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### hw16

Course: PHYS 1200, Spring 2008
School: RPI
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Word Count: 350

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PHYSICS PHYS-1200 II HOMEWORK SOLUTIONS CLASS 16. 16.42, 16.84, 17.41, 17.111; Q17.09, Q17.10 16.42. SPRING 2006 The string will be flat twice each period, so T = 2(0.50 s) = 1.0 s. Then, = vT = (10 cm/s)(1.0 s) = 10 cm = 0.10 m 16.84. Since the two waves are given as cosines, they can be combined with the identity, cos + cos = 2 cos( + )cos( ) Then the formula for the standing wave is, y = y1 + y2 = [2ym...

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PHYSICS PHYS-1200 II HOMEWORK SOLUTIONS CLASS 16. 16.42, 16.84, 17.41, 17.111; Q17.09, Q17.10 16.42. SPRING 2006 The string will be flat twice each period, so T = 2(0.50 s) = 1.0 s. Then, = vT = (10 cm/s)(1.0 s) = 10 cm = 0.10 m 16.84. Since the two waves are given as cosines, they can be combined with the identity, cos + cos = 2 cos( + )cos( ) Then the formula for the standing wave is, y = y1 + y2 = [2ym cos kx] cos t y = [2(0.050) cos x] cos 4 t = [(0.10) cos x] cos 4 t a) For a node the amplitude must be zero. That occurs when cos x = 0. The smallest positive value of x that gives that result is x = , so x = 0.50 m b) The velocity of a particle of the string at x is given by, dy = [(0.10) cos x][ 4sin 4 t] u dt This will be zero at t = 0, , etc., so, the first will occur at: c) The second time that u will be zero is: d) The third time that u will be zero is: t=0 t = 0.25 s t = 0.50 s 17.41. The well acts the same as a closed pipe, so its fundamental (lowest) frequency will be, v B 1 B 1 B f . Then, since v , we have , f so L 4L 4L 4f L 1 1.33 105 Pa 4(7.00 Hz) 1.10 kg /m3 L 12.4 m 17.111. a) f f 1.2 1 v 1 . Therefore, 2 2 1.2 2L 2L f1 1 1 1 1 2L f2 = 482 Hz 1 2 2L f 2 1.2 f1 1.2 (440 Hz) 1 2 L2 b) f2 f1 1 2 L1 L1 L1 3 L2 2 2 L1 3 f2 3 3 f1 (440 Hz) 2 2 f2 = 660 Hz Q17.09. The higher the component of the velocity of the source toward the detector, the higher the frequency that is detected. Therefore, the order will be: 1>4>3>2 Q17.10. a) The merry-go-round with the greatest linear speed will produce the greatest frequency. Therefore, 3 is the fastest, while 1 and 2 are tied for second. 3 > 1 = 2. b) The greatest will result in the shortest period for the observed variation. Therefore, 1 has the greatest , while 2 and 3 are tied for second. 1 > 2 = 3. c) v = r, so r = v/3 has the largest v and smallest , so it has the largest r. 2 has a lower v but the same , so it has a smaller r. 1 has the same v as 2, but a greater , so its r is smaller than 2's. The ranking is: 3 > 2 > 1.
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