SM_chapter16.pdf

16 Wave Motion CHAPTER OUTLINE 16.1 16.2 16.3 16.5 16.6 ANSWERS TO QUESTIONS Propagation of a Disturbance The Traveling Wave Model The Speed of Waves on Strings Rate of Energy Transfer by Sinusoidal Waves on Strings The Linear Wave Equation *Q16.3 (i) (ii) Q16.1 As the pulse moves down the string, the particles of the string itself move side to side. Since the medium—here, the string—moves perpendicular to the direction of wave propagation, the wave is transverse by definition. Q16.2 To use a slinky to create a longitudinal wave, pull a few coils back and release. For a transverse wave, jostle the end coil side to side. Look at the coefficients of the sine and cosine functions: 2, 4, 6, 8, 8, 7. The ranking is d = e > f > c > b > a. Look at the coefficients of x. Each is the wave number, 2πⲐλ , so the smallest k goes with the largest wavelength. The ranking is d > a = b = c > e > f. (iii) Look at the coefficients of t. The absolute value of each is the angular frequency ω = 2π f. The ranking is f > e > a = b = c = d. (iv) Period is the reciprocal of frequency, so the ranking is the reverse of that in part iii: d = c = b = a > e > f. (v) From v = f λ = ω Ⲑk, we compute the absolute value of the ratio of the coefficient of t to the coefficient of x in each case. From a to f respectively the numerical speeds are 5, 5, 5, 7.5, 5, 4. The ranking is d > a = b = c = e > f. *Q16.4 From v = T , we must increase the tension by a factor of 4 to make v double. Answer (b). µ *Q16.5 Answer (b). Wave speed is inversely proportional to the square root of linear density. *Q16.6 (i) (ii) Answer (a). Higher tension makes wave speed higher. Answer (b). Greater linear density makes the wave move more slowly. Q16.7 It depends on from what the wave reflects. If reflecting from a less dense string, the reflected part of the wave will be right side up. Q16.8 Yes, among other things it depends on. The particle speed is described by vy,max = ω A = 2π fA = Here v is the speed of the wave. 2π vA . λ 427 13794_16_ch16_p427-448.indd 427 12/11/06 5:03:19 PM 428 Chapter 16 *Q16.9 (a) through (d): Yes to all. The maximum particle speed and the wave speed are related by 2π vA vy,max = ω A = 2π fA = . Thus the amplitude or the wavelength of the wave can be adjusted λ to make either vy,max or v larger. Q16.10 Since the frequency is 3 cycles per second, the period is Q16.11 1 second = 333 ms. 3 Each element of the rope must support the weight of the rope below it. The tension increases with T height. (It increases linearly, if the rope does not stretch.) Then the wave speed v = increases µ with height. *Q16.12 Answer (c). If the frequency does not change, the amplitude is increased by a factor of 2. The wave speed does not change. *Q16.13 (i) (ii) Answer a. As the wave passes from the massive string to the less massive string, the wave T speed will increase according to v = . µ Answer c. The frequency will remain unchanged. However often crests come up to the boundary they leave the boundary. (iii) Answer a. Since v = f λ , the wavelength must increase. Q16.14 Longitudinal waves depend on the compressibility of the fluid for their propagation. Transverse waves require a restoring force in response to shear strain. Fluids do not have the underlying structure to supply such a force. A fluid cannot support static shear. A viscous fluid can temporarily be put under shear, but the higher its viscosity the more quickly it converts input work into internal energy. A local vibration imposed on it is strongly damped, and not a source of wave propagation. Q16.15 Let ∆t = t s − t p represent the difference in arrival times of the two waves at a station at distance −1 ⎛1 1⎞ d = vs t s = v p t p from the hypocenter. Then d = ∆t ⎜ − ⎟ . Knowing the distance from the ⎝ vs v p ⎠ first station places the hypocenter on a sphere around it. A measurement from a second station limits it to another