Question

Show that only odd harmonics are present in an air column vibrating in a pipe closed at one end.

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

Answer 1

Consider a narrow cylindrical pipe of length l that is closed on one end. When sound waves travel down the air column in a cylindrical pipe that is closed at one end, they are reflected with a phase reversal at the closed end and not at the open end. Under appropriate conditions, interference between incident and reflected waves produces stationary waves in the air column. To account for the end correction e at the open end, the resonating length of the air column is L = l + e.

Let v represent the speed of sound in air. The simplest mode of vibration [Fig. (a)] contains a node at the closed end and an antinode at the open end. The distance between a node and the next antinode is 'λ/4', where λ represents the wavelength of sound. The wavelength (λ) and frequency (n) are:

λ = 4 L 

n = `v/λ`

= `v/(4 L)`

= `v/(4(l + e))`   ...(1)

This gives the fundamental frequency of vibration, which is known as the fundamental mode or first harmonic.

Figure (b) shows the formation of two nodes and two antinodes in the first overtone, the higher mode of vibration. The equivalent wavelength (λ₁) and frequency (n₁) are:

`λ_1 = (4 L)/3`

∴ n = `v/λ_1`

= `(3v)/(4L)`

= `(3v)/(4(l + e))`

= 3n    ...(2)

As a result, the first overtone’s frequency is three times that of the fundamental frequency, indicating that it is the third harmonic.

The second overtone consists of three nodes and three antinodes (Fig. c). The wavelength (λ₂) and frequency (n₂) are:

`λ_2 = (4 L)/5`

`n_2 = v/λ_2`

= `(5v)/(4L)`

= `(5v)/(4(l + e))`

= 5n   ...(3)

which is the fifth harmonic.

Therefore, in general, the frequency of the p^th overtone (p = 1, 2, 3, ...) is:

n_p = (2p + 1) n   ...(4) 

i.e. the pth overtone is the (2p + 1)^th harmonic.

Equations (1), (2), and (3) show that the ratio of the frequencies produced is:

n : n₁ : n₂ =  1 : 3 : 5

Since all resultant frequencies are odd multiples of the fundamental frequency, it is mathematically proven that only odd harmonics are present.