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## 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

## Homework Equations

λ = nL/4

f

_{n}= 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

f

_{extended}= (1)(343 m/s)/4(1.205m)

f

_{extended}= 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?