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?

Options

A. A-R,B-S,C-P,D-S
B. A-R,B-P,C-P,D-S
C. A-R,B-P,C-Q,D-Q
D. A-R,B-P,C-Q,D-S

Answer

C

Explanation

A. The fundamental frequency of the closed organ pipe=`V//4l`. In closed organ pipe only odd harmonics are present.
Second overtone of pipe`=5v//4L`
Given `5v//4L=440`
On solving, we get `L=(5v)/(4xx440)=(5xx330)/(4xx440)=(15)/(16)m=0.9375m=93.75cm`
B. The equation of variation of presure amplitude at any distance x from the node is
`DeltaP=DeltaP_(0)coskx`
Pressure variation is maximum at a node and minimum (zero) at antinode
Distance of centre C from `N_(2)` is `(lamda)/(8)`
`DeltaP=DeltaP_(0)"cos"(2pi)/(lamda)xx(lamda)/(8)=DeltaP_(0)(pi)/(4)=(DeltaP_(0))/(sqrt(2))`

C & D : At antinode, the pressure variation is minimum (zero), therefore at antinode pressure remains equal to
`P_(0)` (Always).
Therefore, at antinode `P_("max")=P_("min")=P_(0)`.