sphere, which intersects with the first in a circle. Data from a third non-collinear station will generally limit the possibilities to a point. Q16.16 The speed of a wave on a “massless” string would be infinite! SOLUTIONS TO PROBLEMS Section 16.1 P16.1 Propagation of a Disturbance Replace x by x − vt = x − 4.5t to get y = 13794_16_ch16_p427-448.indd 428 6 ⎡⎣( x − 4.5t )2 + 3 ⎤⎦ 12/9/06 12:46:48 PM Wave Motion 429 *P16.2 (a) (b) FIG. P16.2 The graph (b) has the same amplitude and wavelength as graph (a). it differs just by being shifted toward larger x by 2.40 m. The wave has traveled 2.40 m to the right. P16.3 (a) The longitudinal P wave travels a shorter distance and is moving faster, so it will arrive at point B first. (b) The wave that travels through the Earth must travel a distance of 2 R sin 30.0° = 2 ( 6.37 × 10 6 m ) sin 30.0° = 6.37 × 10 6 m at a speed of 7 800 m /s Therefore, it takes 6.37 × 10 6 m = 817 s 7 800 m s The wave that travels along the Earth’s surface must travel π a distance of s = Rθ = R ⎛ rad⎞ = 6.67 × 10 6 m ⎝3 ⎠ at a speed of 4 500 m /s 6.67 × 10 6 = 1 482 s 4 500 Therefore, it takes The time difference is P16.4 13794_16_ch16_p427-448.indd 429 665 s = 11.1 min The distance the waves have traveled is d = ( 7.80 km s ) t = ( 4.50 km s ) ( t + 17.3 s ) where t is the travel time for the faster wave. Then, ( 7.80 − 4.50 ) ( km s ) t = ( 4.50 km s ) (17.3 s ) or t= and the distance is d = ( 7.80 km s ) ( 23.6 s ) = 184 km . km s ) (17.3 s ) = 23.6 s ( 7.80 − 4.50 ) km s ( 4.50 12/9/06 12:46:49 PM 430 P16.5 Chapter 16 (a) (b) (c) (d) Section 16.2 *P16.6 P16.7 P16.8 Let u = 10π t − 3π x + du dx = 10π − 3π = 0 at a point of constant phase dt dt dx 10 = = 3.33 m s dt 3 The velocity is in the positive x -direction . π y ( 0.100, 0 ) = ( 0.350 m ) sin ⎛ −0.300π + ⎞ = −0.054 8 m = −5.48 cm ⎝ 4⎠ 2π k= = 3π : λ = 0.667 m ω = 2π f = 10π : f = 5.00 Hz λ π ∂y ⎛ vy = = ( 0.350 ) (10π ) cos 10π t − 3π x + ⎞ v y, max = (10π ) ( 0.350 ) = 11.0 m s ⎝ 4⎠ ∂t The Traveling Wave Model (a) a wave (b) later by TⲐ4 (c) A is 1.5 times larger (d) λ is 1.5 times larger (e) λ is 2Ⲑ3 as large 40.0 vibrations 4 425 cm = Hz v= = 42.5 cm s 30.0 s 3 10.0 s v 42.5 cm s λ= = 4 = 31.9 cm = 0.319 m f 3 Hz f = Using data from the observations, we have Therefore, P16.9 π 4 y = ( 0.020 0 m ) sin ( 2.11x − 3.62t ) in SI units A = 2.00 cm k = 2.11 rad m λ= v = fλ = f = 8.00 ⎞ v = λ f = (1.20 m ) ⎛ = 0.800 m s ⎝ 12.0 s ⎠ ω = 3.62 rad s 13794_16_ch16_p427-448.indd 430 λ = 1.20 m and 8.00 12.0 s 2π = 2.98 m k ω f = = 0.576 Hz 2π ω 2π 3.62 = = 1.72 m s 2π k 2.11 12/9/06 12:46:50 PM Wave Motion P16.10 431 v = f λ = ( 4.00 Hz ) ( 60.0 cm ) = 240 cm s = 2.40 m s *P16.11 (a) (b) (c) ω = 2π f = 2π ( 5 s −1 ) = 31.4 rad s λ= v 20 m s = = 4.00 m f 5 s −1 k= 2π 2π = = 1.57 rad m λ 4m In y = A sin ( kx − ω t + φ ) we take A = 12 cm. At x = 0 and t = 0 we have y = (12 cm ) sin φ . To make this fit y = 0, we take φ = 0. Then y = (12.0 cm ) sin ((1.57 rad m ) x − ( 31.4 rad s ) t ) (d) (e) The transverse velocity is ∂y = − Aω cos ( kx − ω t ) ∂t Its maximum magnitude is Aω = 12 cm ( 31.4 rad s ) = 3.77 m s ay = ∂v y ∂t = ∂ ( − Aω cos ( kx − ω t )) = − Aω 2 sin ( kx − ω t ) ∂t The maximum value is P16.12 2 At time t, the phase of y = (15.0 cm ) cos ( 0.157 x − 50.3t ) at coordinate x is π φ = ( 0.157 rad cm ) x − ( 50.3 rad s ) t. Since 60.0° = rad, the requirement for point B is that 3 π φ B = φ A ± rad, or (since x A = 0), 3 ( 0.157 This reduces to x B = P16.13 Aω 2 = ( 0.12 m ) ( 31.4 s −1 ) = 118 m s 2 rad cm ) x B − ( 50.3 rad s ) t = 0 − ( 50.3 rad s ) t ± π rad 3 ±π rad = ±6.67 cm . 3 ( 0.157 rad cm ) y = 0.250 sin ( 0.300 x − 40.0t ) m Compare this with the general expression y = A sin ( kx − ω t ) 13794_16_ch16_p427-448.indd 431 (a) A = 0.250 m (b) ω = 40.0 rad s (c) k = 0.300 rad m (d) λ= (e) ω 40.0 rad s ⎞ v = fλ = ⎛ ⎞ λ = ⎛ ( 20.9 m ) = 133 m s ⎝ 2π ⎠ ⎝ ⎠ 2π (f) The wave moves to the right, in +x direction . 