Show that all harmonics are present in the case of an air column vibrating in a pipe open at both the ends.

Answer:

In a pipe open at both ends, the air at both ends is free to vibrate, so stationary waves have an antinode at each end. Let the length of the pipe be \(L\) (end correction neglected) and the speed of sound \(v\).

[Figure: three open pipes showing the first three modes: (a) antinodes at both ends with one node in the middle; (b) antinodes at both ends with two nodes; (c) antinodes at both ends with three nodes.]

Fundamental mode: one node at the centre, \(L=\dfrac{\lambda_1}{2}\Rightarrow\lambda_1=2L\), \(n_1=\dfrac{v}{2L}\).

Second mode (first overtone): two nodes, \(L=\lambda_2\), \(n_2=\dfrac{v}{L}=2\cdot\dfrac{v}{2L}=2n_1\).

Third mode (second overtone): three nodes, \(L=\dfrac{3\lambda_3}{2}\), \(n_3=\dfrac{3v}{2L}=3n_1\).

In general, for the \(p\)th mode, \(L=\dfrac{p\lambda_p}{2}\) and \(n_p=\dfrac{pv}{2L}=p\,n_1\), \(p=1,2,3,\dots\)

The frequencies are \(n_1:n_2:n_3:\dots=1:2:3:\dots\). So both odd and even multiples of the fundamental are produced: all harmonics are present in a pipe open at both ends.