Question

An organ pipe of length L is open at one end and closed at other end. The air column in the pipe is vibrating in second overtone. The minimum distance from the open end where the pressure amplitude is half of the maximum value is​

Answer 1

Hello Student,

Please find the answer to your question

(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