The air column in a pipe closed at one end is made to vibrate in its second overtone by a tuning fork of frequency 440 Hz. The speed of sound in air is 330 ms -1 . End corrections may be neglected. Let P 0 denote the mean pressure at any point in the pipe & 
P 0 the maximum amplitude of pressure variation.
(i) Find the length L of the air column.
(ii) What is the amplitude of pressure variation at the middle of the column?
(iii) What are the maximum & minimum pressures at the open end of the pipe.
(iv) What are the maximum & minimum pressures at the closed end of the pipe?
Text Solution
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[(i) L = 
m
(ii) 
(iii) P max = P min = P 0
(iv) P max = P 0 + 
P 0 , P min = P 0 - 
P 0 ]
(i) 
Frequency of second overtone of the closed pipe
= 5 
= 440 H (Given)
∴ L = 
m
Substituting V = speed of sound in air = 330 m/s
L = 
= 
m
λ = 
= 
= 
m
(ii) Open end is displacement antinode. Therefore, it would be a pressure node
Or at x = 0; Δ P = 0
Pressure amplitude at x = x, can be written as
Δ P = 
Δ P 0 sin Kx
Where K = 
= 
= 
m -1 Therefore, pressure amplitude at x (= 
= 
m) will be
Δ P = 
Δ P 0 sin 
= 
Δ P 0 sin 
Δ P = 
(iii) Open end is pressure node i.e. Δ P = 0
Hence P max = P min = Mean pressure (P 0 )
(iv) Closed end is a displacement node or pressure antinode.
Therefore P max = P 0 + Δ P 0
And P min = P 0 – Δ P 0
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