CHAPTER 15 Wave Motion. 1. The speed of the wave is

Transcription

1 CHAPTER 15 Wave Motion 1. The speed of the wave is v = fλ = λ/t = (9.0 m)/(4.0 s) = 2.3 m/s.

2 7. We find the tension from the speed of the wave: v = [F T /(m/l)] 1/2 ; (4.8 m)/(0.85 s) = {F T /[(0.40 kg)/(4.8 m)]} 1/2, which gives F T = 2.7 N.

3 9. (a) Because the pulse travels up and back, the speed is v = 2L/t = 2(600 m)/(16 s) = 75 m/s. (b) The mass density of the cable is µ = m/l = ρal/l = ρa. We find the tension from v = (F T /µ) 1/2 = (F T /ρa) 1/2 ; 75 m/s = [F T /( kg/m 3 )π( m) 2 ] 1/2, which gives F T = N.

4 12. Because the speed, frequency, and medium are the same for the two waves, the intensity depends on the amplitude only: I D M 2. For the ratio of intensities we have I 2 /I 1 = (D M2 /D M1 ) 2 ; 2 = (D M2 /D M1 ) 2, which gives D M2 /D M1 = 1.41.

5 14. We assume that the wave spreads out uniformly in all directions. (a) The intensity will decrease as 1/r 2, so the ratio of intensities is I 2 /I 1 = (r 1 /r 2 ) 2 ; I 2 /( W/m 2 ) = [(100 km)/(4.0 km)] 2, which gives I 2 = W/m 2. (b) We can take the intensity to be constant over the small area, so we have P 2 = I 2 S = ( W/m 2 )(5.0 m 2 ) = W.

6 20. The traveling wave is D = (0.48 m) sin [(5.6 m 1 )x + (84 s 1 )t]. (a) We find the wavelength from the coefficient of x: (5.6 m 1 )x = 2πx/λ, which gives λ = 1.12 m. (b) We find the frequency from the coefficient of t: (84 s 1 )t = 2πft, which gives f = 13 Hz. (c) From the positive sign between the x and t terms, the wave is traveling in the x direction, with speed v = fλ = (13.4 Hz)(1.12 m) = 15 m/s (toward negative x). (d) The amplitude is the coefficient of the sine function: D M = 0.48 m. (e) The speed of a particle is u = D/ t = (0.48 m)(84 s 1 ) cos [(5.6 m 1 )x + (84 s 1 )t] = (40 m/s) cos [(5.6 m 1 )x + (84 s 1 )t]. Thus we have u max = (40 m/s)(1) = 40 m/s; u min = (40 m/s)(0) = 0.

7 21. The traveling wave is D = (0.026 m) sin [(45 m 1 )x (1570 s 1 )t ], with ω = 1570 s 1. (a) Each point has simple harmonic motion, so the maximum velocity is u max = D M ω = (0.026 m)(1570 s 1 ) = 41 m/s. (b) The maximum acceleration is a max = D M ω 2 = (0.026 m)(1570 s 1 ) 2 = m/s 2. (c) For any point on the string we have u = D/ t = D M ω cos [(45 m 1 )x (1570 s 1 )t ]; a = 2 D/ t 2 = D M ω 2 sin [(45 m 1 )x (1570 s 1 )t ]. For the given point and time we get u = (41 m/s) cos [(45 m 1 )(1.00 m) (1570 s 1 )(2.0 s) ] = 41 m/s; a = ( m/s 2 ) sin [(45 m 1 )(1.00 m) (1570 s 1 )(2.0 s) ] = m/s 2.

8 31. (a) The speed of the wave in a string is v = [F T /µ] 1/2. Because the tensions must be the same anywhere along the string, for the ratio of velocities we have v 2 /v 1 = (µ 1 /µ 2 ) 1/2. (b) Because the motion of one string is creating the motion of the other, the frequencies must be the same. For the ratio of wavelengths we have λ 2 /λ 1 = v 2 /v 1 = (µ 1 /µ 2 ) 1/2. (c) From the result for part (b) we see that, if µ 2 > µ 1, we have λ 2 < λ 1, so the lighter cord will have the greater wavelength.

