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 `330ms^(-1)`. End corrections may be neglected. Let `P_(0)` denote the mean pressure at any point in the pipe, and `DeltaP` the maximum amplitude of pressure variation.
(a) What 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?

Explanation

(a) The frequency of second overtone in closed pipe
`= (2 xx 2 + 1) (v)/(4L) = (5v)/(4L)`
`(5v)/(4L) = 440 rArr (5 xx 330)/(4L) = 440`
`L = (15)/(16) m`
(b) In terms of pressure, at the position of displacement node, there is pressure node and vice versa. The variation of pressure amplitude of standing pressure waves along the length of the column with `x = 0` at its open end will be
`p = Delta p_(0) sin kx = Delta p_(0) sin((2pi)/(lambda) x)`
For second overtone, `L = (5)/(4) lambda`
At middle of pipe, `x = (L)/(2) = (5 lambda)/(8)`
`p = Delta p_(0) sin((2pi)/(lambda) . (5 lambda)/(8)) = Delta p_(0) sin ((5pi)/(4))`
`= Delta p_(0) sin(pi + (pi)/(4))`
`= - Delta p_(0) sin.(pi)/(4) = - (Delta p_(0))/(sqrt(2))`
`|p| = (Delta p_(0))/sqrt(2)`

(c) At open end, `x = 0,p = 0` i.e. `Delta p_(0) = 0`
`p_(max) = p_(min) = p_(0) +- 0 = p_(0)`
(d) At closed end, `x = (5 lambda)/(4)`
`p = Delta p_(0) sin((2pi)/(lambda) . (5 lambda)/(4)) = Delta p_(0) sin((5pi)/(2)) = Delta p_(0)`
`p_(max) = p_(0) + Delta p_(0), p_(min) = p_(0) - Delta p_(0)`