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 $V_0$ 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. $\frac{1}{2\pi }\sqrt{\frac{M{V}_0}{A\gamma P_0}}$
• B. $\frac{1}{2\pi }\frac{A\gamma P_0}{2\pi V_{0}M}$
• C. $\frac{1}{2\pi }\frac{V_{0}M{P}_0}{A^2\gamma }$
• D. $\frac{1}{2\pi }\sqrt{\frac{A^2\gamma P_0}{M{V}_0}}$

Answer 1

Answer: (D)

$\frac{1}{2\pi }\sqrt{\frac{A^2\gamma P_0}{M{V}_0}}$

Pressure applied by piston, $\frac{Mg}{A}=p_0$

$Mg=p_{0}A$ __(i)

As no exchange of heat, so process is adiabatic

$p_0{V}_0^{\gamma }=p{V}^{\gamma },p_{0}A{x}_0^{\gamma }=pA(x_0-x{)}^{\gamma }$

$p=\frac{p_0{x}_0^{\gamma }}{(x_0-x{)}^{\gamma }}$

Let piston be displaced by x

$Mg=\left[\frac{p_0{x}_0^{\gamma }}{(x_0-x{)}^{\gamma }}\right)=A=f_{restoring}$

$p_{0}A(1-\frac{x_0^{\gamma }}{(x_0-x{)}^{\gamma }})=F_{restoring}$ $[\because x_0-x\approx x_0]$

$F=-\frac{\gamma p_{0}Ax}{x_0}$

$\therefore f=\frac{1}{2\pi }\sqrt{\frac{\gamma p_{0}A}{x_{0}M}}=\frac{1}{2\pi }\sqrt{\frac{\gamma p_0{A}^2}{M{V}_0}}$