# Frequencies of standing wave in pipe

1. Apr 13, 2013

### susdu

1. The problem statement, all variables and given/known data

A speaker is placed near the open end of a 0.5m pipe (closed on the other end).
what is the speed of sound if the pipe resonances at 600Hz for n=2?

2. Relevant equations

fn=(2n+1)*V/4L
Or?
fn=(2n-1)*V/4L

3. The attempt at a solution

I saw both versions of the formulas above, which one is correct?

2. Apr 13, 2013

### TSny

Both formulas are correct. The first one assumes that the possible values of n are 0, 1, 2, 3, .... The second formula assumes the values of n are 1, 2, 3, .... So both formulas will produce the same set of resonant frequencies.

However, since the formulas produce a different frequency for the same n, you would need to know which formula the problem wants you to use. It's probably the same formula that was used in class or in your textbook.

3. Apr 13, 2013

### susdu

Thanks for the clarification!

4. Apr 13, 2013

### cepheid

Staff Emeritus
Well, the bottom version of the formula obviously doesn't work for n = 0, whereas the top one does, so I'm assuming the two versions are just dependent on where you start counting from. They both generate identical harmonics if you let n be 0, 1, 2, 3, ... for the top one and 1, 2, 3, ... for the bottom one.

EDIT: I was sitting there too long with my compose window open while doing something else and got beaten to the punch!

5. Apr 13, 2013

### susdu

Thanks anyway cepheid.