Question

The pth overtone of an organ pipe open at both ends has a frequency n. When one end of the pipe is closed, the qth overtone has a frequency N. Show that N = (2q + 1)n/2(p+1)

Answer 1

Let L be the length of an organ pipe and v be the speed of sound in air.

When the pipe has both its ends open, the frequency of the pth overtone, ignoring the end correction, is (p + 1) \(\frac v{2L}\),

∴ n = (p + 1) \(\frac v{2L}\) 

OR

\(\frac v{2L}\) = \(\frac n{p+1}\)..........(1)

When one end of the pipe is closed, the frequency of the qth overtone of a pipe of length L and closed at one end is (2q + 1) \(\frac v{4L}\)

∴ N = (2q + 1) \(\frac v{4L}\) = \(\frac{2q+1}2\).\(\frac v{2L}\)

∴ N = \(\frac{(2q + 1)n}{2(p+1)}\).......[from Eq (1)]

which is the required expression.