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) In caseof closed organ pipe asfundamental frequency is (v/4L) and only odd harmonics are present, second overtone will mean fifth harmonic and so
`f = (5v)/(4L)` = 440 Hz
`L = (5 xx 330)/(4 xx 440) = (15)/(16)`
(b) In terms of pressure as at the position of displacement antinode there is pressure node and vice-versa, the variation of pressureamplitude of standingpressurewaves alongthe length ofthe column with x = 0 at its open end will be
`p = Deltap_(0) sin kx = Deltap_(0) sin ((2pi)/(lambda)x)` Now as for second overtone `L = (5/4)lambda`, so at the middle.
`x = (L)/(2) = (5)/(8)lambda`
`p = Deltap_(0) sin" (2pi)/(lambda)((5)/(8)lambda) = Deltap_(0) sin((5)/(4)pi)`
`|p| = Deltap_(0) xx (1)/(sqrt(2)) = (Deltap_(0))/(sqrt(2))`
(C) For free end as x = 0, p = 0, i.e the amlitude of pressure wave is zero (as it is a node), so `p_(max) = p_(min) = p_(0) pm 0 = p_(0)`
(d) For closed end `x + (5/4)lambda` so the amplitude of pressure wave `|p| = |Deltap_(0)sin"(2pi)/(lambda)((5)/4lambda)| = Deltap_(0)`
Thus maximum and minimum pressure are given as
`p_(max) = p_(0) + Deltap_(0)` and `P_(min) = p_(0) - Deltap_(0)`