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

Explanation

To solve the problem of the piston executing simple harmonic motion in an ideal gas, we can follow these steps:

### Step 1: Understand the equilibrium condition
At equilibrium, the forces acting on the piston are balanced. The pressure force exerted by the gas inside the cylinder must balance the weight of the piston and the atmospheric pressure acting on the piston.

**Equation at equilibrium:**
\[ P_0 A = mg + P_{\text{atm}} A \]

### Step 2: Displace the piston
When the piston is displaced by a small distance \( x \) from its equilibrium position, the pressure inside the gas changes. The new pressure can be expressed as:
\[ P = P_0 + \Delta P \]
where \( \Delta P \) is the change in pressure due to the displacement.

### Step 3: Apply the adiabatic condition
Since the system is isolated, the process is adiabatic. For an adiabatic process, we use the relation:
\[ P V^\gamma = \text{constant} \]
Differentiating this gives:
\[ P \, dV + V \, dP = 0 \]
From this, we can express \( dP \):
\[ dP = -\frac{P}{V} \, dV \]

### Step 4: Relate volume change to displacement
The volume change \( dV \) due to the displacement \( x \) of the piston can be expressed as:
\[ dV = A \, dx \]
Substituting this into the expression for \( dP \):
\[ dP = -\frac{P_0}{V_0} A \, dx \]

### Step 5: Set up the equation of motion
Now, we can set up the equation of motion for the piston. The net force acting on the piston when displaced is:
\[ F_{\text{net}} = P A - P_{\text{atm}} A - mg \]
Substituting the expressions we have:
\[ (P_0 + dP) A - P_{\text{atm}} A - mg = m a \]
Substituting \( dP \):
\[ (P_0 - \frac{P_0 A}{V_0} x) A - P_{\text{atm}} A - mg = m a \]

### Step 6: Simplify the equation
After substituting and simplifying, we find that:
\[ -\frac{P_0 A^2}{V_0} x = m a \]
This implies:
\[ a = -\frac{P_0 A^2}{m V_0} x \]

### Step 7: Identify the form of SHM
The equation \( a = -\omega^2 x \) indicates simple harmonic motion, where:
\[ \omega^2 = \frac{P_0 A^2}{m V_0} \]

### Step 8: Calculate the frequency
The frequency \( f \) of the oscillation is given by:
\[ f = \frac{\omega}{2\pi} = \frac{1}{2\pi} \sqrt{\frac{P_0 A^2}{m V_0}} \]

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