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Unformatted text preview: x = 0 . 20 m, x = 0 . 60 m, and x = 1 . 0 m. (b) The waves in both pipes have the same wavespeed (sound in air) and frequency, so the standing waves in both pipes have the same wavelength (0 . 80 m). Therefore, using Eq. 1838 for pipe B , we Fnd L = 3 λ/ 4 = 0 . 60 m. (c) Using v = 343 m/s, we Fnd f 3 = v/λ = 429 Hz. Now, we Fnd the fundamental resonant frequency by dividing by the harmonic number, f 1 = f 3 / 3 = 143 Hz....
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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.
 Fall '08
 SPRUNGER
 Physics

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