9 32. (a) For the traveling wave in the lighter cord, D 1 = (0.050 m) sin [(6.0 m 1 )x (12.0 s 1 )t], we find the wavelength from the coefficient of x: (6.0 m 1 )x = 2πx/λ 1, which gives λ 1 = 1.05 m. (b) We find the tension from the velocity: v 1 = ω/k 1 = (F T /µ 1 ) 1/2 ; (12.0 s 1 )/(6.0 m 1 ) = [F T /(0.10 kg/m)] 1/2, which gives F T = 0.40 N. (c) The tension and frequency do not change, so we have v 2 = ω/k 2 = (F T /µ 2 ) 1/2, or k 1 /k 2 = λ 2 /λ 1 = (µ 1 /µ 2 ) 1/2 ; λ 2 /(1.05 m) = [(0.10 kg/m)/(0.20 kg/m)] 1/2, which gives λ 2 = 0.74 m.

10 34. (a) (b) (c) Because all particles of the string are at equilibrium positions, there is no potential energy. Particles of the string will have transverse velocities, so they have kinetic energy.

11 35. (a) For the sum of the two waves we have D = D 1 + D 2 = D M sin (kx ωt) + D M sin (kx ωt + φ) = 2D M sin (kx ωt + kx ωt + φ) cos (kx ωt kx + ωt φ) = 2D M cos ( φ) sin (2kx 2ωt + φ) = 2D M cos (φ) sin (kx ωt + φ). (b) The amplitude is the coefficient of the sine function: 2D M cos (φ). The variation in x and t is purely sinusoidal. (c) If φ = 0, 2π, 4π, ; φ = 0, π, 2π, ; so cos (φ) = ± 1. Thus the amplitude is maximum and we have complete constructive interference. If φ = π, 3π, 5π, ; φ = π/2, 3π/2, 5π/2, ; so cos (φ) = 0. Thus the amplitude is zero and we have destructive interference. (d) If φ = π/2, we have D = 2D M cos (π/4) sin (kx ωt + π/4) = 2D M sin (kx ωt + π/4). The resultant wave has amplitude D M 2, travels toward + x, and at x = 0, t = 0, the displacement is + D M.

12 36. From the diagram the initial wavelength is 2L, and the final wavelength is 3L/2. The tension has not changed, so the velocity has not changed: v = f 1 λ 1 = f 2 λ 2 ; (294 Hz)(2L) = f 2 (3L/2), which gives f 2 = 392 Hz. L Unfingered Fingered

13 42. We find the speed of the wave from v = [F T /(m/l 0 )] 1/2 = {(520 N)/[( kg)/(0.900 m)]} 1/2 = 361 m/s. The wavelength of the fundamental for a string is λ 1 = 2L. We find the fundamental frequency from f 1 = v/λ 1 = (361 m/s)/2(0.60 m) = 300 Hz. All harmonics are present so the first overtone is the second harmonic: f 2 = (2)f 1 = (2)300 Hz = 600 Hz. The second overtone is the third harmonic: f 3 = (3)f 1 = (3)300 Hz = 900 Hz.

14 44. The hanging weight creates the tension in the string: F T = mg. The speed of the wave depends on the tension and the mass density: v = (F T /µ) 1/2 = (mg/µ) 1/2. The frequency is fixed by the vibrator, so the wavelength is λ = v/f = (1/f )(mg/µ) 1/2. With a node at each end, each loop corresponds to λ/2. (a) For one loop, we have λ 1 /2 = L, or 2L = v 1 /f = (1/f )(m 1 g/µ) 1/2 ; 2(1.40 m) = (1/60 Hz)[m 1 (9.80 m/s 2 )/( kg/m)] 1/2, which gives m 1 = 1.4 kg. (b) For two loops, we have λ 2 /2 = L/2, or L = v 2 /f = (1/f )(m 2 g/µ) 1/2 ; 1.40 m = (1/60 Hz)[m 2 (9.80 m/s 2 )/( kg/m)] 1/2, which gives m 2 = 0.35 kg. (c) For five loops, we have λ 5 /2 = L/5, or 2L/5 = v 5 /f = (1/f )(m 5 g/µ) 1/2 ; 2(1.40 m)/5 = (1/60 Hz)[m 5 (9.80 m/s 2 )/( kg/m)] 1/2, which gives m 5 = kg. The amplitude of the standing wave can be much greater than the vibrator amplitude because of the resonance built up from the reflected waves at the two ends of the string.