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 is B, the time period of the resulting oscillations 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 the ball is depressed by $x$ units. Due to this, the volume of the air chamber would decrease.

Now the decrease in the volume $\Delta V=ax$ where $a$ is the cross-sectional area.

The pressure in the air chamber would also increase due to decrease in its volume.

Let the increase in pressure be $\Delta P$.

Now bulk modulus of air $B=\frac{Stress}{Strain}=-\frac{\Delta P}{\frac{\Delta V}{V}}$

$⟹\Delta P=-\frac{-axB}{V}$

$\therefore \frac{F}{a}=\frac{-axB}{V}$ where $F=m{a}^{\prime }$, $a^{\prime }$ is the acceleration of the ball

$⟹a^{\prime }=-\frac{a^{2}B}{mV}x$ which represents $SHM$

$\therefore w^2=\frac{a^{2}B}{mV}$

Hence time period of the oscillation $T=\frac{2\pi }{w}=2\pi \sqrt{\frac{mV}{B{a}^2}}$