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CHAPTER 16 WAVE MOTION ActivPhysics can help with these problems: 1.25 m apart. What is the speed of the waves on the Activities 10.1, 10.2, 10.7, 10.10 cable? Compare with the speed of light in vacuum. Section 16-2: Wave Properties Solution Problem The distance between adjacent wave crests is one 1. Ocean waves with 18-m wavelength travel at wavelength, so the wave speed in the cable (Equa- 5.3 m/s. What is the time interval between wave tion 16-1) is v = f λ = (145×106 Hz)(1.25 m) = 1.81× crests passing under a boat moored at a ﬁxed 108 m/s = 0.604c, where c = 3×108 m/s is the wave location? speed in vacuum. Solution Problem Wave crests (adjacent wavefronts) take a time of one 6. Calculate the wavelengths of (a) a 1.0-MHz AM period to pass a ﬁxed point, traveling at the wave radio wave, (b) a channel 9 TV signal (190 MHz), speed (or phase velocity) for a distance of one (c) a police radar (10 GHz), (d) infrared radiation wavelength. Thus T = λ/v = 18 m/(5.3 m/s) = 3.40 s. from a hot stove (4.0×1013 Hz), (e) green light (6.0×1014 Hz), and (f) 1.0×1018 Hz X rays. All are Problem electromagnetic waves that propagate at 2. Ripples in a shallow puddle are propagating at 3.0×108 m/s. 34 cm/s. If the wave frequency is 5.2 Hz, what are (a) the period and (b) the wavelength? Solution With Equation 16-1 in an equivalent form, λ = vT = Solution v/f, we ﬁnd: (a) λ = (3×108 m/s)/(106 Hz) = Equation 16-1 gives T = 1/f = 1/5.2 Hz = 0.192 s and 300 m, (b) λ = 1.58 m, (c) λ = 3 cm, (d) λ = 7.5 µm, λ = v/f = vT = (34 cm/s)/(5.2 Hz) = 6.54 cm. (e) λ = 500 nm, (f) λ = 3˚. (See Appendix C on A units.) Problem 3. An 88.7-MHz FM radio wave propagates at the Problem speed of light. What is its wavelength? 7. Detecting objects by reﬂecting waves oﬀ them is eﬀective only for objects larger than about one Solution wavelength. (a) What is the smallest object that From Equation 16-1, λ = v/f = (3×108 m/s)÷ can be seen with visible light (maximum frequency (88.7×106 Hz) = 3.38 m. 7.5×1014 Hz)? (b) What is the smallest object that can be detected with a medical ultrasound unit Problem operating at 5 MHz? The speed of ultrasound 4. One end of a rope is tied to a wall. You shake the waves in body tissue is about 1500 m/s. other end with a frequency of 2.2 Hz, producing waves whose wavelength is 1.6 m. What is their Solution propagation speed? (a) The wavelength of light corresponding to this maximum frequency is λ = c/f = (3×108 m/s)÷ Solution (7.5×1014 Hz) = 400 nm, violet in hue (see Equa- v = f λ = (2.2 Hz)(1.6 m) = 3.52 m/s (see Equa- tion 16-1). (b) The ultrasonic waves described have tion 16-1). wavelength λ = v/f = (1500 m/s)/(5 MHz) = 0.3 mm. Problem Problem 5. A 145-MHz radio signal propagates along a cable. 8. A seismograph located 1200 km from an earthquake Measurement shows that the wave crests are spaced detects waves from the quake 5.0 min after the 2 CHAPTER 16 quake occurs. The seismograph oscillates in step Problem with the waves, at a frequency of 3.1 Hz. Find the 11. An ocean wave has period 4.1 s and wavelength wavelength of the waves. 10.8 m. Find (a) its wave number and (b) its angular frequency. Solution The wave speed can be calculated from the distance Solution and the travel time, which, together with the From Equations 16-3 and 4, (a) k = 2π/10.8 m = frequency and Equation 16-1, gives a wavelength of 0.582 m−1 , and (b) ω = 2π/(4.1 s) = 1.53 s−1 . λ = v/f = (d/t)/f = 1200 km/(5×60×3.1) = 1.29 km. Problem Problem 12. Find (a) the amplitude, (b) the wavelength, 9. In Fig. 16-28 two boats are anchored oﬀshore and (c) the period, and (d) the speed of a wave whose are bobbing up and down on the waves at the rate displacement is given by y = 1.3 cos(0.69x + 31t), of six complete cycles each minute. When one boat where x and y are in cm and t is in seconds. (e) In is up the other is down. If the waves propagate at which direction is the wave propagating? 