An ideal gas enclosed in a 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

Options

A. `(1)/(2pi)(V_(0)MP_(0))/(A^(2)gamma)`
B. `(1)/(2pi)sqrt((A^(2)gammaP_(0))/(MV_(0)))`
C. `(1)/(2pi)sqrt((MV_(0))/(AgammaP_(0)))`
D. `(1)/(2pi)(AgammaP_(0))/(V_(0)M)`

Answer

To find the frequency of the simple harmonic motion (SHM) of the piston in the ideal gas system, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a piston of mass \( M \) in a cylindrical container with an ideal gas. The piston can move freely, and when in equilibrium, the gas has a volume \( V_0 \) and pressure \( P_0 \).

2. **Displacement of the Piston**:
- When the piston is displaced slightly downwards by a distance \( x \), the volume of the gas changes. The change in volume \( dV \) due to the displacement of the piston is given by:
\[ dV = A \cdot dx \]
where \( A \) is the cross-sectional area of the piston.

3. **Change in Pressure**:
- For an ideal gas, the relationship between pressure and volume can be expressed using the adiabatic condition. The change in pressure \( dP \) can be derived from the equation:
\[ dP = -\gamma \frac{P}{V} dV \]
where \( \gamma \) is the adiabatic index. Substituting \( dV \) into this equation gives:
\[ dP = -\gamma \frac{P_0}{V_0} A \cdot dx \]

4. **Force on the Piston**:
- The increase in force \( dF \) acting on the piston due to the change in pressure is:
\[ dF = A \cdot dP = -\gamma \frac{P_0}{V_0} A^2 \cdot dx \]

5. **Equating to Hooke's Law**:
- The force can also be expressed in terms of Hooke's law for SHM:
\[ F = -k \cdot x \]
where \( k \) is the effective spring constant. From our previous equation, we can identify:
\[ k = \gamma \frac{P_0 A^2}{V_0} \]

6. **Finding the Frequency**:
- The frequency \( f \) of SHM is given by:
\[ f = \frac{1}{2\pi} \sqrt{\frac{k}{M}} \]
Substituting the expression for \( k \):
\[ f = \frac{1}{2\pi} \sqrt{\frac{\gamma P_0 A^2}{V_0 M}} \]

### Final Answer:
Thus, the frequency of the simple harmonic motion of the piston is:
\[ f = \frac{1}{2\pi} \sqrt{\frac{\gamma P_0 A^2}{V_0 M}} \]