 m/s. Find the second mass.Assume the string does not stretch appreciably.

SolutionSince the wire is at rest, the tension in the horizontalsection equals the weight attached (provided thepulley is frictionless). Then the wave speeds arev1 =

√m1g/µ and v2 =

√(m1 + m2)g/µ, from which

one finds that m2 = µ(v22 − v2

1)/g = (5.6 g/m)×[(45 m/s)2 − (20 m/s)2]/(9.8 m/s2) = 929g. (If thewire doesn’t stretch, its diameter stays the same and µis constant.)

Problem31. A steel wire can tolerate a maximum tension per

unit cross-sectional area of 2.7 GN/m2 before itundergoes permanent distortion. What is themaximum possible speed for transverse waves in asteel wire if it is to remain undistorted? Steel hasa density of 7.9 g/cm3

.

SolutionThe linear density is the (volume) density times thecross-sectional area (see solution to Problem 27),whereas the maximum tension is 2.7 GN/m2 timesthe same cross-sectional area. Therefore vmax =√

(2.7 GN/m2)/(7.9 g/cm3) = 585 m/s. (Recall that

the prefix giga equals 109 and that 1 g/cm3 =103 kg/m3

.)

Problem32. A uniform cable hangs vertically under its own

weight. Show that the speed of waves on the cableis given by v =

√yg, where y is the distance from

the bottom of the cable.

SolutionThe tension in the cable can be found by integratingNewton’s second law, applied to a small element atrest. With quantities defined in the sketch,0 = T + dT − T − g dm, or dT = g dm. For a uniformcable, dm = µ dy where the linear density µ is aconstant, so T = µgy (the constant of integration iszero for y measured from the bottom of the cable). Itfollows from Equation 16-7 that v =

√T/µ =

√gy.

6 CHAPTER 16

Problem 32 Solution.

Section 16-5: Wave Power and Intensity

Problem33. A rope with 280 g of mass per meter is under

550 N tension. A wave with frequency 3.3 Hz andamplitude 6.1 cm is propagating on the rope.What is the average power carried by the wave?

SolutionThe average power transmitted by transverse travelingwaves in a string is given by Equation 16-8, P =12µω2A2v = 1

2 (0.28 kg/m)(2π×3.3 Hz)2(0.061 m)2×√550 N/(0.28 kg/m) = 9.93 W. (We used Equa-

tion 16-7 for v.)

Problem34. A motor drives a mechanism that produces simple

harmonic motion at one end of a stretched cable.The frequency of the motion is 30 Hz, and themotor can supply energy at an average rate of350 W. If the cable has linear density 450 g/m andis under 1.7 kN tension, (a) what is the maximumwave amplitude that can be driven down thecable? (b) If the motor were replaced by a largerone capable of supplying 700 W, how would themaximum amplitude change?

Solution(a) With all the other quantities in Equation 16-8fixed, the amplitude is proportional to the square rootof the average power transmitted. If there are nolosses, the power transmitted equals the powersupplied by the motor, so the maximum waveamplitude is

√2P /µv/ω =

√2P /

√Fµ/ω =

[2(350 W)/√

(0.45 kg/m)(1.7 kN)]1/2/(2π×30 Hz) =2.67 cm. (Note: µv =

√Fµ from Equation 16-7.)

(b) If the motor’s power is doubled, the maximumamplitude increases by a factor of

√2 to 3.77 cm.

Problem35. 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. Theresulting waves propagate along the Slinky at2.3 m/s. What is the wave amplitude?

SolutionWe assume that the elastic properties of astretched string are shared by the Slinky, soEquation 16-8 applies. Then A =√

2(1.1 W)/(0.06 kg/m)(2.3 m/s)/(2π×1.8 Hz) =35.3 cm.

Problem36. A simple harmonic wave of amplitude 5.0 cm,

wavelength 70 cm, and frequency 14 Hz ispropagating on a wire with linear density 40 g/m.Find the wave energy per unit length of the wire.

