The air column in a pipe closed at one end is made to vibrated in its second overtone by a tuning fork of frequency `440 H_(Z)` . The speed of sound in air is `330 m//s` . End corrections may be neglected. Let `p_(o)` denotes the mean pressure at any point in the pipe, end `Delta p_(o)` the maximum amplitude of pressure variation.
(a) Find the length `L` of the air column.
(b) What is the amplitude of varitation 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 pressure at the closed end of the pipe ?

Explanation

a. The fundamental frequency of the closed organ pipe ` = v// 4l`.
In closed organ pipe only odd harmonics are present .
Second overtone of pipe ` = 5 v// 4L`
Given ` 5v//4L = 440`
On solving , we get
`L = ( 5 v)/( 4 xx 440) = ( 5 xx 330)/( 4 xx 440) = (15)/( 16) m = 0.9375 m = 93.75 cm`
b. The equation of variation of pressure amplitude at any distance `x` from the node is
`Delta P = Delta P_(0) cos kx`
Pressure variation is maximum at a node and minimum (zero) at antinode.
Distance of centre `C from N_(2) is lambda//8)`
`:. DeltaP = Delta P_(0) cos ( 2pi)/( lambda) xx (lambda)/( 8) = DeltaP_(0) (pi)/(4) = ( Delta P_(0))/( sqrt(2))`
c. At antinode , the pressure variation is minimum (zero) , therefore at antinode remains equal to `P_(0)` (always).
Therefore , at antinode `P_(max) = P_(min) = P_(0)`.