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

Explanation

(i) Frequency of second overtone of the closed pipe
`=5((v)/(4L))=440`

`:. L=(5v)/(4xx440)m`
Substituting v = speed of sound in air `= 330 m//s`
`L=(5xx330)/(4xx440)=(15)/(16)m`
`lamda=(4L)/(5)=(4((15)/(16)))/(5)=(3)/(4)m`
(ii) Open end is displacement antinode. Therefore, it would be a pressure node or at x = 0, `Delta P=0`
Pressure amplitude at x = x
can be written as `Delta P = pm Delta P_(0) sin kx`
where `k=(2pi)/(lamda)=(2pi)/(3//4)=(8pi)/(3)m^(-1)`
Therefore, pressure amplitude at
`x=(L)/(2)=(15//16)/(2)m` or `(15//32)m` will be
`DeltaP=pmP_(0)sin((8pi)/(3))((15)/(32))=pm DeltaP_(0)sin((5pi)/(4))`
`rArr Delta P=pm (DeltaP_(0))/(sqrt(2))`
(iii) Open end is a pressure node i.e. `Delta P =0`
Hence, `P_(max)=P_(min)`= Mean pressure `(P_(0))`
(iv) Closed end is a displacement node or pressure antinode
Therefore, `P_(max)=P_(0)+DeltaP_(0) and P_(min)=P_(0)-DeltaP_(0)`.