SolutionUsing the expression found in the solution to thenext problem, we find dE/dx = 1

2 (0.04 kg/m)×(2π×14 Hz)2(0.05 m)2 = 0.387 J/m.

Problem37. Figure 16-32 shows a wave train consisting of two

cycles of a sine wave propagating along a string.Obtain an expression for the total energy in thiswave train, in terms of the string tension F, thewave amplitude A, and the wavelength λ.

λ

A

figure 16-32 Problem 37.

SolutionThe average wave energy, dE, in a small element ofstring of length dx, is transmitted in time, dt, at thesame speed as the waves, v = dx/dt. From Equa-tion 16-8, dE = P dt = 1

2µω2A2v dt = 12µω2A2dx, so

the average linear energy density is dE/dx = 12µω2A2.

The total average energy in a wave train of length` = 2λ is E = (dE/dx)` = 1

2µω2A2(2λ). In terms ofthe quantities specified in this problem (see Equa-tions 16-1 and 7) E = 1

2 (F/v2)(2πv/λ)2A2(2λ) =4π2FA2/λ. (Note: The relation derived can be writtenas P = (dE/dx)v. For a one-dimensional wave, P isthe intensity, so the average intensity equals the

CHAPTER 16 7

average energy density times the speed of wave energypropagation. This is a general wave property, e.g., seethe first unnumbered equation for S in Section 34-10.)

Problem38. A steel wire with linear density 5.0 g/m is under

450 N tension. What is the maximum power thatcan be carried by transverse waves on this wire ifthe wave amplitude is not to exceed 10% of thewavelength?

SolutionEquation 16-8, written in terms of the tension andwavelength, is P = 1

2 (F/v2)(2πv/λ)2A2√

F/µ =2π2F 3/2µ−1/2(A/λ)2. If A/λ < 0.1, thenP < 2π2(450 N)3/2(0.005 kg/m)−1/2(0.1)2 = 26.6 kW.

Problem39. A loudspeaker emits energy at the rate of 50 W,

spread in all directions. What is the intensity ofsound 18 m from the speaker?

SolutionThe wave power is spread out over a sphere of area4πr2, so the intensity is 50 W/4π(18 m)2 =12.3 mW/m2

. (See Equation 16-9.)

Problem40. The light intensity 3.3 m from a light bulb is

0.73 W/m2. What is the power output of the bulb,

assuming it radiates equally in all directions?

SolutionFrom Equation 16-9, P = 4πr2I = 4π(3.3 m)2×(0.73 W/m2) = 99.9 W ≈ 100 W, typical for alightbulb.

Problem41. Use data from Appendix E to determine the

intensity of sunlight at (a) Mercury and (b) Pluto.

SolutionEquation 16-9 gives the ratio of intensities at twodistances from an isotropic source of spherical wavesas I2/I1 = (r1/r2)2. If we use the average intensity ofsunlight given in Table 16-1 and mean orbitaldistances to the sun from Appendix E, we obtain(a) IMerc = IE(rE/rMerc)2 = (1368 W/m2)(150÷57.9)2 = 9.18 kW/m2

, and (b) IPluto = (1368 W/m2)×(150/5.91×103)2 = 0.881 W/m2

. (Alternatively, theluminosity of the sun, P = 3.85×1026 W, fromAppendix E, could be used directly in Equation 16-9,with only slightly different numerical results.)

Problem42. A 9-W laser produces a beam 2 mm in diameter.

Compare its light intensity with that of sunlight atnoon, about 1 kW/m2

.

SolutionFor a beam of constant cross-sectional area, theintensity of the laser beam is 9 W/π(1 mm)2 =2.86 MW/m2

, which is 2.86×103 times the givenintensity of sunlight at the ground.

Problem43. Light emerges from a 5.0-mW laser in a beam

1.0 mm in diameter. The beam shines on a wall,producing a spot 3.6 cm in diameter. What arethe beam intensities (a) at the laser and (b) at thewall?

