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. $\frac{1}{2\pi }\frac{V_{0}M}{A^2\gamma }(P_0+\frac{Mg}{A})$
• B. $\frac{1}{2\pi }\sqrt{\frac{A^2\gamma }{M{V}_0}(P_0+\frac{Mg}{A})}$
• C. $\frac{1}{2\pi }\sqrt{\frac{M{V}_0}{A\gamma (P_0+\frac{Mg}{A})}}$
• D. $\frac{1}{2\pi }\frac{A\gamma }{V_{0}M}(P_0+\frac{Mg}{A})$

Answer 1

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

Let the piston be displaced by small amount $x$ and the increase in pressure be $\Delta P$
Initial pressure $P_1=P_0+\frac{Mg}{A}$
Initial volume $V_1=V_0$
Final pressure $P_2=P_1+\Delta P$
Initial volume $V_2=V_0-\Delta V$



where, $\Delta V=Ax$

Applying Adiabatic law, as system is isolated (no heat transfer)
$P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$

where, $\gamma$ is specific heat ratio.

Substituting the values,
$P_1{V}_1^{\gamma }=(P_1+\Delta p)(V_0-\Delta V{)}^{\gamma }$

$\Rightarrow P_1{V}_0^{\gamma }=P_1{V}_0^{\gamma }(1+\frac{\Delta P}{P_1}){(1-\frac{\Delta V}{V_0})}^{\gamma }$

$\Rightarrow 1=(1+\frac{\Delta P}{P_1})(1-\gamma \frac{\Delta V}{V_0})$

$\Rightarrow 1-\gamma \frac{\Delta V}{V_0}+\frac{\Delta P}{P_1}-\gamma \frac{\Delta P.\Delta V}{P_1{V}_0}=1$

$\Rightarrow -\gamma \frac{\Delta V}{V_0}+\frac{\Delta P}{P_1}-\gamma \frac{\Delta P.\Delta V}{P_1{V}_0}=0$

Neglecting second order terms,
$\Rightarrow \Delta P=\frac{\gamma P_1\Delta V}{V_0}$

Restoring force on piston
$F=-A\Delta P$

$\Rightarrow F=-A\frac{\gamma P_1\Delta V}{V_0}$

Substituting the value of $\Delta V$,
$\Rightarrow F=-\frac{\gamma A{P}_1}{V_0}\left(Ax\right)$

$\Rightarrow F=-\frac{\gamma A^2{P}_1}{V_0}x$

So, 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.

The angular frequency of oscillation will be
$\omega =\sqrt{\left|\frac{a}{x}\right|}$

Substituting the value of $a$,

$\omega =\sqrt{\frac{\gamma A^2{P}_1}{M{V}_0}}$

Substituting the value of $P_1$, frequency $f$ of oscillation

$\therefore f=\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{\gamma A^2}{M{V}_0}(P_0+\frac{Mg}{A})}$

Hence, option (b) is correct answer.