Question

Explain how overtone is produced in a closed organ pipe.

Answer 1

If one end of a pipe is closed, the wave reflected at this closed-end is 180° out of phase with the incoming wave. So, there is no displacement of the particles at the closed end. Hence, nodes are formed at the closed end and anti-nodes are formed at the open end.

No motion of particles which leads to nodes at closed-end and antinodes at the open end (fundamental mode) (N-node, A-antinode)

Consider the simplest mode of vibration of the air column called the fundamental mode. Anti-node is formed at the open end and node at the closed end. From the above figure, let L be the length of the tube and the wavelength of the wave produced. For the fundamental mode of vibration, we have,

– L = `lambda_1/4` (or)

Wave length λ₁ = 4L

The frequency of the note emitted is

`"f"_1 = "v"/lambda_1 = "v"/(4"L")`

which is called the fundamental note.

The frequencies higher than the fundamental frequency can be produced by blowing air strongly at the open end. Such frequencies are called overtones.

The second mode of vibration in open pipes is shown in the figure.

Second mode of vibration having two nodes and two anti-nodes

4L = 3λ₂

L = `(3lambda_2)/4` (or) `lambda_2 = (4"L")/3`

The frequency for this,

`"f"_2 = "v"/lambda_2 = (3"v")/(4"L") = 3 "f"_1`

is called the first overtone, since here, the frequency is three times the fundamental frequency it is called third harmonic. The Figure shows the third mode of vibration having three nodes and three anti-nodes.

Third mode of vibration having three nodes and three anti-nodes

We have, 4L = 5λ₃

L = `(5lambda_3)/4` (or) `lambda_3 = (4"L")/5`

The frequency,

`"f"_3 = "v"/lambda_3 = "5v"/"4L" = 5"f"_1`

is called the second overtone, and since n = 5 here, this is called fifth harmonic. Hence, the closed organ pipe has only odd harmonics and the frequency of the n^th harmonic is f_n = (2n+l)f₁. Hence, the frequencies of harmonics are in the ratio,

f₁ : f₂ : f₃ : f₄ = 1 : 3 : 5 : 7 : …