# Fundamental Frequency of Two Pipe Organs

by Northbysouth
Tags: frequency, fundamental, organs, pipe
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 P: 248 1. The problem statement, all variables and given/known data Two organ pipes, open at one end but closed at the other, are each 1.18 m long. One is now lengthened by 2.50 cm 2. Relevant equations λ = nL/4 fn = nv/4L v = λF 3. The attempt at a solution Here's what I tried First I tried finding the fundamental frequency when their lengths were equal f = (1)(343 m/s)/4(1.18m) f = 72.66949153 Hz I'm assuming that v = 343 m/s. It does not say that this is the case in the problem. Then I tried finding the frequency of the pipe with the extension fextended = (1)(343 m/s)/4(1.205m) fextended = 71.16182573 Hz Having found these two frequencies I then took of the average of them which gave me 71.916 Hz. Unsurprisingly this didn't work. Any suggestions?
 HW Helper P: 4,433 Pipe need not be resonating in the fundamental mode. So take lambda =(2n +1)L/4 and proceed.
 P: 2 If you still need help for this problem, try using this equation fBeat = fa-fb Solve for frequency using f=(nv)/(4L) where fa is the fundamental frequency for the pipe at its original length and fb is the fundamental frequency for the pipe when it is extended. And v=344m/s (speed of sound in air)
 P: 248 Fundamental Frequency of Two Pipe Organs Sorry it's been so long since I've replied, its been a busy week. But yes you're right f_beat = f_a - f_b So I found that if I take f_a to be f_a = (1)(343 m/s)/4(1.18m) f_a = 72.66949153 Hz Then the pipe with the increased length f_b = (1)(343 m/s)/4(1.205 m) f_b = 71.16182573 Hz Then f_beat = 72.66949153 Hz - 71.16182573 Hz f_beat = 1.507 Hz Rounded to 3 sig figs, 1.51 Hz is the correct answer.
 HW Helper P: 4,433 The problem statement is not complete. What is required in the problem?
 P: 248 You're right, it is missing a part; I don't know how I managed that. Sorry to waste your time. The missing part is: a) Find the frequency of the beat they produce when playing together in their fundamental.

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