2π 2π = = 20.9 m k 0.300 rad m 12/9/06 12:46:51 PM 432 Chapter 16 *P16.14 (a) (b) y (cm) See figure at right. 2π 2π = = 0.125 s is the time from one peak ω 50.3 to the next one. T= 10 t (s) 0 0.1 This agrees with the period found in the example in the text. 0.2 –10 FIG. P16.14 P16.15 (a) A = ymax = 8.00 cm = 0.080 0 m k= 2π 2π = = 7.85 m −1 λ ( 0.800 m ) ω = 2π f = 2π ( 3.00 ) = 6.00π rad s y = A sin ( kx + ω t ) Therefore, Or (where y ( 0, t ) = 0 at t = 0) (b) y = ( 0.080 0 ) sin ( 7.85 x + 6π t ) m In general, y = 0.080 0 sin ( 7.85 x + 6π t + φ ) Assuming y ( x, 0 ) = 0 at x = 0.100 m then we require that 0 = 0.080 0 sin ( 0.785 + φ ) or φ = −0.785 y = 0.080 0 sin ( 7.85 x + 6π t − 0.785) m Therefore, P16.16 (a) y (m) 0.2 0.1 0.0 –0.1 –0.2 t=0 0.2 0.4 x (m) FIG. P16.16(a) (b) 2π 2π = = 18.0 rad m λ 0.350 m 1 1 T= = = 0.083 3 s f 12.0 s k= ω = 2π f = 2π 12.0 s = 75.4 rad s v = f λ = (12.0 s ) ( 0.350 m ) = 4.20 m s (c) y = A sin ( kx + ω t + φ ) specializes to y = 0.200 m sin (18.0 x m + 75.4t s + φ ) at x = 0, t = 0 we require −3.00 × 10 −2 m = 0.200 m sin ( +φ ) φ = −8.63° = − 0.151 rad so 13794_16_ch16_p427-448.indd 432 y ( x , t ) = ( 0.200 m ) sin (18.0 x m + 75.4t s − 0.151 rad ) 12/9/06 12:46:52 PM Wave Motion P16.17 433 π y = ( 0.120 m ) sin ⎛ x + 4π t ⎞ ⎝8 ⎠ (a) v= dy : dt π v = ( 0.120 ) ( 4π ) cos ⎛ x + 4π t ⎞ ⎝8 ⎠ v ( 0.200 s, 1.60 m ) = −1.51 m s a= dv : dt π 2 a = ( − 0.120 m ) ( 4π ) sin ⎛ x + 4π t ⎞ ⎝8 ⎠ a ( 0.200 s, 1.60 m ) = 0 (b) π 2π : = 8 λ 2π ω = 4π = : T k= λ = 16.0 m T = 0.500 s v= P16.18 (a) λ 16.0 m = = 32.0 m s T 0.500 s y ( x , t ) = A sin ( kx + ω t + φ ) Let us write the wave function as y ( 0, 0 ) = A sin φ = 0.020 0 m dy = Aω cos φ = −2.00 m s dt 0 , 0 ω= Also, 2π 2π = = 80.0 π s T 0.025 0 s 2 ⎛ 2.00 m s ⎞ v 2 A 2 = xi2 + ⎛ i ⎞ = ( 0.020 0 m ) + ⎜ ⎝ω⎠ ⎝ 80.0 π s ⎟⎠ 2 A = 0.021 5 m (b) A sin φ 0.020 0 = = −2.51 = tan φ A cos φ −2 / 80.0π Your calculator’s answer tan −1 ( −2.51) = −1.19 rad has a negative sine and positive cosine, just the reverse of what is required. You must look beyond your calculator to find φ = π − 1.19 rad = 1.95 rad (c) v y , max = Aω = 0.021 5 m (80.0π s ) = 5.41 m s (d) λ = v xT = ( 30.0 m s ) ( 0.025 0 s ) = 0.750 m k= 2π 2π = = 8.38 m λ 0.750 m ω = 80.0π s y ( x , t ) = ( 0.021 5 m ) sin (8.38 x rad m + 80.0π t rad s + 1.95 rad ) 13794_16_ch16_p427-448.indd 433 12/9/06 12:46:53 PM 434 P16.19 Chapter 16 (a) v (1.00 m s ) = = 0.500 Hz λ 2.00 m f = ω = 2π f = 2π ( 0.500 s ) = 3.14 rad s 2π 2π = = 3.14 rad m λ 2.00 m (b) k= (c) y = A sin ( kx − ω t + φ ) becomes y = ( 0.100 m ) sin ( 3.14 x m − 3.14t s + 0 ) (d) For x = 0 the wave function requires y = ( 0.100 m ) sin ( −3.14t s ) (e) (f) y = ( 0.100 m ) sin ( 4.71 rad − 3.14 t s ) ∂y = 0.100 m ( − 3.14 s ) cos ( 3.14 x m − 3.14t s ) ∂t The cosine varies between +1 and –1, so vy = v y ≤ 0.314 m s P16.20 (a) At x = 2.00 m, y = ( 0.100 m ) sin (1.00 rad − 20.0t ) Because this disturbance varies sinusoidally in time, it describes simple harmonic motion. (b) y = ( 0.100 m ) sin ( 0.500 x − 20.0t ) = A sin ( kx − ω t ) so Section 16.3 P16.21 ω = 20.0 rad s f = ω = 3.18 Hz 2π The Speed of Waves on Strings The down and back distance is 4.00 m + 4.00 m = 8.00 m. The speed is then Now, (a) dtotal 4 (8.00 m ) T = = 40.0 m s = µ t 0.800 s 0.200 kg µ= = 5.00 × 10 −2 kg m 4.00 m v= T = µ v 2 = ( 5.00 × 10 −2 kg m ) ( 40.0 m s ) = 80.0 N 2 So P16.22 and ω = 2π f = 2π ( 500 ) = 3 140 rad s, k = ω 3 140 = = 16.0 rad m v 196 y = ( 2.00 × 10 −4 m ) sin (16.0 x − 3 140t ) (b) v = 196 m s = T 4.10 × 10 −3 kg m T = 158 N 13794_16_ch16_p427-448.indd 434 12/9/06 12:46:54 PM Wave Motion P16.23 v= T 1 350 kg ⋅ m s 2 = = 520 m s µ 5.00 × 10 −3 kg m P16.24 v= T µ 435 T = µ v 2 = ρ Av 2 = ρπ r 2 v 2 T = (8 920 kg m 3 ) (π ) ( 7.50 × 10 −4 m ) ( 200 m s ) 2 2 T = 631 N P16.25 P16.26 T = µ T = Mg is the tension; v= Then, MgL L2 = 2 m t and g= Mg = mⲐ L