Question

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

• A. $\frac{1}{2 \pi} \frac{A_{\gamma}p_0}{V_0M}$
• B. $\frac{2}{2\pi}\frac{V_0MP_0}{A^2 \gamma}$
• C. $\frac{2}{2\pi}\sqrt{\frac{A^2\gamma P_0}{MV_0}}$
• D. $\frac{1}{2 \pi} \sqrt{\frac{MV_0}{A_{\gamma}P_0}}$

Answer 1

Answer: (C)

$\frac{2}{2\pi}\sqrt{\frac{A^2\gamma P_0}{MV_0}}$

Answer 2

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,
$\, \, \, \, \, \, \, \, \, \, \, \, F=(dp)A=-\bigg(\frac{\gamma p_0}{v_0}\bigg)(A)(Ax)$
$\, \, \, \, \, \, \, \, \, \, F ? -x$
Therefore, motion is simple harmonic comparing with
F = - kx we have
$\, \, \, \, \, \, \, \, \, k=\frac{\gamma p_0A^2}{V_0}$
$\therefore \, \, \, \, \, \, f=\frac{1}{2}\pi \sqrt{\frac{k}{m}}=\frac{1}{2 \pi} \sqrt{\frac{\gamma p_0A^2}{MV_0}}$