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?