2.2 m/s, what is the minimum distance between the boats? Solution (a) A = ymax = 1.3 cm, (b) λ = 2π/k = 2π÷ Solution 0.69 cm−1 = 9.11 cm, (c) T = 2π/ω = 2π/31 s−1 = The boats are 180◦ = π rad out of phase, so the 0.203 s−1 , and (d) v = ω/k = 44.9 cm/s. (e) A phase minimum distance separating them is half a of the form kx + ωt describes a wave propagating in wavelength. (In general, they could be an odd number the negative x direction. of half-wavelengths apart.) The frequency is 6/60 s = 1 0.1 Hz, so 2 λ = 1 v/f = 1 (2.2 m/s)/(0.1/s) = 11 m. 2 2 Problem (Fig. 16-28 shows the answer, not the question.) 13. A simple harmonic wave of wavelength 16 cm and amplitude 2.5 cm is propagating along a string in the negative x direction at 35 cm/s. Find (a) the angular frequency and (b) the wave number. (c) Write a mathematical expression describing the displacement y of this wave (in centimeters) as a function of position and time. Assume the displacement at x = 0 is a maximum when t = 0. Solution figure 16-28 Problem 9 Solution. (b) Equation 16-4 gives k = 2π/16 cm = 0.393 cm−1 , and (a) Equation 16-6 gives ω = kv = (0.393 cm−1 )× (35 cm/s) = 13.7 s−1 . (c) Equation 16-5, for a wave Section 16-3: Mathematical Description of moving in the negative x direction, becomes Wave Motion y(x, t) = (2.5 cm) cos[(0.393 cm−1 )x + (13.7 s−1 )t]. Problem Problem 10. Ultrasound used in a particular medical imager 14. Figure 16-29 shows a simple harmonic wave at has frequency 4.8 MHz and wavelength 0.31 mm. time t = 0 and later at t = 2.6 s. Write a Find (a) the angular frequency, (b) the wave mathematical description of this wave. number, and (c) the wave speed. Solution Solution Inspection of Fig. 16-29 shows that the wavelength is (a) Equation 16-3 gives ω = 2π(4.8 MHz) = 3.02× 8 cm, the amplitude is 1.5 cm, and the velocity is 107 s−1 . (b) Equation 16-4 gives k = 2π/0.31 mm = v = ∆x/∆t = 2 cm/2.6 s = 0.769 cm/s. The phase 2.03×104 m−1 . (c) Together, these give v = f λ = constant is zero (since y = A at t = 0 and x = 0) and ω/k = 1.49 km/s (see Equation 16-1 or 6). the wave is traveling in the positive x direction. Thus, k = 2π/λ = 0.785 cm−1 , ω = kv = 0.604 s−1 , and the CHAPTER 16 3 waveform is y(x, t) = (1.5 cm) cos[(0.785 cm−1 )x − positive x direction with speed v, can be obtained by (0.604 s−1 )t]. replacing x by x − vt, y(x, t) = f (x − vt). For the given f (x) and v, y(x, t) = 2[(x − 3t)4 + 1]−1 , with x and y in cm and t in s. Problem 18. Plot the answer to the previous problem as a function of position x for the two cases t = 0 and t = 4 s, and verify that your plots are consistent with the pulse speed of 3 cm/s. figure 16-29 Problem 14 Solution. Solution (Between 0 and 4 s, the pulse has moved Problem (3 cm/s)(4 s) = 12 cm to the right.) 15. What are (a) the amplitude, (b) the frequency in hertz, (c) the wavelength, and (d) the speed of a water wave whose displacement is y = 0.25 sin(0.52x − 2.3t), where x and y are in meters and t in seconds? Solution Comparison of the given displacement with Equa- tion 16-5 reveals that (a) A = 0.25 m, (b) f = ω/2π = (2.3 s−1 )/2π = 0.366 Hz, (c) λ = 2π/k = Problem 18 Solution. 