Daily_002.SOLN

# Daily_002.SOLN - n = 1250 Hz then clearly corresponds to n...

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© 2011 E. Szarmes PHYS 274 DAILY ASSIGNMENT 2 — DUE FRIDAY, JAN 21/11 PROF. SZARMES Q1S.2 For the fundamental mode with two ends fixed, we have where L = 25 in = 0.635 m and μ = 0.2 g/m = 0.0002 kg/m. a) If f 1 = 329 Hz, then we require a tension of F = (2L f 1 ) 2 μ = 34.92 N. b) Note that with L and μ fixed, we have f 1 , and so Q1S.7 For an organ pipe closed at one end and open at the other, the resonant frequencies are given by where the speed of sound is v = 343 m/s and, in this problem, L = 0.343 m. The fundamental pitch (frequency) is then The normal mode whose frequency is f
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Unformatted text preview: n = 1250 Hz then clearly corresponds to n = 5, since Therefore, there are 5 quarter-wavelengths within length L, and the positions of the nodes are as shown (note that λ 5 = v/f 5 = [343 m/s]/[1250 Hz] = 0.2744 m, and λ 5 /4 = 0.0686 m): f 1 = v 2L (1) = 1 2L F μ F f 1 new f 1 old = F new F old , yielding F new F old = f 1 new f 1 old 2 = 392 Hz 329 Hz 2 = 1.420 f n = n odd v 4L , f 1 = 1 343 m/s 4 (0.343 m) = 250 Hz . f n = n odd f 1 = n odd 250 Hz = 1250 Hz for n odd = 5 ....
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## This note was uploaded on 03/07/2011 for the course PHYS 274 taught by Professor Ericszarmes during the Spring '11 term at University of Hawaii, Manoa.

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