An ideal gas is enclosed in a vertical cylindrical container and supports a freely moving piston of mass M. The piston and the cylinder have an equal cross-sectional area A. Atmospheric pressure is `p_(0)` and when the piston is in equilibrium, the volume of the gas is `V_(0)`. The piston is now displaced slightly from its equilibrium position. Assuming that the system, is completely isolated from, its surroundings, what is the frequency of oscillation.

Options

A. `f=(1)/(2pi)sqrt((gamma(p_(0)A^(2)+MgA))/(V_(0)M))`
B. `f=(1)/(2pi)sqrt((1)/(gamma)((p_(0)A^(2)+MgA))/(V_(0)M))`
C. `f=(1)/(2pi)sqrt(((p_(0)A^(2)+MgA))/(V_(0)M))`
D. `f=(1)/(2pi)sqrt((A(p_(0)A^(2)+MgA))/(V_(0)M))`

Answer

To find the frequency of oscillation of the piston in the given system, we can follow these steps:

### Step 1: Understand the equilibrium condition
At equilibrium, the forces acting on the piston are balanced. The upward force due to the gas pressure (P1) and the atmospheric pressure (P0) must equal the downward force due to the weight of the piston (Mg).

\[ P_1 A = P_0 A + Mg \]

### Step 2: Displace the piston
When the piston is displaced by a small distance \(x\), the pressure inside the cylinder changes. The new pressure can be expressed as:

\[ P = P_1 + \Delta P \]

where \(\Delta P\) is the change in pressure due to the displacement.

### Step 3: Apply the force balance after displacement
The net force acting on the piston after displacement can be expressed as:

\[ F_{net} = (P_1 + \Delta P) A - P_0 A - Mg \]

This can be simplified to:

\[ F_{net} = \Delta P A \]

### Step 4: Relate pressure change to volume change
Using the adiabatic condition for an ideal gas, we know that:

\[ P V^\gamma = \text{constant} \]

Differentiating this gives us:

\[ \Delta P = -\gamma \frac{P}{V} \Delta V \]

The change in volume \(\Delta V\) can be expressed as \(A \cdot x\), where \(x\) is the displacement of the piston. Thus,

\[ \Delta P = -\gamma \frac{P_1 A}{V_0} x \]

### Step 5: Substitute \(\Delta P\) into the force equation
Substituting \(\Delta P\) into the net force equation gives:

\[ F_{net} = -\gamma \frac{P_1 A^2}{V_0} x \]

Using Newton's second law, \(F = ma\), we have:

\[ M \frac{d^2x}{dt^2} = -\gamma \frac{P_1 A^2}{V_0} x \]

### Step 6: Identify the form of simple harmonic motion
This equation resembles the standard form of simple harmonic motion:

\[ \frac{d^2x}{dt^2} + \omega^2 x = 0 \]

where \(\omega^2 = \frac{\gamma P_1 A^2}{M V_0}\).

### Step 7: Find the frequency of oscillation
The frequency \(f\) is related to \(\omega\) by:

\[ \omega = 2\pi f \]

Thus, we can express the frequency as:

\[ f = \frac{1}{2\pi} \sqrt{\frac{\gamma P_1 A^2}{M V_0}} \]

### Step 8: Final expression for frequency
Substituting back into the equation, we find:

\[ f = \frac{1}{2\pi} \sqrt{\frac{\gamma (Mg + P_0 A)}{M V_0}} \]

This gives us the frequency of oscillation of the piston.