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Unformatted text preview: homework 35 PAPAGEORGE, MATT Due: Apr 24 2008, 4:00 am 1 Question 1, chap 16, sect 2. part 1 of 3 10 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 = 3 . 4 cm, A 2 = 4 . 9 cm, k 1 = 6 cm 1 , k 2 = 4 cm 1 , 1 = 5 rad / s, 2 = 3 rad / s, y and x are in centimeters, and t is in seconds. Find the superposition of the waves y 1 + y 2 at the position x 1 = 1 cm and time t 1 = 1 s. Correct answer: 5 . 96024 cm (tolerance 1 %). Explanation: At this point we have y 1 = (3 . 4 cm) cos bracketleftBig (6 cm 1 ) (1 cm) (5 rad / s) (1 s) bracketrightBig = 1 . 83703 cm y 2 = (4 . 9 cm) sin bracketleftBig (4 cm 1 ) (1 cm) (3 rad / s) (1 s) bracketrightBig = 4 . 12321 cm , so y 1 + y 2 = 5 . 96024 cm . Question 2, chap 16, sect 2. part 2 of 3 10 points Find the superposition of the waves y 1 + y 2 at the position x 2 = 1 cm and time t 2 = 0 . 4 s. Correct answer: . 580946 cm (tolerance 1 %). Explanation: At this point we have y 1 = (3 . 4 cm) cos bracketleftBig (6 cm 1 ) (1 cm) (5 rad / s) (0 . 4 s) bracketrightBig = 2 . 22239 cm y 2 = (4 . 9 cm) sin bracketleftBig (4 cm 1 ) (1 cm) (3 rad / s) (0 . 4 s) bracketrightBig = 1 . 64144 cm , so y 1 + y 2 = . 580946 cm . Question 3, chap 16, sect 2. part 3 of 3 10 points Find the superposition of the waves y 1 + y 2 at the position x 3 = 0 . 6 cm and time t 3 = 61 s. Correct answer: 8 . 23271 cm (tolerance 1 %). Explanation: At this point we have y 1 = (3 . 4 cm) cos bracketleftBig (6 cm 1 ) (0 . 6 cm) (5 rad / s) (61 s) bracketrightBig = 3 . 33694 cm y 2 = (4 . 9 cm) sin bracketleftBig (4 cm 1 ) (0 . 6 cm) (3 rad / s) (61 s) bracketrightBig = 4 . 89577 cm , so y 1 + y 2 = 8 . 23271 cm . Question 4, chap 16, sect 3. part 1 of 1 10 points Consider two organ pipes. The first pipe is open at both ends and its 2 . 026 m long. The second pipe is open at one end and closed at the other end; its 3 . 09 m long. When both pipes are played together, the first overtone the lowest harmonic above the fundamental frequency of the second pipe produces beats agains the fundamental harmonic of the first pipe. What is the fre quency of these beats? Take the sound speed in air to be 336 m / s. homework 35 PAPAGEORGE, MATT Due: Apr 24 2008, 4:00 am 2 Correct answer: 1 . 36862 Hz (tolerance 1 %)....
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This note was uploaded on 03/02/2009 for the course PHY 58235 taught by Professor Kleinman during the Spring '09 term at University of Texas at Austin.
 Spring '09
 KLEINMAN
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