An air chamber of volume V has a neck area of cross section A into which a ball of mass m just fits and can move up and down without any friction, figure. Show that when the ball is pressed down a little and released, it executes SHM. Obtain an expression for the time period of oscillations assuming pressure volume variations of air to be isothermal.

Options

A. `T=2pisqrt((Ba^(2))/(mV))`
B. `T=2pisqrt((BV)/(ma^(2)))`
C. `T=2pisqrt((mB)/(Va^(2)))`
D. `T=2pisqrt((mV)/(Ba^(2)))`

Answer

D

Explanation

The situation is as shown in the figure. Let P be pressure of air in the chamber. When the ball is pressed down a distance x, the volume of air decreases from V to say `V-DeltaV`. Hence the pressure increases from P to `P+DeltaP`. the change in volume is `DeltaV=ax`
The excess pressure `DeltaP` is related to the bulk modulus B as
`DeltaP=-B(DeltaV)/(V)`
Restoring force on ball=excess pressure`xx`cross-sectional
or `F=-(Ba)/(V)DeltaV` or `F=-(Ba^(2))/(V)x" "(becauseDeltaV=ax)`
or `F=-kx,` where `k=(Ba^(2))/(V)" i.e., "Fprop-x`
Hence, the motion of the ball is simple harmonic. if m is the ball, the time period of the SHM is
`T=2pisqrt((m)/(k))" or "T=2pisqrt((mV)/(Ba^(2)))`