Q. An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass $M$. The piston and the cylinder have equal cross-sectional area $A$. When the piston is in equilibrium, the volume of the gas is $V_0$ and its pressure is $P_0$. The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency
Solution:
In equilibrium,
$ \, \, \, \, \, \, \, \, \, p_0A =Mg \, \, \, \, \, \, \, \, \, \, \, \, $ ...(i)
when slightly displaced downwards.
$dp=-\gamma\bigg(\frac{p_0}{V_0}\bigg)dV\bigg( As \, in \, adiabatic \, process , \frac{dp}{dV}=-\gamma \frac{p}{V}\bigg)$
$\therefore \, \, $ Restoring force,
$ \, \, \, \, \, \, \, \, \, \, \, \, F=(dp)A=-\bigg(\frac{\gamma p_0}{v_0}\bigg)(A)(Ax)$
$ \, \, \, \, \, \, \, \, \, \, F \propto -x$
Therefore, motion is simple harmonic comparing with
F = - kx we have
$ \, \, \, \, \, \, \, \, \, k=\frac{\gamma p_0A^2}{V_0}$
$\therefore \, \, \, \, \, \, f=\frac{1}{2}\pi \sqrt{\frac{k}{m}}=\frac{1}{2 \pi} \sqrt{\frac{\gamma p_0A^2}{MV_0}}$
