Question

explain the vibrations in a closed organ pipe and show that the frequency of third harmonic is 5V/4L?

Answer 1

Answer:

If we use y for the wavelength.

The first harmonic occurs at y/4 since the closed end is a node and

the open end is an anti-node.   Call this N-A

The next harmonic occurs at N-A-N-A   since the distance between two anti-nodes is 1/2 y and the open end must be an anti-node. So for the second harmonic you have L = y/4 + y/2 = 3 y / 4

Note that for successive anti-nodes, 1 is a rarefaction and 1 is a compression

so that for a full wavelength you must have A-N-A-N-A. Another way of saying this is that the distance between a node and anti-node is 1/4 wavelength.

Now we said the second harmonic occurred at N-A-N-A so the next harmonic (the third) will occur at N-A-N-A-N-A which is L = 5 y /4

Since the speed of sound V = f * y and y = 4 L /5

you have f = V / y = V / (4 L / 5) = 5 V / (4 L)