This problem involves an ideal gas supporting a piston that executes simple harmonic motion when displaced slightly. Let's break down the concept step by step.

Step 1: Understand the Equilibrium Condition
In equilibrium, the piston is at rest. The forces acting on the piston are:

• Downward: Weight of the piston, $\mathrm{M\; g}$
• Upward: Force due to gas pressure, $P_{0}A$

Step 2: Analyze a Small Displacement
When the piston is displaced downward by a small distance $x$, the volume decreases by $\mathrm{A\; x}$ (since cross-sectional area is A). The new volume is $V_0-\mathrm{A\; x}$.

Since the system is isolated (adiabatic condition), the process is adiabatic. For an adiabatic process in an ideal gas: $P{V}^{\gamma }=\text{constant}$ where $\gamma$ is the adiabatic index.

Let the new pressure be $P$. Then: $P{(V_0-\mathrm{A\; x})}^{\gamma }=P_0{V_0}^{\gamma }$

For small $x$, we can use binomial approximation: ${(V_0-\mathrm{A\; x})}^{\gamma }\approx {V_0}^{\gamma }(1+\frac{\gamma Ax}{V_0})$ (Note: Actually, ${(1+y)}^n\approx 1+\mathrm{n\; y}$ for small $y$. Here, $y=-\frac{Ax}{V_0}$, so ${(1+y)}^{\gamma }\approx 1+\mathrm{\gamma \; y}=1-\frac{\gamma Ax}{V_0}$)

Thus, $P\approx P_0{V_0}^{\gamma }/\left[{V_0}^{\gamma }(1-\frac{\gamma Ax}{V_0})\right]=P_0(1-\frac{\gamma Ax}{V_0}){}^{-1}$ Using binomial approximation again: $P\approx P_0(1+\frac{\gamma Ax}{V_0})$

Step 3: Find the Restoring Force
The net force on the piston after displacement is: $F_{\text{net}}=PA-\mathrm{M\; g}$ But from equilibrium, $\mathrm{M\; g}=P_{0}A$, so: $F_{\text{net}}=PA-P_{0}A=A(P-P_0)$

Substitute $P$ from above: $P-P_0\approx P_0\frac{\gamma Ax}{V_0}$ So, $F_{\text{net}}\approx A·P_0\frac{\gamma Ax}{V_0}=\frac{P_0\gamma A^2}{V_0}x$

This net force is restoring (opposite to displacement) and proportional to $x$. So, it represents a simple harmonic motion with spring constant: $k=\frac{P_0\gamma A^2}{V_0}$

Step 4: Find the Frequency
The angular frequency $\omega$ for SHM is given by: $\omega =\sqrt{\frac{k}{M}}=\sqrt{\frac{P_0\gamma A^2}{V_{0}M}}$

The frequency $f$ is: $f=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{P_0\gamma A^2}{V_{0}M}}$

Comparing with the options, this matches the fourth option: $\frac{1}{2\pi }\sqrt{\frac{A^2\gamma P_0}{M{V}_0}}$

Final Answer: The piston executes simple harmonic motion with frequency $\frac{1}{2\pi }\sqrt{\frac{A^2\gamma P_0}{M{V}_0}}$

Related Topics and Formulae

Simple Harmonic Motion (SHM): A type of periodic motion where the restoring force is directly proportional to the displacement and acts in the direction opposite to that of displacement. The general form is $F=-kx$, leading to angular frequency $\omega =\sqrt{\frac{k}{m}}$ and frequency $f=\frac{\omega }{2\pi }$.

Adiabatic Process: A thermodynamic process in which no heat is exchanged with the surroundings. For an ideal gas, $P{V}^{\gamma }=\text{constant}$, where $\gamma$ is the ratio of specific heats.

Binomial Approximation: For small $x$, ${(1+x)}^n\approx 1+nx$. This is crucial for linearizing equations in physics problems involving small displacements.