SolutionIf we assume that the power output of the laser isspread uniformly over the cross-sectional area of itsbeam, then I = P / 1

4πd2. (a) When the beam emerges,I = 5 mW/1

4π(1 mm)2 = 6.37 kW/m2, while (b) after

its diameter has expanded by 36 times, at the wall,I ′ = I(1/36)2 = 4.91 W/m2

.

Problem44. A large boulder drops from a cliff into the ocean,

producing circular waves. A small boat 18 m fromthe impact point measures the wave amplitude at130 cm. At what distance will the amplitude be50 cm?

SolutionThe intensity of a surface wave decreases inverselywith the distance from the source (see diagram), andis proportional to the square of the amplitude. ThenA2 ∼ 1/r or, at two distances from the source,(A/A′)2 = r′/r. Thus, r′ = (130/50)2(18 m) = 122 mfor the wave in this problem.

Problem 44 Solution.

8 CHAPTER 16

Problem45. Use Table 16-1 to determine how close to a rock

band you should stand for it to sound as loud as ajet plane at 200 m. Treat the band and the planeas point sources. Is this assumption reasonable?

SolutionTo have the same loudness, the soundwave intensitiesshould be equal, i.e., Iband(r) = Ijet(200 m). Regardedas isotropic point sources, use of Equation 16-9 givesPband/r2 = Pjet/(200 m)2. The average power of eachsource can be found from Table 16-1 and a secondapplication of Equation 16-9, Pband =4π(4 m)2(1 W/m2) and Pjet = 4π(50 m)2(10 W/m2).Then r2 = (Pband/Pjet)(200 m)2 = (200 m)2(4 m)2×(1 W/m2)/(50 m)2(10 W/m2), or r = 5.06 m. The sizeof a rock band is several meters, nearly equal to thisdistance, so a point source is not a goodapproximation. Besides, the acoustical output of arock band usually emanates from an array of speakers,which is not point-like. Moreover, the size of a jetplane is also not very small compared to 50 m.

Section 16-6: The Superposition Principle andWave Interference

Problem46. Consider two functions f(x± vt) and g(x± vt)

that both satisfy the wave equation (Equa-tion 16-12). Show that their sum also satisfies thewave equation.

SolutionThe derivative of a sum equals the sum of thederivatives, i.e., ∂2(f + g)/∂x2 = ∂2f/∂x2 + ∂2g/∂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 thefunction and its derivatives, so any linear combinationof solutions af ± bg, is itself a solution.)

Problem47. Two wave pulses are described by

y1(x, t) =2

(x− t)2 + 1, y2(x, t) =

−2(x− 5 + t)2 + 1

,

where x and y are in cm and t in seconds.(a) What is the amplitude of each pulse?(b) At t = 0, where is the peak of each pulse, andin what direction is it moving? (c) At what timewill the two pulses exactly cancel?

Solution(a) The absolute value of the maximum displacementfor each pulse is 2 cm, a value attained when thedenominators are minimal (x− t = 0 for the first pulseand x− 5 + t = 0 for the second). (b) At t = 0, thepeak of the first pulse is at x = 0 moving in thepositive x direction. (x− t = 0 represents the peak, soif t increases so does x. This is why a wave traveling inthe positive x direction is represented by a function ofx− vt.) For the second pulse, the peak is at x = 5,moving in the negative x direction, when t = 0(x− 5 + t = 0 implies x = 5− t and dx/dt = −1 < 0).(c) y1(x, t) + y2(x, t) = 0 for all values of x implies(x− t)2 = (x− 5 + t)2. This is true for all x, only if(x− t) = +(x− 5 + t) or at t = 5

2 = 2.5 s. (The otherroot, (x− t) = −(x− 5 + t), shows that x = 2.5 cm isalways a node, i.e., the net displacement there is zeroat all times.)

