P18_040

# P18_040 - x = 0 . 20 m, x = 0 . 60 m, and x = 1 . 0 m. (b)...

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40. We observe that “third lowest . .. frequency” corresponds to harmonic number n = 3 for a pipe open at both ends. Also, “second lowest . .. frequency” corresponds to harmonic number n = 3 for a pipe closed at one end. (a) Since λ =2 L/n for pipe A ,whe re L =1 . 2m ,then λ =0 . 80 m for this mode. The change from node to antinode requires a distance of λ/ 4 so that every increment of 0 . 20 m along the x axis
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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. 18-38 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.

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