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# ch16 - Chapter 16 1 Let y1 = 2.0 mm(corresponding to time...

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685 Chapter 16 1. Let y 1 = 2.0 mm (corresponding to time t 1 ) and y 2 = –2.0 mm (corresponding to time t 2 ). Then we find kx + 600 t 1 + = sin 1 (2.0/5.0) and kx + 600 t 2 + = sin 1 (–2.0/5.0) . Subtracting equations gives 600( t 1 t 2 ) = sin 1 (2.0/5.0) – sin 1 (–2.0/5.0). Thus we find t 1 t 2 = 0.0014 s (or 1.4 ms). 2. (a) The speed of the wave is the distance divided by the required time. Thus, 853 seats 21.87 seats/s 22 seats/s 39 s v . (b) The width w is equal to the distance the wave has moved during the average time required by a spectator to stand and then sit. Thus, (21.87 seats/s)(1.8 s) 39 seats w vt . 3. (a) The angular wave number is 1 2 2 4.19m . 1.50m k (b) The speed of the wave is 1.50m 110rad s 26.3m s. 2 2 v f   4. The distance d between the beetle and the scorpion is related to the transverse speed t v and longitudinal speed v as t t d v t v t   where t t and t are the arrival times of the wave in the transverse and longitudinal directions, respectively. With 50 m/s t v and 150 m/s v , we have

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CHAPTER 16 686 150 m/s 3.0 50 m/s t t t v t v . Thus, if 3 3 3.0 2.0 4.0 10 s 2.0 10 s , t t t t t t t t   then 3 (150 m/s)(2.0 10 s) 0.30 m 30 cm. d v t   5. (a) The motion from maximum displacement to zero is one-fourth of a cycle. One- fourth of a period is 0.135 s, so the period is T = 4(0.135 s) = 0.540 s. (b) The frequency is the reciprocal of the period: 1 1 1.85Hz. 0.540s f T (c) A sinusoidal wave travels one wavelength in one period: 1.40m 2.59m s. 0.540s v T 6. The slope that they are plotting is the physical slope of the sinusoidal waveshape (not to be confused with the more abstract “slope” of its time development; the physical slope is an x -derivative, whereas the more abstract “slope” would be the t -derivative). Thus, where the figure shows a maximum slope equal to 0.2 (with no unit), it refers to the maximum of the following function: sin( ) cos( ) m m dy d y kx t y k kx t dx dx . The problem additionally gives t = 0, which we can substitute into the above expression if desired. In any case, the maximum of the above expression is y m k , where 2 2 15.7 rad/m 0.40 m k . Therefore, setting y m k equal to 0.20 allows us to solve for the amplitude y m . We find 0.20 0.0127 m 1.3 cm 15.7 rad/m m y . 7. (a) Recalling from Chapter 12 the simple harmonic motion relation u m = y m , we have
687 16 400rad/s. 0.040 Since = 2 f , we obtain f = 64 Hz. (b) Using v = f , we find = 70/64 = 1.1 m. (c) The amplitude of the transverse displacement is 2 4.0 cm 4.0 10 m. m y (d) The wave number is k = 2 / = 5.7 rad/m. (e) The angular frequency, as obtained in part (a), is 2 16/ 0.040 4.0 10 rad/s. (f) The function describing the wave can be written as 0.040sin 5.7 400 y x t where distances are in meters and time is in seconds. We adjust the phase constant to satisfy the condition y = 0.040 at x = t = 0. Therefore, sin

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