Problem48. The triangular wave of Fig. 16-33 can be described

by the following sum of simple harmonic terms:

y(x) =8π2

(sin x

12− sin 3x

32+

sin 5x

52− · · ·

).

Plot the sum of the first three terms in this seriesfor x ranging from 0 to 2π, and compare with thefirst cycle shown in Fig. 16-33. (See alsoActivPhysics Activity 10.7.)

y

1

–1

0 x2 4ππ

figure 16-33 Problem 48.

SolutionThe amplitudes of the first three harmoniccomponents are 8/π2 = 0.81057, 8/9π2 = 0.09006, and8/25π2 = 0.03243, and their wavelengths areλ1 = 3λ3 = 5λ5. The phases alternate by 180◦. Asketch of the components, their superposition, and thefirst cycle of the triangular wave is shown.

Problem49. You’re in an airplane whose two engines are

running at 560 rpm and 570 rpm. How often doyou hear the sound intensity increase as a result ofwave interference?

CHAPTER 16 9

Problem 48 Solution.

SolutionAs mentioned in the text, pilots of twin-engineairplanes use the beat frequency to synchronize therpm’s of their engines. The beat frequency is simplythe difference of the two interfering frequencies,fbeat = (570− 560)/60 s = 1

6 s−1, so you would hearone beat every six seconds.

Problem50. Two waves have the same angular frequency ω,

wave number k, and amplitude A, but theydiffer in phase: y1 = A cos(kx− ωt) and y2 =A cos(kx− ωt + φ). Show that their superpositionis also a simple harmonic wave, and determine itsamplitude As as a function of the phasedifference φ.

SolutionUsing the identity cos α + cos β = 2 cos 1

2 (α− β)×cos 1

2 (α + β), we find y1 + y2 = 2A cos 12φ cos(kx−

ωt + 12φ) ≡ As cos(kx− ωt + φs). This shows that the

amplitude is As = 2A cos 12φ, (and also φs = 1

2φ).

Problem51. What is the wavelength of the ocean waves in

Example 16-5 if the calm water you encounter at33 m is the second calm region on your voyagefrom the center line?

SolutionThe second node occurs when the path difference isthree half-wavelengths, or AP −BP ≡ ∆r = 3

2λ2. (Aphase difference of k2∆r = (2π/λ2)∆r = 3π, or an odd

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

Problem52. The two loudspeakers shown in Fig. 16-34 emit

identical 500-Hz sound waves. Point P is on thefirst nodal line of the interference pattern. Use thenumbers shown to calculate the speed of the soundwaves.

figure 16-34 Problem 52.

SolutionThe path difference between the two loudspeakersand a point on the first nodal line is one half-wavelength, ∆r = 1

2λ = 12v/f, or v = 2f ∆r, where

f is the frequency. From Fig. 16-34 and use of thePythagorean theorem,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.

Section 16-7: The Wave Equation

Problem53. The following equation arises in analyzing the

behavior of shallow water:∂2y

dx2− 1

gh

∂2y

dt2= 0,

where h is the equilibrium depth and y thedisplacement from equilibrium. Give an expressionfor the speed of waves in shallow water. (Hereshallow means the water depth is much less thanthe wavelength.)

SolutionThe equation given is in the standard form for the one-dimensional linear wave equation (Equation 16-12), so

10 CHAPTER 16

the wave speed is the reciprocal of the square root ofthe quantity multiplying ∂2y/∂t2. Thus v =

√gh.

Problem54. Use the chain rule for differentiation to show

explicitly that any function of the form f(x± vt)satisfies the wave equation (Equation 16-12).

SolutionLet primes denote differentiation with respect to thewhole argument φ = (x± vt). Then the chain rulegives ∂f/∂x = (df/dφ)(∂φ/∂x) = f ′, and ∂2f÷∂x2 = f ′′. Similarly, ∂f/∂t = f ′(±v), and ∂2f/∂t2 =v2f ′′. Therefore, any function of φ satisfies the waveequation ∂2f/∂x2 − (1/v2)∂2f/∂t2 = f ′′ − (1/v2)×v2f ′′ = 0.

