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 $$M_0$$. 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
[Assume the system is in space.]

We have a vertical cylinder of uniform cross-sectional area $$A$$ containing an ideal gas. A light-tight piston of mass $$M$$ can slide without friction. Because the entire arrangement is floating in space, the weight of the piston is zero, so in the equilibrium position the only force acting on the piston is produced by the gas itself. At this position the gas occupies the volume $$V_0$$ and its pressure is $$P_0$$.

Now the piston is displaced very slightly by a distance $$x$$ (positive outward) and released. During this motion the system is perfectly insulated, therefore the gas changes state adiabatically. For an adiabatic process of an ideal gas, we first state the well-known relation

$$P\,V^{\gamma}= \text{constant},$$

where $$\gamma$$ is the ratio of the two heat capacities $$C_p/C_v.$$ Immediately after the displacement the new volume of the gas is

$$V = V_0 + A\,x,$$

because the piston moves through the distance $$x$$ and every point of its face sweeps out a prism of base area $$A$$ and height $$x$$.

Substituting this new volume in the adiabatic relation, we write the new pressure $$P$$ as

$$P = P_0\left(\frac{V_0}{V}\right)^{\gamma} = P_0\left(\frac{V_0}{V_0 + A\,x}\right)^{\gamma}.$$

Since the displacement is very small, the ratio $$\dfrac{A\,x}{V_0}$$ is a small quantity. We therefore expand the bracket by the first-order binomial approximation $$(1+\varepsilon)^{-\gamma}\approx 1-\gamma\varepsilon,$$ with $$\varepsilon = \dfrac{A\,x}{V_0}.$$ Thus

$$P \;\approx\; P_0\Bigl[1 - \gamma\,\frac{A\,x}{V_0}\Bigr].$$

The change in pressure is therefore

$$\Delta P = P - P_0 = -\,\gamma\,P_0\frac{A\,x}{V_0}.$$

The force exerted by the gas on the piston is the pressure multiplied by the area of the piston. Consequently, the change in that force is

$$\Delta F = A\,\Delta P = -\,\gamma\,P_0\,\frac{A^2}{V_0}\,x.$$

So, for small displacements, the net force acting on the piston is

$$F = -\Bigl(\gamma\,P_0\,\frac{A^2}{V_0}\Bigr)\,x.$$

This expression is exactly of the simple harmonic form $$F = -k\,x$$ with an effective force constant

$$k = \gamma\,P_0\,\frac{A^2}{V_0}.$$

Newton’s second law, $$M\,\dfrac{d^2x}{dt^2}=F,$$ therefore gives

$$M\,\frac{d^2x}{dt^2} = -\,\gamma\,P_0\,\frac{A^2}{V_0}\,x.$$

Comparing with the standard SHM equation $$\dfrac{d^2x}{dt^2} + \omega^2 x = 0,$$ we recognise

$$\omega = \sqrt{\frac{\gamma\,P_0\,A^2}{M\,V_0}}.$$

The frequency $$f$$ is related to the angular frequency by $$f=\dfrac{\omega}{2\pi},$$ so

$$f = \frac{1}{2\pi}\sqrt{\frac{A^2\,\gamma\,P_0}{M\,V_0}}.$$

This expression matches Option A.

Hence, the correct answer is Option A.