2π/(0.52 m−1 ) = 12.1 m, and v = ω/k = (2.3 s−1 )/0.52 m−1 = 4.42 m/s. (Note: The presence of a phase constant of φ = −π/2 in the expression for Problem y(x, t) = A sin(kx − ωt) = A cos(kx − ωt + φ) does not 19. Figure 16-30a shows a wave plotted as a function aﬀect any of the quantities queried in this problem.) of position at time t = 0, while Fig. 16-30b shows the same wave plotted as a function of time at Problem position x = 0. Find (a) the wavelength, (b) the 16. A sound wave with frequency 256 Hz (the musical period, (c) the wave speed, and (d) the direction note middle C) is propagating in air at 343 m/s. of propagation. How far apart are two points on the wave that diﬀer in phase by π/2 or 90◦ ? Solution Two points in space separated by one wavelength diﬀer in phase by 2π (or one cycle). Therefore, a phase diﬀerence of π/2 (one quarter of a cycle) corresponds to a separation of λ/4 = (v/f )/4 = (343 m/s)÷ (4×256 Hz) = 33.5 cm. Problem 17. At time t = 0, the displacement in a transverse wave pulse is described by y = 2(x4 + 1)−1 , with both x and y in cm. Write an expression for the pulse as a function of position x and time t if it is propagating in the positive x direction at 3 cm/s. Solution figure 16-30 Problem 19. From the shape of the pulse at t = 0, y(x, 0) = f (x), a pulse with the same waveform, traveling in the 4 CHAPTER 16 Solution Problem (a) The wavelength is the distance between successive 23. A transverse wave with 3.0-cm amplitude and maxima at the same time (say t = 0), so Fig. 16-30a 75-cm wavelength is propagating on a stretched gives λ = 3 m. (b) The period is the time interval spring whose mass per unit length is 170 g/m. If between successive maxima (or some other speciﬁc the wave speed is 6.7 m/s, ﬁnd (a) the spring phases diﬀering by 2π) at the same point (say x = 0), tension and (b) the maximum speed of any point so Fig. 16-30b gives T = 1.5 s. (c) v = ω/k = λ/T = on the spring. 2 m/s. (d) Fig. 16-30b shows that as t increases from 0, the displacement at x = 0 ﬁrst becomes negative. Solution The waveform in Fig. 16-30a must therefore move to (a) Equation 16-7 gives F = µv 2 = (0.17 kg/m)× the right, in the positive x direction. [For the (6.7 m/s)2 = 7.63 N. (b) The unnumbered equation for sinusoidal wave pictured, y(x, 0) = A sin kx and the vertical velocity of the medium in Section 16.5 y(0, t) = −A sin ωt = A sin(−ωt), so y(x, t) = gives umax = (dy/dt)max = ωA = (2πv/λ)A = A sin(kx − ωt).] 2π(6.7 m/s)(3 cm)/(75 cm) = 1.68 m/s. Problem Problem 20. Write a mathematical description of the wave in 24. A rope is stretched between supports 12 m apart; the preceding problem. its tension is 35 N. If one end of the rope is tweaked, the resulting disturbance reaches the Solution other end 0.45 s later. What is the total mass of From Fig. 16-30, the amplitude is probably 2 m, so the rope? y(x, t) = (2 m)sin[(2π/3 m)x − (2π/1.5 s)t] = (2 m)sin[(2.09 m−1 )x − (4.19 s−1 )t]. (See solution to Solution preceding problem.) The wave speed is v = /t = F/(m/ ), where = 12 m is the length and t = 0.45 s is the travel Section 16-4: Waves on a String time. Therefore, the mass is m = F t2 / = (35 N)× (0.45 s)2 /(12 m) = 591 g. Problem 21. The main cables supporting New York’s George Problem Washington Bridge have a mass per unit length of 25. A 3.1-kg mass hangs from a 2.7-m-long string 4100 kg/m and are under tension of 250 MN. At whose total mass is 0.62 g. What is the speed of what speed would a transverse wave propagate on transverse waves on the string? Hint: You can these cables? ignore the string mass in calculating the tension but not in calculating the wave speed. Why? Solution v = F/µ = (2.5×108 N)/(4100 kg/m) = 247 m/s Solution (from Equation 16-7). The tension in the string is approximately equal to the weight of the 3.1 kg mass (since the weight of the Problem string is only 2% of this). Thus, v = F/µ = 22. A transverse wave 1.2 cm in amplitude is 2 propagating on