Paired Problems

Problem55. A wave on a taut wire is described by the equation

y = 1.5 sin(0.10x− 560t), where x and y are in cmand t is in seconds. If the wire tension is 28 N,what are (a) the amplitude, (b) the wavelength,(c) the period, (d) the wave speed, and (e) thepower carried by the wave?

SolutionThe wave has the form of Equation 16-5, with a phaseconstant of −π

2 = −90◦, y(x, t) = A sin(kx− ωt) =A cos(kx− ωt− π

2 ). Comparison reveals thatk = 0.1 cm−1, ω = 560 s−1, and (a) A = 1.5 cm(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-tion 16-6). (e) P = 1

2µω2A2v = 12 (ωA)2(F/v) =

12 (560 s−1×0.015 m)2(28 N)/(56 m/s) = 17.6 W(Equation 16-8, and Equation 16-7 to eliminate µ).

Problem56. A wave given by y = 23 cos(0.025x− 350t), with x

and y in mm and t in seconds, is propagating on acable with mass per unit length 410 g/m. Find(a) the amplitude, (b) the wavelength, (c) thefrequency in Hz, (d) the wave speed, and (e) thepower carried by the wave.

SolutionInspection of the given function and use of Equa-tions 16-3 through 8 gives (a) A = 23 mm; (b) λ =2π/k = 2π/(0.025 mm−1) = 25.1 cm; (c) f = ω/2π =350 s−1/2π = 55.7 Hz; (d) v = ω/k = (350 s−1)÷(0.025 mm−1) = 14 m/s; (e) P = 1

2µω2A2v =12 (0.41 kg/m)(350×23 mm/s)2(14 m/s) = 186 W.

Problem57. A spring of mass m and spring constant k has an

unstretched length `0. Find an expression for thespeed of transverse waves on this spring when ithas been stretched to a length `.

SolutionThe spring may be regarded as a stretched stringwith tension, F = k(`− `0), and linear mass densityµ = m/`. Equation 16-7 gives the speed of transversewaves as v =

√k`(`− `0)/m.

Problem58. When a 340-g spring is stretched to a total length

of 40 cm, it supports transverse waves propagatingat 4.5 m/s. When it’s stretched to 60 cm, thewaves propagate at 12 m/s. Find (a) theunstretched length of the spring and (b) its springconstant.

SolutionFrom the solution to the previous problem, mv2 =k`(`− `0). (a) With v1 and v2 given for `1 and `2, kmay be eliminated by division, before solving for`0: (v2/v1)2 = `2(`2 − `0)/`1(`1 − `0) or

`0 =`21(v2/v1)2 − `22`1(v2/v1)2 − `2

=(40)2(12/4.5)2 − (60)2

(40)(12/4.5)2 − (60)cm = 34.7 cm.

(b) From either pair of values of wave speed andlength,

k = mv2/`(`− `0) =(0.34 kg)(4.5 m/s)2

(0.4 m)(0.4 m− 0.347 m)

=(0.34 kg)(12 m/s)2

(0.6 m)(0.6 m− 0.347 m)= 322 N/m.

Problem59. At a point 15 m from a source of spherical

sound waves, you measure a sound intensity of750 mW/m2

. How far do you need to walk,directly away from the source, until the intensityis 270 mW/m2?

SolutionThe intensity of spherical waves from a point sourceis given by Equation 16-9. At a distance r1, I1 =P /4πr2

1, while after increasing the radial distance byd, I2 = P /4π(r1 + d)2. Dividing and solving for d, onefinds d = r1(

√I1/I2 − 1) = (15 m)(

√(750/270)− 1) =

10.0 m.

