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Unformatted text preview: portillo (gdp347) oldhomework 38 Turner (58185) 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 = 2 cm, A 2 = 4 . 5 cm, k 1 = 4 cm 1 , k 2 = 6 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 2 s. Correct answer: 4 . 2379 cm. Explanation: Let : x 1 = 1 cm and t 1 = 2 s . At this point we have y 1 = (2 cm) cos bracketleftBig (4 cm 1 ) (1 cm) (3 rad / s) (2 s) bracketrightBig = . 832294 cm and y 2 = (4 . 5 cm) sin bracketleftBig (6 cm 1 ) (1 cm) (1 rad / s) (2 s) bracketrightBig = 3 . 40561 cm , so y 1 + y 2 = 4 . 2379 cm . 002 (part 2 of 3) 10.0 points Find the superposition of the waves y 1 + y 2 at the position 2 cm and time 0 . 4 s. Correct answer: 1 . 96393 cm. Explanation: Let : x 2 = 2 cm and t 2 = 0 . 4 s . At this point we have y 1 = (2 cm) cos bracketleftBig (4 cm 1 ) (2 cm) (3 rad / s) (0 . 4 s) bracketrightBig = 1 . 73879 cm and y 2 = (4 . 5 cm) sin bracketleftBig (6 cm 1 ) (2 cm) (1 rad / s) (0 . 4 s) bracketrightBig = 3 . 70273 cm , so y 1 + y 2 = 1 . 96393 cm . 003 (part 3 of 3) 10.0 points Find the superposition of the waves y 1 + y 2 at the position 0 . 6 cm and time 70 s. Correct answer: 3 . 71245 cm. Explanation: Let : x 3 = 0 . 6 cm and t 3 = 70 s . At this point we have y 1 = (2 cm) cos bracketleftBig (4 cm 1 ) (0 . 6 cm) (3 rad / s) ( 70 s) bracketrightBig = 0 . 671678 cm and y 2 = (4 . 5 cm) sin bracketleftBig (6 cm 1 ) (0 . 6 cm) (1 rad / s) ( 70 s) bracketrightBig = 4 . 38412 cm , so y 1 + y 2 = 3 . 71245 cm . 004 10.0 points The distance between two successive maxima of a certain transverse wave is 1 . 26 m. Eight crests, or maxima, pass a given point along the direction of travel every 11 s. Calculate the wave speed. Correct answer: 0 . 916364 m / s. portillo (gdp347) oldhomework 38 Turner (58185) 2 Explanation: Let : = 1 . 26 m , t = 11 s , and n = 8 . The frequency of the transverse wave is f = n t = 8 11 s = 0 . 727273 Hz . The wave speed then is v = f = (1 . 26 m)(0 . 727273 Hz) = . 916364 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 . 2602 s....
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This note was uploaded on 01/25/2012 for the course CHEM 302 taught by Professor Mccord during the Spring '10 term at University of Texas at Austin.
 Spring '10
 McCord

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