P18_031

# P18_031 - 2 L A = 3 v sound 4 L B = ⇒ L B = 3 4 L A(a...

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31. At the beginning of the exercises and problems section in the textbook, we are told to assume v sound = 343 m/s unless told otherwise. The second harmonic of pipe A is found from Eq. 18-39 with n =2and L = L A , and the third harmonic of pipe B is found from Eq. 18-41 with n =3and L = L B . Since these frequencies are equal, we have 2 v sound
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Unformatted text preview: 2 L A = 3 v sound 4 L B = ⇒ L B = 3 4 L A . (a) Since the fundamental frequency for pipe A is 300 Hz, we immediately know that the second harmonic has f = 2(300) = 600 Hz. Using this, Eq. 18-39 gives L A = (2)(343) / 2(600) = 0 . 572 m. (b) The length of pipe B is L B = 3 4 L A = 0 . 429 m....
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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