CHAPTER 16 11

Problem60. Figure 16-35 shows two observers 20 m apart, on a

line that connects them and a spherical lightsource. If the observer nearest the source measuresa light intensity 50% greater than the otherobserver, how far is the nearest observer from thesource?

20 m x = ?

figure 16-35 Problem 60.

SolutionIf we assume the source emits spherical waves, theratio of the intensities for the two observers isI1/I2 = (x2/x1)2 (Equation 16-9), where the closerobserver is at x1. Then I1 = 1.5I2 and x2 = x1 + 20 m,so this equation becomes (x1 + 20 m)2 = 1.5x2

1. Thepositive solution of this quadratic (when bothobservers are on the same side of the source) isx1 = 2(20 m)(1 +

√3/2) = 89.0 m. (The negative root,

2(20 m)(1−√

3/2) = −8.99 m, corresponds toobservers on opposite sides of the source, i.e., with thelamp in Fig. 16-35 between the two observers.)

Problem61. Two motors in a factory produce sound waves

with the same frequency as their rotation rates. Ifone motor is running at 3600 rpm and the other at3602 rpm, how often will workers hear a peak inthe sound intensity?

SolutionThe beat frequency equals the difference in themotors’ rpm’s, so the period of the beats is Tbeat =1/fbeat = 1/(3602− 3600) min−1 = 30 s. (See alsoProblem 49.)

Problem62. Two radio waves with frequencies of

approximately 50 MHz interfere. The compositewave is detected and fed to a loudspeaker, whichemits audible sound at 500 Hz. What is thepercentage difference between the frequencies ofthe two radio waves?

SolutionThe difference in the frequencies (really its absolutevalue) equals the beat frequency, ∆f = fbeat =

500 Hz, so the percent difference is (100)∆f/f =(100)(500 Hz/50 MHz) = 10−3%.

Supplementary Problems

Problem63. For a transverse wave on a stretched string, the

requirement that the string be nearly horizontalis met if the amplitude is much less than thewavelength. (a) Show this by drawing anappropriate sketch. (b) Show that, under thisapproximation that A ¿ λ, the maximum speed uof the string must be considerably less than thewave speed v. (c) If the amplitude is not to exceed1% of the wavelength, how large can the stringspeed u be in relation to the wave speed v?

Solution(a) The relative “flatness” or “peakedness” of asinusoidal waveform is determined by its maximumslope, |dy/dx|max = |∂/∂x[A cos(kx− ωt)]|max =kA = 2π(A/λ). If A ¿ λ (or kA ¿ 1), the slope isnearly horizontal. (b) In terms of the speeds, kA =ωA/v = umax/v, so the string is nearly flat ifumax ¿ v. (c) If A/λ < 1%, then umax/v = 2π(A/λ) <2π(1%) = 6.3%.

Problem 63 Solution.

Problem64. A 64-g spring has unstretched length 25 cm. With

a 940-g mass attached, the spring undergoessimple harmonic motion with angular frequency6.1 s−1. What will be the speed of transversewaves on this spring when it’s stretched to a totallength of 40 cm?

SolutionWhen used as a nearly ideal mass-spring system(since the spring’s mass is much less than the attachedmass), ω2 = k/m, which allows the spring constant tobe determined, k = (0.94 kg)(6.1 s−1)2. When used ina different way to support transverse waves (seeProblem 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 =5.73 m/s.

12 CHAPTER 16

Problem65. An ideal spring is compressed until its total length

is `1, and the speed of transverse waves on thespring is measured. When it’s compressed furtherto a total length `2, waves propagate at the samespeed. Show that the uncompressed spring lengthis just `1 + `2.

SolutionThe tension in a compressed spring has magnitudek(`0 − `) while its linear mass-density is µ = m/`.Therefore, the speed of transverse waves is v =√

F/µ =√

k`(`0 − `)/m (as in Problem 57 for astretched spring). If v1 = v2 for two differentcompressed lengths, then `1(`0 − `1) = `2(`0 − `2) or(`1 − `2)`0 = `21 − `22 = (`1 − `2)(`1 + `2). Since `1 6= `2,division by `1 − `2 gives `0 = `1 + `2.