a string; the wave frequency is (3.1 kg)(9.8 m/s )(2.7 m)/(0.62 g) = 364 m/s. 44 Hz. The string is under 21 N tension and has (0.62 g is small compared to 3.1 kg, but not small mass per unit length of 15 g/m. Determine (a) the compared to zero!) wave speed and (b) the maximum speed of a point on the string. Problem 26. Transverse waves propagate at 18 m/s on a string Solution whose tension is 14 N. What will be the wave (a) The wave speed is v = F/µ = speed if the tension is increased to 40 N? (21 N)/(0.015 kg/m) = 37.4 m/s. (b) From the second equation in Section 16.5, umax = ωA = Solution 2 2π(44 Hz)(1.2 cm) = 3.32 m/s. Since µ = F1 /v1 , v2 = F2 /µ = F2 /F1 v1 = 40/14(18 m/s) = 30.4 m/s is the speed at the increased tension. CHAPTER 16 5 Problem pulley and tied to a wall as shown in Fig. 16-31. 3 27. The density of copper is 8.29 g/cm . What is the The speed of transverse waves on the horizontal tension in a 1.0-mm-diameter copper wire that section of wire is observed to be 20 m/s. If a propagates transverse waves at 120 m/s? second mass m2 is added to the ﬁrst, the wave speed increases to 45 m/s. Find the second mass. Solution Assume the string does not stretch appreciably. The linear mass density of copper wire with diameter 3 d is µ = m/ = ρ 1 πd2 = (8.29 g/cm ) 1 π(1 mm)2 = Solution 4 4 6.51×10−3 kg/m, so F = µv 2 = (6.51×10−3 kg/m)× Since the wire is at rest, the tension in the horizontal (120 m/s)2 = 93.8 N. section equals the weight attached (provided the pulley is frictionless). Then the wave speeds are Problem v1 = m1 g/µ and v2 = (m1 + m2 )g/µ, from which 2 2 one ﬁnds that m2 = µ(v2 − v1 )/g = (5.6 g/m)× 28. A 100-m-long wire has a mass of 130 g. A sample 2 2 2 of the wire is tested and found to break at a [(45 m/s) − (20 m/s) ]/(9.8 m/s ) = 929g. (If the tension of 150 N. What is the maximum wire doesn’t stretch, its diameter stays the same and µ propagation speed for transverse waves on this is constant.) wire? Problem Solution 31. A steel wire can tolerate a maximum tension per 2 unit cross-sectional area of 2.7 GN/m before it undergoes permanent distortion. What is the vmax = Fmax /µ = 150 N/(0.13 kg/100 m) maximum possible speed for transverse waves in a = 340 m/s. steel wire if it is to remain undistorted? Steel has 3 a density of 7.9 g/cm . Problem 29. A 25-m-long piece of 1.0-mm-diameter wire is put Solution under 85 N tension. If a transverse wave takes The linear density is the (volume) density times the 0.21 s to travel the length of the wire, what is the cross-sectional area (see solution to Problem 27), 2 density of the material comprising the wire? whereas the maximum tension is 2.7 GN/m times the same cross-sectional area. Therefore vmax = Solution 2 3 (2.7 GN/m )/(7.9 g/cm ) = 585 m/s. (Recall that From the length of wire, travel time, and Equa- 3 the preﬁx giga equals 109 and that 1 g/cm = tion 16-7, v = 25 m/0.21 s = 85 N/µ, so µ = 6.00× 3 103 kg/m .) 10−3 kg/m. But for a uniform wire of length and 1 diameter d, ρ = µ/ 4 πd2 = (6.00×10−3 kg/m)÷ 1 2 3 Problem 4 π(1 mm) = 7.64 g/cm (see solution to Problem 27). 32. A uniform cable hangs vertically under its own weight. Show that the speed of waves on the cable √ Problem is given by v = yg, where y is the distance from the bottom of the cable. 30. A mass m1 is attached to a wire of linear density 5.6 g/m, and the other end of the wire run over a Solution The tension in the cable can be found by integrating Newton’s second law, applied to a small element at rest. With quantities deﬁned in the sketch, 0 = T + dT − T − g dm, or dT = g dm. For a uniform cable, dm = µ dy where the linear density µ is a constant, so T = µgy (the constant of integration is m1 zero for y measured from the bottom of the cable). It √ follows from Equation 16-7 that v = T /µ = gy. figure 16-31 Problem 30. 