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Unformatted text preview: Chapter 17 3. (a) The time for the sound to travel from the kicker to a spectator is given by d/v, where d is the distance and v is the speed of sound. The time for light to travel the same distance is given by d/c, where c is the speed of light. The delay between seeing and hearing the kick is Δt = (d/v) – (d/c). The speed of light is so much greater than the speed of sound that the delay can be approximated by Δt = d/v. This means d = v Δt. The distance from the kicker to spectator A is dA = v ΔtA = (343 m/s)(0.23 s) = 79 m. (b) The distance from the kicker to spectator B is dB = v ΔtB = (343 m/s)(0.12 s) = 41 m. (c) Lines from the kicker to each spectator and from one spectator to the other form a right triangle with the line joining the spectators as the hypotenuse, so the distance between the spectators is 2 2 D = d A + dB = ( 79 m ) 2 + ( 41m ) = 89 m . 2 10. (a) The amplitude of a sinusoidal wave is the numerical coefficient of the sine (or cosine) function: pm = 1.50 Pa. (b) We identify k = 0.9π and ω = 315π (in SI units), which leads to f = ω/2π = 158 Hz. (c) We also obtain λ = 2π/k = 2.22 m. (d) The speed of the wave is v = ω/k = 350 m/s. 16. Let the separation between the point and the two sources (labeled 1 and 2) be x1 and x2, respectively. Then the phase difference is ⎛x ⎞ ⎛x ⎞ 2π ( x1 − x2 ) 2π (4.40 m − 4.00 m) Δφ = φ1 − φ2 = 2π ⎜ 1 + ft ⎟ − 2π ⎜ 2 + ft ⎟ = = = 4.12 rad. λ (330 m/s) / 540 Hz ⎝λ ⎠ ⎝λ ⎠ 19. Let L1 be the distance from the closer speaker to the listener. The distance from the other speaker to the listener is L2 =
2 L1 + d 2 , where d is the distance between the speakers. The phase difference at the listener is φ = 2π(L2 – L1)/λ, where λ is the wavelength. For a minimum in intensity at the listener, φ = (2n + 1)π, where n is an integer. Thus, λ = 2(L2 – L1)/(2n + 1). The frequency is f= v = λ2 ( (2n + 1)v
2 L1 + d 2 − L1 )(
2 = (2n + 1)(343m/s) (3.75m) 2 + (2.00 m) 2 − 3.75m ) = (2n + 1)(343Hz). Now 20,000/343 = 58.3, so 2n + 1 must range from 0 to 57 for the frequency to be in the audible range. This means n ranges from 0 to 28. (a) The lowest frequency that gives minimum signal is (n = 0) f min,1 = 343 Hz. (b) The second lowest frequency is (n = 1) f min,2 = [2(1) + 1]343 Hz = 1029 Hz = 3 f min,1. Thus, the factor is 3. (c) The third lowest frequency is (n=2) f min,3 = [2(2) + 1]343 Hz = 1715 Hz = 5 f min,1. Thus, the factor is 5. For a maximum in intensity at the listener, φ = 2nπ, where n is any positive integer. Thus λ = (1/ n ) ( 2 L1 + d 2 − L1 and ) f= v = λ nv L + d − L1
2 1 2 = n(343m/s) (3.75m) + (2.00 m) 2 − 3.75m
2 = n(686 Hz). Since 20,000/686 = 29.2, n must be in the range from 1 to 29 for the frequency to be audible. (d) The lowest frequency that gives maximum signal is (n =1) f max,1 = 686 Hz. (e) The second lowest frequency is (n = 2) f max,2 = 2(686 Hz) = 1372 Hz = 2 f max,1. Thus, the factor is 2. (f) The third lowest frequency is (n = 3) f max,3 = 3(686 Hz) = 2058 Hz = 3 f max,1. Thus, the factor is 3. 24. (a) Since intensity is power divided by area, and for an isotropic source the area may be written A = 4πr2 (the area of a sphere), then we have I= P 1.0 W = = 0.080 W/m 2 . 2 A 4π(1.0 m) (b) This calculation may be done exactly as shown in part (a) (but with r = 2.5 m instead of r = 1.0 m), or it may be done by setting up a ratio. We illustrate the latter approach. Thus, I ′ P / 4π( r′) 2 ⎛ r ⎞ = =⎜ ⎟ I P / 4 πr 2 ⎝ r′ ⎠
