Question

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass $M$. The piston and 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. $\frac{1}{2\pi }\frac{V_{0}M(P_0+\frac{Mg}{A})}{A^2\gamma }$
• B. $\frac{1}{2\pi }\sqrt{\frac{A^2\gamma (P_0+\frac{Mg}{A})}{M{V}_0}}$
• C. $\frac{1}{2\pi }\sqrt{\frac{M{V}_0}{A\gamma (P_0+\frac{Mg}{A})}}$
• D. $\frac{1}{2\pi }\frac{A\gamma (P_0+\frac{Mg}{A})}{V_{0}M}$

Answer 1

The correct option is B $\frac{1}{2\pi }\sqrt{\frac{A^2\gamma (P_0+\frac{Mg}{A})}{M{V}_0}}$

Let the piston be displaced by small amount X and the increase in pressure be $△p$.
Initial pressure $P_1=P_0+\frac{Mg}{A}$
Volume $V_1=V_0$
Final pressure $P_2=P_1+△p$
Volume $V_2=V_0-△V,△V=Ax$
Applying Poisson's equation
$P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$
$P_1{V}_1^{\gamma }=(P_1+△p)(V_0-△V{)}^{\gamma }$
$=P_1{V}_0^{\gamma }(1+\frac{△P}{P_1}){(1-\frac{△V}{V_0})}^{\gamma }$
$1=(1+\frac{△P}{P_1})(1-\gamma \frac{△V}{V_0})$
$=1-\gamma \frac{△V}{V_0}+\frac{△P}{P_1}-\gamma \frac{△P\cdot △V}{P_1{V}_0}$
Neglecting second order terms we have
$△p=\frac{\gamma P_1△V}{V_0}$
Restoring force on piston
$F=-A△P$
$=-\frac{\gamma A{P}_1}{V_0}\cdot \left(Ax\right)$
$=-\frac{\gamma A^2{P}_1}{V_0}x$
Acceleration of piston
$a=\frac{F}{M}=-\left(\frac{\gamma A^2{P}_1}{M{V}_0}\right)x$
Acceleration is proportional to the displacement, therefore motion is simple harmonic.
angular frequency of oscillation
$\omega =\sqrt{\frac{\gamma A^2{P}_1}{M{V}_0}}$
$\therefore f=\frac{1}{2\pi }\sqrt{\frac{\gamma A^2}{M{V}_0}(P_0+\frac{Mg}{A})\cdot }$