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hw 35 - homework 35 HETTIGER CHRISTOF Due 4:00 am Question...

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homework 35 – HETTIGER, CHRISTOF – 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 - ω 1 t ) y 2 = A 2 sin( k 2 x - ω 2 t ) where A 1 = 4 . 3 cm, A 2 = 1 . 4 cm, k 1 = 7 cm 1 , k 2 = 3 cm 1 , ω 1 = 3 rad / s, ω 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: - 2 . 81067 cm (tolerance ± 1 %). Explanation: At this point we have y 1 = (4 . 3 cm) cos bracketleftBig (7 cm 1 ) (1 cm) - (3 rad / s) (1 s) bracketrightBig = - 2 . 81067 cm y 2 = (1 . 4 cm) sin bracketleftBig (3 cm 1 ) (1 cm) - (3 rad / s) (1 s) bracketrightBig = 0 cm , so y 1 + y 2 = - 2 . 81067 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 = 0 . 6 cm and time t 2 = 0 . 5 s. Correct answer: - 3 . 47378 cm (tolerance ± 1 %). Explanation: At this point we have y 1 = (4 . 3 cm) cos bracketleftBig (7 cm 1 ) (0 . 6 cm) - (3 rad / s) (0 . 5 s) bracketrightBig = - 3 . 88751 cm y 2 = (1 . 4 cm) sin bracketleftBig (3 cm 1 ) (0 . 6 cm) - (3 rad / s) (0 . 5 s) bracketrightBig = 0 . 413728 cm , so y 1 + y 2 = - 3 . 47378 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 . 7 cm and time t 3 = 77 s. Correct answer: 3 . 68917 cm (tolerance ± 1 %). Explanation: At this point we have y 1 = (4 . 3 cm) cos bracketleftBig (7 cm 1 ) (0 . 7 cm) - (3 rad / s) (77 s) bracketrightBig = 4 . 28075 cm y 2 = (1 . 4 cm) sin bracketleftBig (3 cm 1 ) (0 . 7 cm) - (3 rad / s) (77 s) bracketrightBig = - 0 . 591578 cm , so y 1 + y 2 = 3 . 68917 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 it’s 1 . 657 m long. The second pipe is open at one end and closed at the other end; it’s 2 . 465 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 339 m / s.
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homework 35 – HETTIGER, CHRISTOF – Due: Apr 24 2008, 4:00 am 2 Correct answer: 0 . 850715 Hz (tolerance ± 1 %). Explanation: The first pipe is open at both ends, so a resonating standing wave in it has pressure nodes at both ends. This implies L o = λ 2 , 2 λ 2 , 3 λ 2 , 4 λ 2 , . . .
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