Problem66. An ideal spring is stretched to a total length `1.

When that length is doubled, the speed oftransverse waves on the spring triples. Find anexpression for the unstretched length of the spring.

SolutionUtilizing 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 = 5

7`1.

Problem67. A 1-megaton nuclear explosion produces a shock

wave whose amplitude, measured as excess airpressure above normal atmospheric pressure, is1.4×105 Pa (1 Pa = 1 N/m2) at a distance of1.3 km from the explosion. An excess pressure of3.5×104 Pa will destroy a typical woodframehouse. At what distance from the explosion willsuch houses be destroyed? Assume the wavefrontis spherical.

SolutionThe intensity of a spherical wavefront varies inverselywith the square of the distance from the central source(see Fig. 16-18b). In general, the intensity isproportional to the amplitude squared, so A ∼ 1/r fora spherical wave. (This can be proved rigorously bysolution of the spherical wave equation, ageneralization of Equation 16-12.) Therefore A1/A2 =r2/r1, or the overpressure reaches the stated limit at adistance r2 = (1.4×105 Pa/3.5×104 Pa)(1.3 km) =5.2 km from the explosion.

Problem68. Show that the time it takes a wave to propagate

up the cable in Problem 32 is t = 2√

`/g, where `is the cable length.

SolutionThe wave speed in the cable of Problem 32 wasv = dy/dt =

√gy, where y is the distance from the

bottom of the cable. The time for a transverse wavesignal to propagate from the bottom to the top (y = 0to `) is

t =∫

dt =∫ `

0

dy

v=

∫ `

0

dy√gy

=1√g2√

y

∣∣∣∣∣

`

0

= 2

√`

g.

Problem69. In Example 16-5, how much farther would you

have to row to reach a region of maximum waveamplitude?

figure 16-36 Problem 69 Solution.

SolutionIn general, the interference condition for waves in thegeometry of Example 16-5 is AP −BP = nλ/2, wheren is an odd integer for destructive interference (a node)and n is an even integer for constructive interference(a maximum amplitude). (In Example 16-5, n = 1gave the first node and in Problem 51, n = 3 gave

CHAPTER 16 13

the second node.) If d = 20 m is the distance betweenthe openings, ` = 75 m is the perpendicular distancefrom the breakwater, and x is the distance parallelto the breakwater measured from the midpointof the openings, the interference condition is√

`2 + (x + 12d)2 −

√`2 + (x− 1

2d)2 = nλ/2 (seeFig. 16-36). In this problem, we wish to find x for thefirst maximum, n = 2, and the wavelength calculatedin Example 16-5, λ = 16.01 m. Solving for x, we find:

x2 =[`2 + ( 1

2d)2 − ( 14nλ)2]

(2d/nλ)2 − 1

=[(75)2 + (10)2 − (8.005)2] m2

(40/32.02)2 − 1= (100.5 m)2.

This is 100.5 m− 33 m = 67.5 m farther than the firstnode in Example 16-5. (Note: We rounded off to threefigures; if you round off to two figures, the answer is67 m. Also, if x = 33 m is substituted into the generalinterference condition, one can recapture the

wavelengths of the first and second nodes, for n = 1and 3, calculated in Example 16-5 and Problem 51,respectively.)

Problem70. Suppose the wavelength of the ocean waves in

Example 16-5 were 8.4 m. How far would youhave to row from the center line, staying 75 mfrom the breakwater, in order to find (a) the firstand (b) the second region of relative calm?

SolutionFrom the general solution for x in the previousproblem, (a) x = 16.2 m to the first node (n = 1),and (b) x = 61.2 m to the second node (n = 3). (Use` = 75 m, d = 20 m, and λ = 8.4 m for the othervalues.)