2 leads to I′ = (0.080 W/m2)(1.0/2.5)2 = 0.013 W/m2. 28. (a) The intensity is given by I = P/4πr2 when the source is “point‐like.” Therefore, at r = 3.00 m, I= 1.00 × 10−6 W = 8.84 × 10−9 W/m 2 . 4π(3.00 m) 2 (b) The sound level there is
⎛ 8.84 × 10−9 W/m 2 ⎞ = 39.5dB. β = 10 log ⎜ −12 2⎟ ⎝ 1.00 × 10 W/m ⎠ 42. At the beginning of the exercises and problems section in the textbook, we are told to assume vsound = 343 m/s unless told otherwise. The second harmonic of pipe A is found from Eq. 17–39 with n = 2 and L = LA, and the third harmonic of pipe B is found from Eq. 17–41 with n = 3 and L = LB. Since these frequencies are equal, we have 2vsound 3vsound 3 = ⇒ LB = LA . 2LA 4 LB 4 (a) Since the fundamental frequency for pipe A is 300 Hz, we immediately know that the second harmonic has f = 2(300 Hz) = 600 Hz. Using this, Eq. 17–39 gives LA = (2)(343 m/s)/2(600 s−1) = 0.572 m. 3 (b) The length of pipe B is LB = 4 LA = 0.429 m. 51. Let the period be T. Then the beat frequency is 1/ T − 440 Hz = 4.00 beats/s. Therefore, T = 2.25 × 10–3 s. The string that is “too tightly stretched” has the higher tension and thus the higher (fundamental) frequency. 55. In the general Doppler shift equation, the trooper’s speed is the source speed and the speeder’s speed is the detector’s speed. The Doppler effect formula, Eq. 17–47, and its accompanying rule for choosing ± signs, are discussed in §17‐10. Using that notation, we have v = 343 m/s, vD = vS = 160 km/h = (160000 m)/(3600 s) = 44.4 m/s, and f = 500 Hz. Thus, ⎛ 343 m/s − 44.4 m/s ⎞ f ′ = (500 Hz) ⎜ ⎟ = 500 Hz ⇒ Δf = 0. ⎝ 343 m/s − 44.4 m/s ⎠ 61. We denote the speed of the French submarine by u1 and that of the U.S. sub by u2. (a) The frequency as detected by the U.S. sub is ⎛ v + u2 ⎞ ⎛ 5470 km/h + 70.00 km/h ⎞ 3 3 f1′ = f1 ⎜ ⎟ = (1.000 ×10 Hz) ⎜ ⎟ = 1.022 × 10 Hz. 5470 km/h − 50.00 km/h ⎠ v − u1 ⎠ ⎝ ⎝ (b) If the French sub were stationary, the frequency of the reflected wave would be fr = f1(v+u2)/(v – u2). Since the French sub is moving towards the reflected signal with speed u1, then (v + u1 )(v + u2 ) (1.000 × 103 Hz)(5470 + 50.00)(5470 + 70.00) ⎛ v + u1 ⎞ f r′ = f r ⎜ = f1 = ⎟ v (v − u2 ) (5470)(5470 − 70.00) ⎝v⎠ = 1.045 × 103 Hz. 69. (a) The half angle θ of the Mach cone is given by sin θ = v/vS, where v is the speed of sound and vS is the speed of the plane. Since vS = 1.5v, sin θ = v/1.5v = 1/1.5. This means θ = 42°. (b) Let h be the altitude of the plane and suppose the Mach cone intersects Earth's surface a distance d behind the plane. The situation is shown on the diagram below, with P indicating the plane and O indicating the observer. The cone angle is related to h and d by tan θ = h/d, so d = h/tan θ. The shock wave reaches O in the time the plane takes to fly the distance d: t= d h 5000 m = = = 11 s . v v tan θ 1.5(331 m/s)tan42° 71. (a) Incorporating a term (λ/2) to account for the phase shift upon reflection, then the path difference for the waves (when they come back together) is L2 + (2d)2 − L + λ/2 = Δ(path) . Setting this equal to the condition needed to destructive interference (λ/2, 3λ/2, 5λ/2 …) leads to d = 0, 2.10 m, … Since the problem explicitly excludes the d = 0 possibility, then our answer is d = 2.10 m. ...
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 Spring '10
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