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Unformatted text preview: Physics 214, Spring 2006 Solution to Assignment #5 by Yor Limkumnerd 1. Standing waves for sound (a) According to Figure 1, the oboe fits 2 1 / 4 wavelengths, or = 4 / 9 L , where L is the length of the oboe. From v = f , knowing the wavelength and the speed of sound in air, one can figure out the frequency of the standing wave: f = v = 9 v 4 L = 9 350 m/s 4 . 85 m = 926 Hz (1) (b) To find all the standing wave frequencies that are smaller than the one found in (a), lets work out the general formula for the frequencies of the standing wave that can fit into an openclosed pipe. Note that in the fundamental mode, the pipe can fit only a quarter of the wavelength. To get to a higher mode, one has to add a half wavelength. Therefore for the n th mode, L = n 4 + ( n 1) n 2 n = 4 L 2 n 1 . (2) With this, the frequency of the n th harmonic is f n = (2 n 1) v 4 L . (3) Figure 1 shows the n = 5 mode, therefore the frequencies of the standing wave that are less than this are found by substituting n = 1 to 4 into (3). This gives the frequencies of 103, 309, 515, 721 Hz respectively. 2. Show that ( B ) = ( B ) 2 B Let me just do it the brute force way 1 . We are asked to prove: ( B ) = ( B ) 2 B (4) We can do this by first writing B as B = fl fl fl fl fl fl fl fl x y z x y z B 1 B 2 B 3 fl fl fl fl fl fl fl fl = B 3 y B 2 z B 1 z B 3 x B 2 x B 1 y . (5) 1 There is a more elegant way of doing this using tensor analysis. Introducing it now just to solve this problem, however, may not be worth the effort. 1 Applying another curl to this gives ( B ) = fl fl fl fl fl fl fl fl fl fl x y z x y z B 3 y B 2 z B 1 z B 3 x B 2 x B 1 y fl fl fl fl fl fl fl fl fl fl = y B 2 x B 1 y  z B 1 z B 3 x z B 3 y B 2 z  x B 2 x B 1 y x B 1 z...
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This note was uploaded on 09/25/2008 for the course PHYS 2214 taught by Professor Giambattista,a during the Spring '07 term at Cornell University (Engineering School).
 Spring '07
 GIAMBATTISTA,A
 Physics

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