Question

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 $P_0$. 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

• A. $\frac{1}{2 \pi} \frac{A_{\gamma}p_0}{V_0M}$
• B. $\frac{2}{2\pi}\frac{V_0MP_0}{A^2 \gamma}$
• C. $\frac{2}{2\pi}\sqrt{\frac{A^2\gamma P_0}{MV_0}}$
• D. $\frac{1}{2 \pi} \sqrt{\frac{MV_0}{A_{\gamma}P_0}}$

Answer 1

The Correct Option is C

To determine the frequency of simple harmonic motion (SHM) executed by the piston, we need to analyze the system where an ideal gas is enclosed in a cylindrical container with a piston of mass \(M\). The equilibrium conditions and properties of the gas, along with the dynamics of the piston, will help us derive the required expression.

1. Initially, the system is in equilibrium. In this state, the upward force exerted by the gas pressure on the piston balances the weight of the piston. Hence, the equilibrium condition is given by: \(P_0 \cdot A = Mg\) where \(A\) is the cross-sectional area, \(P_0\) is the pressure, and \(g\) is the acceleration due to gravity.
2. Upon a small displacement from equilibrium, the piston will execute simple harmonic motion. In SHM, the force is directly proportional to the displacement and is given by Hooke's law. For this gas-piston system, the restoring force due to change in pressure can be found using the adiabatic condition (since the system is isolated): \(PV^{\gamma} = \text{constant}\) where \(\gamma\) is the adiabatic index.
3. A small displacement \(x\) leads to a small change in volume (\(\Delta V = A \cdot x\)) and pressure, and the restoring force \(F\) can then be expressed using: \(F = -A \cdot \Delta P = -A \cdot \left(-\frac{\gamma P_0 A}{V_0}x\right) = \frac{\gamma P_0 A^2}{V_0}x\)
4. This is a standard form of SHM with effective spring constant \(k = \frac{\gamma P_0 A^2}{V_0}\). Using the formula for the frequency of SHM, \(f = \frac{1}{2\pi} \sqrt{\frac{k}{M}}\), we substitute the value of \(k\): \(f = \frac{1}{2\pi} \sqrt{\frac{\gamma P_0 A^2}{M V_0}}\)
5. The correct frequency matches the third option given: \(\frac{2}{2\pi}\sqrt{\frac{A^2\gamma P_0}{MV_0}}\)

Thus, the correct answer is \(\frac{2}{2\pi}\sqrt{\frac{A^2\gamma P_0}{MV_0}}\), confirming that the piston indeed executes an SHM with this frequency under the given conditions.