6 CHAPTER 16 Problem 35. A 600-g Slinky is stretched to a length of 10 m. You shake one end at the frequency of 1.8 Hz, applying a time-average power of 1.1 W. The resulting waves propagate along the Slinky at 2.3 m/s. What is the wave amplitude? Solution We assume that the elastic properties of a stretched string are shared by the Slinky, so Equation 16-8 applies. Then A = Problem 32 Solution. 2(1.1 W)/(0.06 kg/m)(2.3 m/s)/(2π×1.8 Hz) = 35.3 cm. Section 16-5: Wave Power and Intensity Problem Problem 36. A simple harmonic wave of amplitude 5.0 cm, 33. A rope with 280 g of mass per meter is under wavelength 70 cm, and frequency 14 Hz is 550 N tension. A wave with frequency 3.3 Hz and propagating on a wire with linear density 40 g/m. amplitude 6.1 cm is propagating on the rope. Find the wave energy per unit length of the wire. What is the average power carried by the wave? Solution Solution Using the expression found in the solution to the The average power transmitted by transverse traveling ¯ next problem, we ﬁnd dE/dx = 1 (0.04 kg/m)× ¯ waves in a string is given by Equation 16-8, P = 2 1 2 2 1 2 2 (2π×14 Hz)2 (0.05 m)2 = 0.387 J/m. 2 µω A v = 2 (0.28 kg/m)(2π×3.3 Hz) (0.061 m) × 550 N/(0.28 kg/m) = 9.93 W. (We used Equa- Problem tion 16-7 for v.) 37. Figure 16-32 shows a wave train consisting of two cycles of a sine wave propagating along a string. Problem Obtain an expression for the total energy in this 34. A motor drives a mechanism that produces simple wave train, in terms of the string tension F, the harmonic motion at one end of a stretched cable. wave amplitude A, and the wavelength λ. The frequency of the motion is 30 Hz, and the motor can supply energy at an average rate of λ 350 W. If the cable has linear density 450 g/m and A is under 1.7 kN tension, (a) what is the maximum wave amplitude that can be driven down the cable? (b) If the motor were replaced by a larger one capable of supplying 700 W, how would the figure 16-32 Problem 37. maximum amplitude change? Solution Solution ¯ The average wave energy, dE, in a small element of (a) With all the other quantities in Equation 16-8 string of length dx, is transmitted in time, dt, at the ﬁxed, the amplitude is proportional to the square root same speed as the waves, v = dx/dt. From Equa- of the average power transmitted. If there are no ¯ ¯ 1 tion 16-8, dE = P dt = 1 µω 2 A2 v dt = 2 µω 2 A2 dx, so losses, the power transmitted equals the power 2 ¯ 1 the average linear energy density is dE/dx = 2 µω 2 A2 . supplied by the motor, so the maximum wave ¯ ¯ √ amplitude is 2P /µv/ω = 2P / F µ/ω = The total average energy in a wave train of length ¯ ¯ = 2λ is E = (dE/dx) = 1 µω 2 A2 (2λ). In terms of [2(350 W)/ (0.45 kg/m)(1.7 kN)]1/2 /(2π×30 Hz) = √ 2 the quantities speciﬁed in this problem (see Equa- 2.67 cm. (Note: µv = F µ from Equation 16-7.) ¯ tions 16-1 and 7) E = 1 (F/v 2 )(2πv/λ)2 A2 (2λ) = 2 (b) If the motor’s power is doubled, the maximum √ 4π 2 F A2 /λ. (Note: The relation derived can be written amplitude increases by a factor of 2 to 3.77 cm. ¯ ¯ ¯ as P = (dE/dx)v. For a one-dimensional wave, P is the intensity, so the average intensity equals the CHAPTER 16 7 average energy density times the speed of wave energy Problem propagation. This is a general wave property, e.g., see 42. A 9-W laser produces a beam 2 mm in diameter. the ﬁrst unnumbered equation for S in Section 34-10.) Compare its light intensity with that of sunlight at 2 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

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