Question

An air chamber of volume V has a neck of cross – sectional area a into which a light ball of mass m can move without friction. The diameter of the ball is equal to that of the neck of the chamber. The ball is pressed down a little and released. If the bulk modulus of air in B, the time period of the resulting oscillation of the ball is given by

• A. $T=2\pi \sqrt{\frac{B{a}^2}{mV}}$
• B. $T=2\pi \sqrt{\frac{BV}{m{a}^2}}$
• C. $T=2\pi \sqrt{\frac{mB}{V{a}^2}}$
• D. $T=2\pi \sqrt{\frac{mV}{B{a}^2}}$

Answer 1

The correct option is D $T=2\pi \sqrt{\frac{mV}{B{a}^2}}$

Let P be the pressure of air in the chamber. When the ball is pressed down a distance x, the volume of air decreases from V to Sat V − $\Delta$ V. Hence the pressure increases from P to P+ $\Delta$ P. The change in volume is $\Delta V=ax$
The excess pressure $\Delta$ P is related to the bulk modulus B
As $\Delta$ P =$-B\frac{\Delta V}{V}$
Now restoring force on ball = excess pressure $\times$ cross-sectional area
or, $F=\frac{Ba}{V}\Delta V$
or,$F=-\frac{B{a}^2}{V}x(\because \Delta V=ax)$
or, $F=-kx$
where $k=-\frac{B{a}^2}{V}$
i.e., $F\propto -x$
Hence the motion of the ball is simple harmonic. If m is the mass of the ball, the time period of the SHM is

$T=2\pi \sqrt{\frac{m}{k}}$
Or, $T=2\pi \sqrt{\frac{mV}{B{a}^2}}$