oldhw38 - oldhomework 38 Turner (58185) This print-out...

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This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – fnd 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 - ω 1 t ) y 2 = A 2 sin( k 2 x - ω 2 t ) where A 1 = 4 . 8 cm, A 2 = 3 . 4 cm, k 1 = 4 cm 1 , k 2 = 2 cm 1 , ω 1 = 1 rad / s, ω 2 = 3 rad / s, y and x are in centimeters, and t is in seconds. ±ind the superposition oF the waves at the position 2 cm and time 0 . 7 s. Correct answer: 5 . 74259 cm. Explanation: Let : x 1 = 2 cm and t 1 = 0 . 7 s . At this point we have y 1 = (4 . 8 cm) cos b (4 cm 1 ) (2 cm) - (1 rad / s) (0 . 7 s) B = 2 . 52517 cm and y 2 = (3 . 4 cm) sin b (2 cm 1 ) (2 cm) - (3 rad / s) (0 . 7 s) B = 3 . 21742 cm , so y 1 + y 2 = 5 . 74259 cm . 002 (part 2 oF 3) 10.0 points ±ind the superposition oF the waves y 1 + y 2 at the position 1 cm and time 0 . 8 s. Correct answer: - 6 . 11584 cm. Explanation: Let : x 2 = 1 cm and t 2 = 0 . 8 s . At this point we have y 1 = (4 . 8 cm) cos b (4 cm 1 ) (1 cm) - (1 rad / s) (0 . 8 s) B = - 4 . 79181 cm and y 2 = (3 . 4 cm) sin b (2 cm 1 ) (1 cm) - (3 rad / s) (0 . 8 s) B = - 1 . 32402 cm , so y 1 + y 2 = - 6 . 11584 cm . 003 (part 3 oF 3) 10.0 points ±ind the superposition oF the waves y 1 + y 2 at the position 0 . 7 cm and time 21 s. Correct answer: 7 . 02905 cm. Explanation: Let : x 3 = 0 . 7 cm and t 3 = 21 s . At this point we have y 1 = (4 . 8 cm) cos b (4 cm 1 ) (0 . 7 cm) - (1 rad / s) (21 s) B = 3 . 82249 cm and y 2 = (3 . 4 cm) sin b (2 cm 1 ) (0 . 7 cm) - (3 rad / s) (21 s) B = 3 . 20656 cm , so y 1 + y 2 = 7 . 02905 cm . 004 10.0 points The distance between two successive maxima oF a certain transverse wave is 1 . 86 m. Eight crests, or maxima, pass a given point along the direction oF travel every 17 . 4 s. Calculate the wave speed. Correct answer: 0 . 855172 m / s. oldhomework 38 – Turner – (58185) 1
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hinojosa (jlh3938) – oldhomework 38 – Turner – (58185) 2 Explanation: Let : λ = 1 . 86 m , t = 17 . 4 s , and n = 8 . The frequency of the transverse wave is f = n t = 8 17 . 4 s = 0 . 45977 Hz . The wave speed then is v = λ f = (1 . 86 m)(0 . 45977 Hz) = 0 . 855172 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 0 . 1775 s.
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oldhw38 - oldhomework 38 Turner (58185) This print-out...

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