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Unformatted text preview: hernandez (ejh742) oldhomework 33 Turner (58120) 1 This printout should have 14 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 points Two waves in one string are described by the relationships y 1 = A 1 cos( k 1 x 1 t ) y 2 = A 2 sin( k 2 x 2 t ) where A 1 = 4 . 2 cm, A 2 = 4 . 9 cm, k 1 = 6 cm 1 , k 2 = 5 cm 1 , 1 = 3 rad / s, 2 = 1 rad / s, y and x are in centimeters, and t is in seconds. Find the superposition of the waves at the position 1 cm and time 1 s. Correct answer: 7 . 8663 cm. Explanation: Let : x 1 = 1 cm and t 1 = 1 s . At this point we have y 1 = (4 . 2 cm) cos bracketleftBig (6 cm 1 ) (1 cm) (3 rad / s) (1 s) bracketrightBig = 4 . 15797 cm and y 2 = (4 . 9 cm) sin bracketleftBig (5 cm 1 ) (1 cm) (1 rad / s) (1 s) bracketrightBig = 3 . 70833 cm , so y 1 + y 2 = 7 . 8663 cm . 002 (part 2 of 3) 10.0 points Find the superposition of the waves y 1 + y 2 at the position 0 . 6 cm and time 0 . 7 s. Correct answer: 3 . 95105 cm. Explanation: Let : x 2 = 0 . 6 cm and t 2 = 0 . 7 s . At this point we have y 1 = (4 . 2 cm) cos bracketleftBig (6 cm 1 ) (0 . 6 cm) (3 rad / s) (0 . 7 s) bracketrightBig = 0 . 297096 cm and y 2 = (4 . 9 cm) sin bracketleftBig (5 cm 1 ) (0 . 6 cm) (1 rad / s) (0 . 7 s) bracketrightBig = 3 . 65396 cm , so y 1 + y 2 = 3 . 95105 cm . 003 (part 3 of 3) 10.0 points Find the superposition of the waves y 1 + y 2 at the position 0 . 4 cm and time 28 s. Correct answer: 4 . 81531 cm. Explanation: Let : x 3 = 0 . 4 cm and t 3 = 28 s . At this point we have y 1 = (4 . 2 cm) cos bracketleftBig (6 cm 1 ) (0 . 4 cm) (3 rad / s) ( 28 s) bracketrightBig = 0 . 0260483 cm and y 2 = (4 . 9 cm) sin bracketleftBig (5 cm 1 ) (0 . 4 cm) (1 rad / s) ( 28 s) bracketrightBig = 4 . 84135 cm , so y 1 + y 2 = 4 . 81531 cm . 004 10.0 points The distance between two successive maxima of a certain transverse wave is 1 . 22 m. Eight hernandez (ejh742) oldhomework 33 Turner (58120) 2 crests, or maxima, pass a given point along the direction of travel every 13 . 2 s. Calculate the wave speed. Correct answer: 0 . 739394 m / s. Explanation: Let : = 1 . 22 m , t = 13 . 2 s , and n = 8 . The frequency of the transverse wave is f = n t = 8 13 . 2 s = 0 . 606061 Hz . The wave speed then is v = f = (1 . 22 m)(0 . 606061 Hz) = . 739394 m / s . 005 (part 1 of 3) 10.0 points The time needed for a water wave to change from the equilibrium level to the crest is . 5741 s....
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 Spring '08
 Turner
 Physics, Work

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