MgL L = is the wave speed. m t −3 Lm 1.60 m ( 4.00 × 10 kg ) 2 = 2 = 1.64 m s Mt 2 3.00 kg ( 3.61 × 10 −2 s ) The period of the pendulum is T = 2π L g Let F represent the tension in the string (to avoid confusion with the period) when the pendulum is vertical and stationary. The speed of waves in the string is then: F = µ v= Mg = mⲐ L MgL m Since it might be difficult to measure L precisely, we eliminate L = T g 2π so Tg Mg T g = 2π m 2π v= P16.27 Since µ is constant, µ = M m T2 T1 = and v22 v12 2 2 ⎛v ⎞ ⎛ 30.0 m s ⎞ T2 = ⎜ 2 ⎟ T1 = ⎜ ( 6.00 N ) = 13.5 N ⎝ 20.0 m s ⎟⎠ ⎝ v1 ⎠ P16.28 From the free-body diagram mg = 2T sin θ mg T= 2 sin θ The angle θ is found from 3 LⲐ8 3 = L Ⲑ2 4 ∴θ = 41.4° (a) v= T µ or (b) 13794_16_ch16_p427-448.indd 435 v = 60.0 = 30.4 m and cos θ = v= FIG. P16.28 ⎛ ⎞ mg 9.80 m s 2 =⎜ ⎟ m −3 2 µ sin 41.4° ⎝ 2 (8.00 × 10 kg m ) sin 41.4° ⎠ ⎛ m s⎞ v = ⎜ 30.4 m kg ⎟⎠ ⎝ m = 3.89 kg 12/9/06 12:46:55 PM 436 P16.29 Chapter 16 If the tension in the wire is T, the tensile stress is T Stress = so A The speed of transverse waves in the wire is v= T = µ T = A ( stress ) Stress A ( Stress ) = = mⲐ L mⲐ AL Stress Stress = ρ mⲐVolume where ρ is the density. The maximum velocity occurs when the stress is a maximum: vmax = f has units Hz = 1 s, so T = *P16.30 (a) with units 1 has units of seconds, s . For the other T we have T = µ v 2, f kg m 2 kg ⋅ m = 2 = N . m s2 s The first T is period of time; the second is force of tension. (b) P16.31 2.70 × 108 Pa = 185 m s 7 860 kg m 3 The total time is the sum of the two times. L µ t= =L In each wire v T Let A represent the cross-sectional area of one wire. The mass of one wire can be written both as m = ρV = ρ AL and also as m = µ L . πρd 2 4 Then we have µ = ρA = Thus, ⎛ πρd 2 ⎞ t = L⎜ ⎝ 4T ⎟⎠ For copper, ⎡ (π ) (8 920 ) (1.00 × 10 −3 )2 ⎤ ⎥ t = ( 20.0 ) ⎢ ( 4 ) (150 ) ⎥⎦ ⎢⎣ = 0.137 s For steel, ⎡ (π ) ( 7 860 ) (1.00 × 10 −3 )2 ⎤ ⎥ t = ( 30.0 ) ⎢ ( 4 ) (150 ) ⎥⎦ ⎢⎣ = 0.192 s The total time is 0.137 + 0.192 = 0.329 s 12 12 12 Section 16.5 P16.32 f = Rate of Energy Transfer by Sinusoidal Waves on Strings v 30.0 = = 60.0 Hz λ 0.500 P = 1 µω 2 13794_16_ch16_p427-448.indd 436 2 A2 v = ω = 2π f = 120π rad s 1 ⎛ 0.180 ⎞ (120π )2 ( 0.100 )2 ( 30.0 ) = 1.07 kW 2 ⎝ 3.60 ⎠ 12/9/06 12:46:56 PM Wave Motion P16.33 437 Suppose that no energy is absorbed or carried down into the water. Then a fixed amount of power is spread thinner farther away from the source. It is spread over the circumference 2π r of an expanding circle. The power-per-width across the wave front P 2π r is proportional to amplitude squared so amplitude is proportional to P 2π r P16.34 (a) T 1 ; P = µω 2 A2 v 2 µ If L is doubled, v remains constant and (b) If A is doubled and ω is halved, P ⬀ ω 2 A 2 remains constant . (c) If λ and A are doubled, the product ω 2 A 2 ⬀ T = constant; v = A2 remains constant, so P remains constant . λ2 1 If L and λ are halved, then ω 2 ⬀ 2 is quadrupled, so P is quadrupled . λ (Changing L doesn’t affect P .) (d) P16.35 A = 5.00 × 10 −2 m = µ = 4.00 × 10 −2 kg m v= Therefore, P P is constant . 1 2 2 µω A v : 2 P = 300 W T = 100 N T = 50.0 m s µ 2 ( 300 ) 2P ω 2 = _____ 2 = − 2 µ A v ( 4.00 × 10 ) ( 5.00 × 10 −2 )2 ( 50.0 ) ω = 346 rad s ω f = = 55.1 Hz 2π P16.36 µ = 30.0 g m = 30.0 × 10 −3 kg m λ = 1.50 m f = 50.0 Hz: ω = 2π f = 314 s −1 2 A = 0.150 m: A = 7.50 × 10 −2 m (a) 2π y = A sin ⎛ x − ωt ⎞ ⎝ λ ⎠ y = ( 7.50 × 10 −2 ) sin ( 4.19 x − 314t ) (b) 13794_16_ch16_p427-448.indd 437 P = FIG. P16.36 2 1 1 314 ⎞ µω 2 A 2 v = ( 30.0 × 10 −3 ) ( 314 )2 ( 7.50 × 10 −2 ) ⎛ W ⎝ 2 2 4.19 ⎠ P = 625 W 12/9/06 12:46:57 PM 438 Chapter 16 P16.37 v = fλ = (b) λ= 2π 2π = m = 7.85 m k 0.800 (c) f = 50.0 = 7.96 Hz 2π (d) P16.38 ω 2π ω 50.0 = = m s = 62.5 m s 2π k k 0.800 (a) = P 1 1 µω 2 A 2 v = (12.0 × 10 −3 ) ( 50.0 )2 ( 0.150 )2 ( 62.5) W = 21.1 W 2 2 π Comparing y = 0.35 sin ⎛ 10π t − 3π x + ⎞ with y = A sin ( kx − ω t + φ ) = A sin (ω t − kx − φ + π ) ⎝ 4⎠ we have k = (a) λ ω 10π s 3π , ω = 10π s, A = 0.35 m. Then v = f λ = 2π f = = = 3.33 m s . 