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 ?

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)`