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Question

With a neat labelled diagram, show that all harmonics are present in an air column contained in a pipe open at both the ends. Define end correction.

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Solution

Let V be the speed of sound in air. In the simplest mode of vibrating fig. (a). The fundamental mode on first harmonic, there is a node midway between the two antinodes at the open ends. Teh distance between two consecutive antinodes is λ2, where λ is the wavelength of sound. The corresponding wavelength λ and the fundamental frequency n are
λ=2L and n=Vλ=V2L ...(1)
In the next higher mode, the first overtone there are two nodes and three antinodes [fig.(b)].
The corresponding wavelength λ1 and frequency n1 are
λ1=L and n1=Vλ1=VL=2n ...(2)
i.e., twice the fundamental. Therefore the first overtone is the second harmonic.
In the second overtone, there are three nodes and four antinodes [fig.(c)].
The corresponding wavelength λ2 and frequency n2 are
λ2=2L3 and n2=Vλ2=3V2L=3n ....(3)
or thrice the fundamental. Therefore the second overtone is the third harmonic.
Therefore, in general, the frequency of the pth overtone (P=1,2,3,...) is
nP=(P+1) ....(4)
i.e., the Pth overtone is the (P+1)th harmonic.
Equation 1,2,3 show that allowed frequencies in an air column in a pipe open at both ends are n,2n,3n,.... That is all, the harmonics are present as overtones.
Stationary waves formed in the air column of a narrow tube or pipe closed at one end have a node at the closed end and an antinode at the open end. For a pipe open at both ends, there is an antinode at each end.
However, an antinode is formed slightly beyond the rim of the pipe at the open end. The distance of the antinode from the rim is approximately 30% of the inner diameter of a cylindrical pipe. This distance must be taken into account in accurate determination of the wavelength of sound. Hence, this distance is called the end-correction.
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