2π k 3π m m The rate of energy transport is P = 1 µω 2 (b) A2 v = 1 ( 75 × 10 −3 kg m ) (10π s )2 ( 0.35 m )2 3.33 m s = 15.1 W 2 The energy per cycle is Eλ = P T = P16.39 2 1 1 2π m 2 µω 2 A 2 λ = ( 75 × 10 −3 kg m ) (10π s ) ( 0.35 m )2 = 3.02 J 2 2 3π Originally, P P P 0 = 1 2 2 µω A v 2 0 = 1 2 2 T µω A µ 2 0 1 = ω 2 A2 Tµ 2 The doubled string will have doubled mass-per-length. Presuming that we hold tension constant, it can carry power larger by 2 times. 1 2P0 = ω 2 A2 T 2 µ 2 *P16.40 As for a string wave, the rate of energy transfer is proportional to the square of the amplitude and to the speed. The rate of energy transfer stays constant because each wavefront carries constant energy and the frequency stays constant. As the speed drops the amplitude must increase. We write P = FvA2 where F is some constant. With no absorption of energy, 2 2 F vbedrock Abedrock = F vmudfill Amudfill vbedrock A = mudfill = vmudfill Abedrock 25vmudfill =5 vmudfill The amplitude increases by 5.00 times. 13794_16_ch16_p427-448.indd 438 12/9/06 12:46:58 PM Wave Motion Section 16.6 P16.41 439 The Linear Wave Equation (a) A = ( 7.00 + 3.00 ) 4.00 yields A = 40.0 (b) In order for two vectors to be equal, they must have the same magnitude and the same direction in three-dimensional space. All of their components must be equal. Thus, 7.00 ˆi + 0 ˆj + 3.00 kˆ = Aˆi + Bˆj + C kˆ requires A = 7.00, B = 0, and C = 3.00 . (c) In order for two functions to be identically equal, they must be equal for every value of every variable. They must have the same graphs. In A + B cos ( Cx + Dt + E ) = 0 + 7.00 mm cos ( 3.00 x + 4.00t + 2.00 ) the equality of average values requires that A = 0 . The equality of maximum values requires B = 7.00 mm . The equality for the wavelength or periodicity as a function of x requires C = 3.00 rad m . The equality of period requires D = 4.00 rad s , and the equality of zero-crossings requires E = 2.00 rad . P16.42 The linear wave equation is ∂2 y 1 ∂2 y = ∂x 2 v 2 ∂t 2 If y = eb( x − vt ) then ∂y = −bveb( x − vt ) ∂t ∂2 y = b 2 v 2 eb( x − vt ) ∂t 2 and ∂y = beb( x − vt ) ∂x ∂2 y = b 2 eb( x − vt ) ∂x 2 ∂2 y ∂2 y = v 2 2 , demonstrating that eb( x − vt ) is a solution. 2 ∂t ∂x Therefore, P16.43 and The linear wave equation is 1 ∂2 y ∂2 y = v 2 ∂t 2 ∂x 2 To show that y = ln [ b ( x − vt )] is a solution, we find its first and second derivatives with respect to x and t and substitute into the equation. 2 ∂y 1 = ( − bv ) ∂t b ( x − vt ) −1( −bv ) ∂2 y v2 = 2 2 = − 2 ∂t b ( x − vt ) ( x − vt ) 2 ∂y −1 = [ b ( x − vt ) ] b ∂x ∂2 y b 1 = − ( x − vt )−2 = − 2 b ( x − vt ) 2 ∂x 2 −v ) 1 ∂ 2 y so the given wave function is a solution. Then 1 ∂ y = 1 ( 2 = − 2 = 2 2 2 v ∂t v ( x − vt ) ( x − vt ) ∂x 2 2 13794_16_ch16_p427-448.indd 439 12/28/06 6:39:46 PM 440 P16.44 Chapter 16 (a) From y = x 2 + v 2t 2, ∂y evaluate = 2x ∂x ∂y = v 2 2t ∂t ∂2 y =2 ∂x 2 ∂2 y = 2v 2 ∂t 2 ∂2 y 1 ∂2 y ? = ∂t 2 v 2 ∂t 2 Does 1 By substitution, we must test 2 = 2 2v 2 and this is true, so the wave function does satisfy v the wave equation. (b) 1 1 1 1 1 1 2 2 Note ( x + vt ) + ( x − vt ) = x 2 + xvt + v 2t 2 + x 2 − xvt + v 2t 2 2 2 2 2 2 2 = x 2 + v 2t 2 as required. So 1 2 f ( x + vt ) = ( x + vt ) 2 (c) 1 2 g ( x − vt ) = ( x − vt ) 2 and y = sin x cos vt makes ∂y = cos x cos vt ∂x ∂2 y = − sin x cos vt ∂x 2 ∂y = − v sin x sin vt ∂t ∂2 y = − v 2 sin x cos vt ∂t 2 Then ∂2 y 1 ∂2 y = ∂x 2 v 2 ∂t 2 −1 2 v sin x cos vt which is true as required. v2 Note sin ( x + vt ) = sin x cos vt + cos x sin vt becomes − sin x cos vt = sin ( x − vt ) = sin x cos vt − cos x sin vt So sin x cos vt = f ( x + vt ) + g ( x − vt ) with f ( x + vt ) = 1 sin ( x + vt ) 2 and g ( x − vt ) = 1 sin ( x − vt ) 2 Additional Problems P16.45 Assume a typical distance between adjacent people ~1 m. Then the wave speed