To find out mode of vibration of air in a resonating pipe

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SUMMARY

The discussion centers on the mode of vibration of air in a cylindrical pipe closed at one end, resonating at a frequency of 864 Hz. The calculated first resonating length is 9.83 cm, while the maximum resonating length is 29.51 cm, indicating that the mode of vibration is 2. However, due to the nature of closed-end pipes, only odd harmonics are allowed, making the second resonance correspond to the third harmonic (n=3). This clarifies that the second mode of vibration is indeed the third harmonic, as a node cannot exist at the open end of the pipe.

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Homework Statement

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.
But the answer given is 3! Please help...
 
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Given a pipe with one open end and one closed, can you have n=2? and, more generally, can you have an even n?
 
how many quarter-waves are in this pipe?
 
bacon said:
Given a pipe with one open end and one closed, can you have n=2? and, more generally, can you have an even n?

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

lightgrav said:
how many quarter-waves are in this pipe?

l2 = 3λ/4
So, there are 3 quarter waves, right?
 
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.
 
bacon said:
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.

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 :smile: :smile:
 
Last edited:
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:

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