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# hw14 - PHYS-1200 PHYSICS II HOMEWORK SOLUTIONS CLASS 14...

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PHYS-1200 PHYSICS II SPRING 2005 HOMEWORK SOLUTIONS CLASS 14. 16.06, 16.19, 16.22, 16.89; Q17.01, Q33.02 16.06. y = (6.0 cm) sin [(0.020 π cm -1 ) x + (4.0 π s -1 ) t ], and we know, y = y m sin ( kx ± ϖ t ). a) By inspection, y m = 6.0 cm ¯¯¯¯¯¯¯¯¯¯ Hz 0 . 2 2 s 0 . 4 2 c) cm 100 cm 020 . 0 2 2 b) 1 - 1 - = = = = = = f f k π λ d) v = f = (2.0 Hz)(100 cm) v = 200 cm/s e) The + sign in the expression tells us that the wave is going in 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, u y t y kx t y m m = = ± cos( ). The maximum value is . Then, u m = y m = (4.0 π s -1 )(6.0 cm) u m = 24 π cm/s = 75 cm/s ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ g) y = (6.0 cm) sin [(0.020 π cm -1 )(3.5 cm) + (4.0 π s -1 )(0.26 s)] y = – 2.0 cm 16.19. a) The amplitude can be read from the graph. y m = 5.0 cm ¯¯¯¯¯¯¯¯¯¯ b) can be read off the graph also. = 40 cm = 0.40 m ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ s / m 4 . 9 s / m 3 m) (0.05 s ) 30 / 1 ( 2 2 (f), 16.06

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hw14 - PHYS-1200 PHYSICS II HOMEWORK SOLUTIONS CLASS 14...

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