is v= ∆x 1 m ~ ~ 10 m s ∆t 0.1 s Model the stadium as a circle with a radius of order 100 m. Then, the time for one circuit around the stadium is 2 2π r 2π (10 ) T= = 63 s ~1 min ~ v 10 m s 13794_16_ch16_p427-448.indd 440 12/9/06 12:47:00 PM Wave Motion *P16.46 (a) From y = 0.150 m sin(0.8x – 50t) we compute dyⲐdt = 0.150 m (–50) cos(0.8x – 50t) and a = d2yⲐdt2 = –0.150 m (–50Ⲑs)2 sin(0.8x – 50t) Then amax = 375 m /s 2 441 For the 1-cm segment with maximum force acting on it, ΣF = ma = [12 gⲐ(100 cm)] 1 cm 375 m /s2 = 0.045 0 N (b) We find the tension in the string from v = f λ = ω Ⲑk = (50Ⲑs)Ⲑ(0.8Ⲑm) = 62.5 m /s = (TⲐµ)1Ⲑ2 T = v2µ = (62.5 m /s)2(0.012 kgⲐm) = 46.9 N. The maximum transverse force is very small compared to the tension, more than a thousand times smaller. P16.47 The equation v = λ f is a special case of speed = (cycle length)(repetition rate) Thus, v = (19.0 × 10 −3 m frame ) ( 24.0 frames s ) = 0.456 m s P16.48 (a) 0.175 m = ( 0.350 m ) sin ⎡⎣( 99.6 rad s ) t ⎤⎦ ∴ sin ⎡⎣( 99.6 rad s ) t ⎤⎦ = 0.5 The smallest two angles for which the sine function is 0.5 are 30° and 150°, i.e., 0.523 6 rad and 2.618 rad. ( 99.6 ( 99.6 rad s ) t1 = 0.523 6 rad , thus t1 = 5.26 ms rad s ) t 2 = 2.618 rad , thus t 2 = 26.3 ms ∆t ⬅ t 2 − t1 = 26.3 ms − 5.26 ms = 21.0 ms (b) P16.49 ⎛ 99.6 rad s ⎞ ω Distance traveled by the wave = ⎛ ⎞ ∆ t = ⎜ ( 21.0 × 10 −3 s ) = 1.68 m . ⎝ k⎠ ⎝ 1.25 rad m ⎟⎠ Energy is conserved as the block moves down distance x: (K + U g + Us ) top ( + ∆ E = K + Ug + Us ) bottom 1 0 + Mgx + 0 + 0 = 0 + 0 + kx 2 2 2 Mg x= k (a) T = kx = 2 Mg = 2 ( 2.00 kg ) ( 9.80 m s 2 ) = 39.2 N (b) L = L0 + x = L0 + (c) v= T TL = µ m v= 39.2 N × 0.892 m 5.0 × 10 −3 kg 2 Mg k 39.2 N L = 0.500 m + = 0.892 m 100 N m v = 83.6 m s 13794_16_ch16_p427-448.indd 441 12/9/06 12:47:01 PM 442 P16.50 Chapter 16 Mgx = 1 2 kx 2 (a) T = kx = 2 Mg (b) L = L0 + x = L0 + (c) v= *P16.51 (a) T TL = = µ m 2 Mg k 2 Mg ⎛ 2 Mg ⎞ L0 + ⎝ m k ⎠ The energy a wave crest carries is constant in the absence of absorption. Then the rate at which energy passes a stationary point, which is the power of the wave, is constant. The power is proportional to the square of the amplitude and to the wave speed. The speed decreases as the wave moves into shallower water near shore, so the amplitude must increase. (b) For the wave described, with a single direction of energy transport, the intensity is the same at the deep-water location 1 and at the place 2 with depth 9 m. To express the constant intensity we write A12 v1 = A22 v2 = A22 gd2 (1.8 m )2 200 m s = A22 ( 9.8 m s2 ) 9 m = A22 9.39 m s ⎛ 200 m s ⎞ A2 = 1.8 ⎜ ⎝ 9.39 m s ⎟⎠ 12 = 8.31 m (c) P16.52 As the water depth goes to zero, our model would predict zero speed and infinite amplitude. In fact the amplitude must be finite as the wave comes ashore. As the speed decreases the wavelength also decreases. When it becomes comparable to the water depth, or smaller, our formula gd for wave speed no longer applies. Assuming the incline to be frictionless and taking the positive x-direction to be up the incline: ∑F x 13794_16_ch16_p427-448.indd 442 = T − Mg sin θ = 0 or the tension in the string is T = Mg sin θ Mg sin θ = mⲐ L MgL sin θ m The speed of transverse waves in the string is then v= T = µ The time interval for a pulse to travel the string’s length is ∆t = L m =L = v MgL sin θ mL Mg sin θ 12/9/06 12:47:02 PM Wave Motion *P16.53 (a) In P = 12 µω 2 A2 v where v is the wave speed, the quantity ωA is the maximum particle speed vymax. We have µ = 0.5 × 10–3 kgⲐm and v = (TⲐµ)1Ⲑ2 = (20 NⲐ0.5 × 10–3 kgⲐm)1Ⲑ2 = 200 m/s P Then (b) (c) 443 = 12 (0.5 × 10–3 kgⲐm) v2ymax (200 m Ⲑs) = (0.050 0 kg/s)v2y,max The power is proportional to the square of the maximum particle speed. In time t = (3 m)Ⲑv = (3 m)Ⲑ(200 m /s) = 1.5 × 10–2 s, all the energy in a 3-m length of string goes past a point. Therefore the amount of this energy is 2 E = P t = (0.05 kgⲐs) v2y,max (0.015 s) = 7.5 × 10−4 kg v y,max The mass of this section is m3 = (0.5 × 10–3 kgⲐm)3 m = 1.5 × 10–3 kg so (1Ⲑ2)m3 = 7.5 × 10−4 kg and E = (1/2) m3v2y,max = Kmax. The string also contains potential energy. We could write its energy as Umax or as Uavg + Kavg P16.54 (d) 2 E = P t = (0.05 kgⲐs) v2y,max (6 s) = 0.300 kg v y,max v= T µv 2 and in this case T = mg; therefore, m = µ g Now v = f λ implies v = ω so that k 2 µ ω 0.250 kg m ⎡ 18π s −1 ⎤ = 14.7 kg m= ⎛ ⎞ = g⎝ k⎠ 9.80 m s 2 ⎢⎣ 0.750π m −1 ⎥⎦ 2 P16.55 Let M = mass of block, m = mass of string. For the block, ∑ F = ma implies T = mvb2 = mω 2 r r The speed of a wave on the string is then v= T = µ t= r 1 = v ω θ = ωt = P16.56 (a) µ= v= Mω 2r M = rω mⲐ r m m M m = M 0.003 2 kg = 0.084 3 rad 0.450 kg dm dx = ρA = ρA dL dx T = µ T = ρA T [ ρ ( ax + b )] = T ⎡⎣ ρ (10 −3 x + 10 −2 ) cm 2 ⎤⎦ With all SI units, v= (b) v x= 0 = v x=10.0 = 13794_16_ch16_p427-448.indd 443 T ms ⎡⎣ ρ (10 −3 x + 10 −2 )10 −4 ⎤⎦ 24.0 = 94.3 m s ⎡⎣( 2 700 ) ( 0 + 10 −2 ) (10 −4 ) ⎤⎦ 24.0 = 66.77 m s ⎡⎣( 2 700 ) (10 + 10 −2 ) (10 −4 ) ⎤⎦ −2 12/9/06 12:47:03 PM 444 P16.57 Chapter 16 v= T where µ T = µ xg, to support the weight of a length x, of rope. v = gx Therefore, But v = dx , so that dt dt = L t= and ∫ 0 P16.58 1 dx = gx g x 1 2 L = 2 0 L g mxg ⎞ At distance x from the bottom, the tension is T = ⎛ + Mg, so the wave speed is: ⎝ L ⎠ v= t (a) Then T TL MgL ⎞ dx = = xg + ⎛ = ⎝ m ⎠ dt µ m L MgL ⎞ ⎤ ⎡ t = ∫ dt = ∫ ⎢ xg + ⎛ ⎝ m ⎠ ⎥⎦ 0 0 ⎣ t= MgL ⎞ 2 ⎡⎛ ⎢ Lg + g ⎣⎝ m ⎠ 12 1 2 x= L 1 ⎡ xg + ( MgL m ) ⎤⎦ t= ⎣ 1 g 2 −1 2 dx MgL ⎞ −⎛ ⎝ m ⎠ 12 ⎤ ⎥ ⎦ When M = 0, as in the previous problem, (c) m As m → 0 we expand m + M = M ⎛ 1 + ⎞ ⎝ M⎠ (a) ( t=2 1 L⎛ M +2 m ⎜ g⎝ t≈2 L g ⎛1 m ⎞ ⎜⎝ 2 M ⎟⎠ = t=2 t=2 (b) to obtain P16.59 dx gx 12 x=0 L ⎛ m+M − M ⎞ ⎟⎠ g ⎜⎝ m L ⎛ m − 0⎞ L = 2 ⎟ ⎜ g⎝ m ⎠ g ⎛ 1 m 1 m 2 …⎞ = M ⎜1 + − + ⎟ ⎝ ⎠ 2 M 8 M2 ) M − 81 ( m 2 M 3 2 ) + … − M ⎞ ⎟ m ⎠ mL Mg The speed in the lower half of a rope of length L is the same function of distance (from the L bottom end) as the speed along the entire length of a rope of length ⎛ ⎞ . ⎝ 2⎠ L L′ Thus, the time required = 2 with L ′ = 2 g and the time required = 2 ⎛ L⎞ L = 0.707 ⎜ 2 2g ⎝ g ⎟⎠ It takes the pulse more that 70% of the total time to cover 50% of the distance. (b) By the same reasoning applied in part (a), the distance climbed in τ is given by d = gτ 2 4 L L , we find the distance climbed = . g 4 1 In half the total trip time, the pulse has climbed of the total length. 4 For τ = 13794_16_ch16_p427-448.indd 444 t = 2 12/9/06 12:47:04 PM Wave Motion P16.60 (a) ∑F y = may 2T sin θ down = θ θ Consider a short section of chain at the top of the loop. A free-body diagram is shown. Its length is s = R(2θ) and its mass is µ R2θ. In the frame of reference of the center of the loop, Newton’s second law is T µ R2θ v02 mv02 down = R R 445 2θ R T FIG. P16.60(a) For a very short section, sin θ = θ and T = µ v02 T = v0 µ (b) The wave speed is v = (c) In the frame of reference of the center of the loop, each pulse moves with equal speed clockwise and counterclockwise. v v0 v0 v v0 FIG. P16.60(c1) In the frame of reference of the ground, once pulse moves backward at speed v0 + v = 2v0 and the other forward at v0 − v = 0 . The one pulse makes two revolutions while the loop makes one revolution and the other pulse does not move around the loop. If it is generated at the six-o’clock position, it will stay at the six-o’clock position. v0 v0 v0 FIG. P16.60(c2) P16.61 T ⲐA , where T is the tension maintained in ∆ LⲐ L the wire and ∆ L is the elongation produced by this tension. Also, the mass density of the wire µ may be expressed as ρ = A The speed of transverse waves in the wire is then Young’s modulus for the wire may be written as Y = v= and the strain in the wire is Y ( ∆ L ⲐL ) T T ⲐA = = µ µⲐ A ρ ∆ L ρv 2 = Y L If the wire is aluminum and v = 100 m s, the strain is 3 3 ∆ L ( 2.70 × 10 kg m ) (100 m s ) = = 