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. $\frac{1}{2\pi }\frac{A_{\gamma }{p}_0}{V_{0}M}$
• B. $\frac{2}{2\pi }\frac{V_{0}M{P}_0}{A^2\gamma }$
• C. $\frac{2}{2\pi }\sqrt{\frac{A^2\gamma P_0}{M{V}_0}}$
• D. $\frac{1}{2\pi }\sqrt{\frac{M{V}_0}{A_{\gamma }{P}_0}}$

Answer 1

Answer: (C)

In equilibrium,
$p_{0}A=Mg$ ...(i)
when slightly displaced downwards.
$dp=-\gamma \left(\frac{p_0}{V_0}\right)dV(Asinadiabaticprocess,\frac{dp}{dV}=-\gamma \frac{p}{V})$
$\therefore$ Restoring force,
$F=(dp)A=-\left(\frac{\gamma p_0}{v_0}\right)(A)(Ax)$
$F\propto -x$
Therefore, motion is simple harmonic comparing with
F = - kx we have
$k=\frac{\gamma p_0{A}^2}{V_0}$
$\therefore f=\frac{1}{2}\pi \sqrt{\frac{k}{m}}=\frac{1}{2\pi }\sqrt{\frac{\gamma p_0{A}^2}{M{V}_0}}$