Show that only odd harmonics are present in the vibrations of air column in a pipe closed at one end.

Answer: In a pipe closed at one end, the closed end is always a node and the open end is always an antinode (neglecting end correction). Let the length of the air column be \(L\) and the speed of sound \(v\).

Fundamental mode: one node at the closed end and one antinode at the open end. \(L = \dfrac{\lambda_1}{4}\Rightarrow \lambda_1 = 4L\), \(n_1 = \dfrac{v}{4L}\).

Second mode (first overtone): two nodes and two antinodes. \(L = \dfrac{3\lambda_2}{4}\), \(n_2 = \dfrac{3v}{4L} = 3n_1\).

Third mode (second overtone): \(L = \dfrac{5\lambda_3}{4}\), \(n_3 = \dfrac{5v}{4L} = 5n_1\).

In general, \(L = (2p - 1)\dfrac{\lambda}{4}\), so

[Figure: three closed pipes showing the standing-wave patterns, with N at the closed end and A at the open end: \(\lambda/4\), \(3\lambda/4\) and \(5\lambda/4\) in the tube.]

Thus the frequencies are in the ratio \(n_1 : n_2 : n_3 = 1 : 3 : 5\); only odd harmonics are present and even harmonics are absent.