Question

An air chamber of volume V has a neck of cross-sectional area a into which a light ball of mass m just fits and can move up and down 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 oscillation of the ball is

• 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

Answer: (D)

$T=2\pi \sqrt{\frac{mV}{B{a}^2}}$

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 $V-△V$. Hence the pressure increases from $PtoP+△P$ . The change in volume is, $△V=ax$
The excess pressure $△P$ is related to the bulk modulus B as
$△P=B\frac{△V}{V}$
Restoring force on ball = excess pressure \times cross-sectional area
or $F=\frac{Ba}{V}△V$
or $F=-\frac{B{a}^2}{V}x$
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 ball, the time period of the SHM is
$T=2\pi \sqrt{\frac{m}{k}}orT=2\pi \sqrt{\frac{mV}{B{a}^2}}$