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?

Answer

(i) `L = (15)/(16)m` , (ii) `(DeltaP_(0))/(sqrt(2))` , (iii)
`P_(max) = P_(min) = P_(0)` , (iv) `P_(max) = P_(0) + DeltaP_(0), P_(min) = P_(0) - DeltaP_(0)`

Explanation

Frequency of second overtone of the closed pipe
`= 5((V)/(4L)) = 440 Hz` (Given)
`L = (5V)/(4 xx 440)m`
Substituting `V =` speed of sound in air `= 330 m//s`
`L = (5 xx 330)/(4 xx 440) = (15)/(16) m`
`lambda = (4L)/(5) = (4(15//16))/(5) = (3)/(4)`
(b) Open end is displacement antinode. Therefore. If would be a pressure node
or at `x = 0, Delta P = 0`
Pressure amplitude at `x = x`, can be written as
`Delta P = +- DeltaP_(0) sin Kx`
where `K = (2pi)/(lambda) = (2pi)/(3//4) = (8pi)/(3) m`
Therfore, prssure amplitude at `x (=(L)/(2) = (15//16)/(2)m)` will be
`DeltaP = +- DeltaP_(0) sin ((8pi)/(3)) ((15)/(32))`
`= +- DeltaP_(0) sin ((5pi)/(4))`
`DeltaP +- (DeltaP_(0))/(sqrt(2))`
(c) Open end is pressure node i.e. `DeltaP = 0`
Hence `P_(max) = P_(min) =` Mean pressure `(P_(0))`
(d) closed end is a displacement node or pressure antinode.
Therefore , `P_(max) = P_(0) + DeltaP_(0)`
and , `P_(min) = P_(0) - DeltaP_(0)`