3.86 × 10 −4 7.00 × 1010 N m 2 L 2 13794_16_ch16_p427-448.indd 445 12/11/06 3:50:29 PM 446 P16.62 Chapter 16 (a) Assume the spring is originally stationary throughout, extended to have a length L much greater than its equilibrium length. We start moving one end forward with the speed v at which a wave propagates on the spring. In this way we create a single pulse of compression that moves down the length of the spring. For an increment of spring with length dx and mass dm, just as the pulse swallows it up, ∑ F = ma kdx = adm becomes or k =a dmⲐdx k dm = µ so a = µ dx dv v a= = when vi = 0 dt t But Also, L = vt, But so a= v2 L Equating the two expressions for a, we have P16.63 P16.64 (b) Using the expression from part (a) v = (a) P ( x ) = 1 µω (b) P (0) = (c) (x) _____ = e−2bx 2 2 A2 v = kL = µ k v2 = µ L kL2 = m (100 or v= kL µ N m ) ( 2.00 m ) = 31.6 m s 0.400 kg 2 ω µω 3 2 −2bx 1 µω 2 A02 e−2bx ⎛ ⎞ = A0 e ⎝ k⎠ 2 2k µω 3 2 A0 2k P P (0) v= 4 450 km = 468 km h = 130 m s 9.50 h d= v 2 (130 m s ) = = 1 730 m g ( 9.80 m s 2 ) 2 13794_16_ch16_p427-448.indd 446 12/9/06 12:47:06 PM Wave Motion P16.65 (a) µ ( x ) is a linear function, so it is of the form µ ( x ) = mx + b To have µ ( 0 ) = µ0 we require b = µ0. Then µ ( L ) = µ L = mL + µ0 so m= (b) µ L − µ0 L µ ( x) = Then 447 ( µ L − µ0 ) x + µ L 0 dx dx , the time required to move from x to x + dx is . The time required to move dt v from 0 to L is From v = L ∆t = ∫ 0 L L 1 dx dx = = v ∫0 T Ⲑ µ T µ ( x )dx ∫ 0 ⎛ ( µ L − µ0 ) x ⎞ ∫0 ⎜⎝ L + µ0 ⎟⎠ L 1 ∆t = T 12 ⎛ µ L − µ0 ⎞ ⎛ L ⎞ ⎜⎝ ⎟ dx L ⎠ ⎜⎝ µ L − µ0 ⎟⎠ ⎞ 1 ⎛ L ⎞ ⎛ ( µ L − µ0 ) x ∆t = + µ0 ⎟ ⎜ ⎟ ⎜ L T ⎝ µ L − µ0 ⎠ ⎝ ⎠ ∆t = ∆t = ∆t = 32 1 L 3 2 0 2L ( µL3 2 − µ03 2 ) 3 T ( µ L − µ0 ) 2L ( µ L − µ0 3 T ( )( µ µ L − µ0 L + µ L µ0 + µ0 )( µ L + µ0 ) ) 2 L ⎛ µ L + µ L µ0 + µ0 ⎞ µ L + µ0 ⎟⎠ 3 T ⎜⎝ ANSWERS TO EVEN PROBLEMS P16.2 See the solution. The graph (b) has the same amplitude and wavelength as graph (a). It differs just by being shifted toward larger x by 2.40 m. The wave has traveled 2.40 m to the right. P16.4 184 km P16.6 See the solution P16.8 0.800 m s P16.10 2.40 m s P16.12 ±6.67 cm P16.14 (a) see the solution (b) 0.125 s, in agreement with the example P16.16 (a) see the solution (b) 18.0 m; 83.3 ms; 75.4 rad s; 4.20 m s (c) ( 0.2 m ) sin (18 x + 75.4t − 0.151) P16.18 (a) 0.021 5 m (b) 1.95 rad (c) 5.41 m s (d) y ( x , t ) = ( 0.021 5 m ) sin (8.38 x + 80.0π t + 1.95) 13794_16_ch16_p427-448.indd 447 12/9/06 12:47:06 PM 448 Chapter 16 P16.20 (a) see the solution (b) 3.18 Hz P16.22 (a) y = ( 0.2 mm ) sin (16 x − 3 140t ) P16.24 631 N P16.26 v= P16.28 ⎛ m ⎞ (a) v = ⎜ 30.4 m s ⋅ kg ⎟⎠ ⎝ P16.30 (a) s and N P16.32 1.07 kW P16.34 (a), (b), (c) P is a constant (d) P is quadrupled P16.36 (a) y = ( 0.075 0 ) sin ( 4.19 x − 314t ) P16.38 (a) 15.1 W (b) 3.02 J P16.40 As for a string wave, the rate of energy transfer is proportional to the square of the amplitude and to the speed. The rate of energy transfer stays constant because each wavefront carries constant energy and the frequency stays constant. As the speed drops the amplitude must increase. It increases by 5.00 times. P16.42 see the solution P16.44 (a) see the solution (b) P16.46 (a) 375 m /s2 (b) 0.0450 N. This force is very small compared to the 46.9-N tension, more than a thousand times smaller. P16.48 (a) 21.0 ms (b) 1.68 m P16.50 (a) 2Mg P16.52 ∆t = P16.54 14.7 Kg P16.56 (a) v = P16.58 See the solution. P16.60 (a) µ v02 P16.62 (a) see the solution (b) 31.6 m /s P16.64 130 m s; 1.73 km 13794_16_ch16_p427-448.indd 448 Tg 2π (b) 158 N M m (b) 3.89 kg (b) The first T is period of time; the second is force of tension. (b) L0 + 2 Mg k (b) 625 W 1 1 ( x + vt )2 + ( x − vt )2 2 2 (c) (c) 1 1 sin ( x + vt ) + sin ( x − vt ) 2 2 2 Mg ⎛ 2 Mg ⎞ L0 + m ⎝ k ⎠ mL Mg sin θ T in SI units (b) 94.3 m s; 66.7 m s ρ (10 x + 10 −6 ) −7 (b) v0 (c) One travels 2 rev and the other does not move around the loop. 12/12/06 2:55:31 PM