# Homework Help: To find out mode of vibration of air in a resonating pipe

1. Jan 18, 2012

### opencircuit

The problem statement, all variables and given/known data

A cyclindrical pipe of length 29.5 cm closed at one end resonates with a tuning fork of frequency 864 Hz. Then mode of vibration of air in the pipe is(Velocity of sound in air=340m/s):

The attempt at a solution

V=Velocity, n=Frequency

If l1 is the first resonating length
then l1 = V/4n = 340/(4X864) = 0.0983m = 9.83 cm

Also, l2 is second resonating length
l2 = 3V/4n = 3 X 9.83 = 29.51 cm

This is the maximum resonating length possible in the pipe. Therefore the mode of vibration is 2.

2. Jan 18, 2012

### bacon

Given a pipe with one open end and one closed, can you have n=2? and, more generally, can you have an even n?

3. Jan 18, 2012

### lightgrav

how many quarter-waves are in this pipe?

4. Jan 18, 2012

### opencircuit

No the frequency(n) is not 2. The mode of vibration is 2. So there's a difference.

l2 = 3λ/4
So, there are 3 quarter waves, right?

5. Jan 18, 2012

### bacon

The mode of vibration is the "harmonic". The second allowed harmonic is not necessarily the second harmonic. In this case, the second allowed harmonic is the "third harmonic" corresponding to n= 3. If n=2 were allowed, the second harmonic, there would be a node at the open end of the pipe, which can't exist.
I hope this makes sense.

6. Jan 19, 2012

### opencircuit

Ok I get it. Pipe closed at one end has only odd harmonics. And the second resonance actually occurs at the third harmonic(i.e. the third mode of vibration).Thanks

Last edited: Jan 19, 2012
7. Jan 19, 2012

### opencircuit

Hey second resonance does occur at the third harmonic but it is still the second mode of vibration( harmonic is not necessarily same as the mode of vibration)

I got this reference in a book:
Second normal mode of vibration : n2 = v/λ2 = 3v/4L = 3n1
(pipe closed at one end)

I think the answer given is wrong....

Last edited: Jan 19, 2012