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Question
Show that only odd harmonics are present as overtones in the case of an air column vibrating in a pipe closed at one end.
Solution
The stationary waves in the air column, in this case, are subject to two boundary conditions that there must be a node at the closed end and an antinode at the open end. In what follows, we shall ignore the end correction.
Let v be the speed of sound in air. In the simplest mode of vibration, as shown in the figure, there is a node at the closed end and an antinode at the open end. The distance between a node and a constructive antinode is `lambda/4`where `lambda` is the wavelength of sound. The corresponding wavelength `lambda` and frequency n are
`lambda = 4L " and " n = v/lambda = v/(4L)` ....(1)
This gives the fundamental frequency of vibration and the mode of vibration is called the fundamental mode or first harmonic.
In the next higher mode of vibration, the first overtune, two nodes and two antinodes are formed as shown in the figure. The corresponding wavelength `lambda_1` and frequency `lambda_1` are `lambda_2` are
`lambda_1 = (4L)/3` and `n_1 = v/lambda_1 = (3v)/(4L) = 3n` .....(2).
Therefore, the frequency in the first overtone is three times the fundamental frequency, i.e., the first overtone is the third hamonic.
In the second overtone, three nodes and three antinodes are formed as shown in the figure. The corresponding wavelength `lambda_2`and frequency `n_2` are `lambda_2 = (4L)/5` and `n_2 = v/lambda_2 = (5v)/(4L) = 5n` ...(3)
which is the fifth harmonic.
Therefore, in general, the frequency of the pth overtone (p = 1, 2, 3, ......) is
np = (2p + 1)n ......(4)
i.e., the pth overtone is the (2p + 1) the harmonic
Equations (1), (2) and (3) show that allowed frequencies in an air column are a pipe closed at one end and n, 3n, 5n, ...... That is, only odd harmonics are present as overtones.
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