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### hw14

Course: PHYS 1200, Spring 2008
School: RPI
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Word Count: 477

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PHYSICS PHYS-1200 II HOMEWORK SOLUTIONS CLASS 14. 16.06, 16.19, 16.89; Q17.01, Q33.02 16.06. y = (6.0 cm) sin [(0.020 cm )x + (4.0 s )t], and we know, y = ym sin (kx t). a) By inspection, b) c) f 2 k 2 2 0.020 cm-1 4.0 s -1 2 -1 -1 SPRING 2006 ym = 6.0 cm 100 cm f 2.0 Hz d) v = f = (2.0 Hz)(100 cm) e) The + sign in the expression tells us that the wave is going in v = 200 cm/s the x direction f) Taking the...

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PHYSICS PHYS-1200 II HOMEWORK SOLUTIONS CLASS 14. 16.06, 16.19, 16.89; Q17.01, Q33.02 16.06. y = (6.0 cm) sin [(0.020 cm )x + (4.0 s )t], and we know, y = ym sin (kx t). a) By inspection, b) c) f 2 k 2 2 0.020 cm-1 4.0 s -1 2 -1 -1 SPRING 2006 ym = 6.0 cm 100 cm f 2.0 Hz d) v = f = (2.0 Hz)(100 cm) e) The + sign in the expression tells us that the wave is going in v = 200 cm/s the x direction f) Taking the derivative of the expression for the displacement of the wave with respect to time gives the speed of a particle in the string. Then, y u y m cos( kx t ). The maximum value is ym . Then, t -1 um= ym = (4.0 s )(6.0 cm) um = 24 cm/s = 75 cm/s -1 g) y = (6.0 cm) sin [(0.020 cm-1)(3.5 cm) + (4.0 s )(0.26 s)] y = 2.0 cm 16.19. a) The amplitude can be read from the graph. b) can be read off the graph also. ym= 5.0 cm = 40 cm = 0.40 m c) v d) v 3.6 N 0.025 kg /m T , so T v 0.40 m 12 m/s 144 m 2 / s 2 1 s 30 ym 2 ym T 2 (0.05 m) 3 m/s (1 / 30) s um v 12 m/s T 0.033 s e) As in problem 16.06 (f), um 9.4 m/s f) k 2 2 0.40 m 2 (1/30) s k = 5.0 m = 1.57 m -1 -1 g) 2 T = 60 s = 188 s -1 -1 h) At x = 0, and t = 0, y = 4.0 cm. That value is given in the book, so y = ym sin (kx + t + ) becomes 4.0 cm = (5.0 cm) sin , so sin = 0.8. That is true for, = 0.93 radians, or 2.2 radians. However, the slope of the sine function is positive for = 0.93 radians and negative for = 2.2 radians. Since the graph shows a positive the slope, correct answer is: = 0.93 radians i) Since the wave is going in the negative x direction, the sign in front of The entire equation for the wave can be written: -1 -1 y = (5.0 cm) sin [(5.0 m ) x + (60 s ) t + 0.93] is negative. 16.89. a) f min c max 3.0 108 m/s 700 10-9 m 3.0 108 m/s 400 10-9 m 4.3 1014 Hz , and 7.5 1014 Hz 4.3 10 Hz 14 f max c min fvisible 7.5 10 Hz 14 b) c min f max c f min 3.0 108 m/s 1.0 m , and 300 106 Hz 3.0 108 m/s 1.5 106 Hz 2.0 102 m 1.0 m radio max 200 m c) f min c max 3.0 108 m/s 5.0 10-9 m 3.0 108 m/s 1.0 10-11 m 6.0 1016 Hz , and 3.0 1019 Hz 6.0 10 Hz 16 f max c min fX-ray 3.0 10 Hz 19 Q17.01. a) The path difference is 4.0 meters. Since the wavelength is 2.0 meters, the path difference is equal to 2.0 wavelengths. Since each wavelength difference is equivalent to a phase difference of 2 , the phase difference is 4 . b) The path difference is 3.0 meters. Since the wavelength is 2.0 meters, the path difference is equal to 1.5 wavelengths. Since each wavelength difference is equivalent to a phase difference of 2 , the phase difference is 3 . c) The first case, a path difference of 2.0 wavelengths, produces constructive interference. The second case, a path difference of 1.5 wavelengths, produces destructive interference. Q33.02. a) The wave travels in the + z direction. b) Since B is in the y direction, and wave travels in the z direction, E is in the x direction.
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