# Beat frequency of organ pipes

1. Feb 2, 2006

### FishieKissie06

A friend in another city tells you that she has a pair of organ pipes, one open at both ends, the other open at one end only. In addition, she has determined that the beat frequency caused by the second-lowest frequency of each pipe is equal to the beat frequency caused by the ninth-lowest frequency of each pipe. Her challenge to you is to calculate the length of the organ pipe that is open at both ends, given that the length of the other pipe is 1.10 m. Note that there are two possible answers to this question. List them both, in the order indicated below.

so in other words fbeat = f1 - f2 where.... fbeat 2= fbeat 9.... where 2f1 - 2f2 = 9f1 - 9f2 ... and i know the length of the closed end pipe which is 1.10 how am i suppose to consider even solving this or where I can get the frequency of a closed end pipe at n=2....

2. Feb 2, 2006

### lightgrav

Draw the wavelengths that resonate in the two types of pipes ...
the calculate the frequencies that these resonances have.

One will be the odd numbers times the fundamental, the other will be the integers times the fundamental (all the even quarter-waves "fit")

So the second-lowest frequencies are 2*f_open and 3*f_closed .
These beat the same as 9*f_open and 17*f_closed .
Re-write this to get f_closed in terms of f_open ,
which you can compute since v = 340 m/s = lambda * f .

3. Feb 2, 2006

### SoccaCrazy24

ok well im looking at this problem and i see confused... and im trying to work it out myself... so what you are saying is that... 2f_open and 1.5f_open beat the same as 9f_open and 8.5f_open?