# Fundamental Frequency of Two Pipe Organs

## Homework Statement

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

λ = nL/4

fn = nv/4L

v = λF

## 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?

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rl.bhat
Homework Helper
Pipe need not be resonating in the fundamental mode. So take lambda =(2n +1)L/4 and proceed.

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)

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.

rl.bhat
Homework Helper
The problem statement is not complete. What is required in the problem?

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.