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. $\frac{1}{2\pi }\sqrt{\frac{M{V}_0}{A\gamma P_0}}$
• B. $\frac{1}{2\pi }\frac{A\gamma P_0}{V_{0}M}$
• 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)

$FBD$ of piston at equilibrium

$P_{atm}A+Mg=P_{0}A$ $\dots \left(i\right)$
$FBD$ of piston when piston is pushed down a distance $x$

$\left(P_0+dp\right)A-\left(p_{atm}A+Mg\right)=M\frac{d^{2}x}{d{t}^2}\dots \left(ii\right)$
As the system is completely isolated from its surrounding therefore the change is adiabatic
For an adiabatic process
$P{V}^{\gamma }$ =constant $\therefore V^{\gamma }dP+P\gamma V^{\gamma -1}dV=0$
or $dP=-\frac{\gamma PdV}{V}$
$\therefore dp=-\frac{\gamma P_0\left(Ax\right)}{V_0}\dots \left(iii\right)$
Using $\left(i\right)$ and $\left(iii\right)$ in $\left(ii\right)$, we get
$M\frac{d^{2}x}{d{t}^2}=-\frac{\gamma P_0{A}^2}{V_0}x$
or $\frac{d^{2}x}{d{t}^2}=-\frac{\gamma P_0{A}^2}{M{V}_0}x$
Comparing it with standard equation of $SHM$,
$\frac{d^{2}x}{d{t}^2}=-{\omega }^{2}x$
We get,${\omega }^2=\frac{\gamma P_0{A}^2}{M{V}_0}$ or $\omega =\sqrt{\frac{\gamma P_0{A}^2}{M{V}_0}}$
Frequency, $\upsilon =\frac{\omega }{2\pi }=\frac{1}{2\pi }\sqrt{\frac{\gamma P_0{A}^2}{M{V}_0}}$