# 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 300 m s -1. End corrections may be neglected. Let P0 denote the mean pressure at any point in the pipe, and ∆P0 the maximum amplitude of pressure variation.(a) Find the length L of the air column. (b) What is the amplitude of pressure variation at the middle of the column?(c) What are the maximum and minimum pressures at the open end of the pipe ?(d) What are the maximum and minimum pressures at the closed end of the pipe ?

10 years ago
Hello Student,
(a) For second overtone as shown,
5ℷ/4 = ℓ ∴ ℷ = 4ℓ/5
Also, v = vℷ
=> 330 = 440 x 4ℓ/5 => ℓ = 15/16 m.
(b) KEY CONCEPT : At any position x, the pressure is given by
∆P = ∆P0 cos kx cos ω t
Here amplitude A = ∆P0 cos kx = ∆P0 cos 2π/ℷ x
For x = 15/2 x 16 = 15/32 m (mid point)
Amplitude = ∆P0 cos [2π/(330 / 440) x 15/32] = ∆P0/√2
(c) At open end of pipe, pressure is always same i.e. equal to mean pressure
∵ ∆P = 0, Pmax = Pmin = P0
(d) At the closed end :
Maximum Pressure = P0 + ∆P0
Minimum Pressure = P0 - ∆P0

Thanks