## In This post you will get answer of Q.

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

Solution:

$FBD$ of piston at equilibrium

$P _{atm}A+Mg=P_{0}A$ $ldotsleft(iright)$

$FBD$ of piston when piston is pushed down a distance $x$

$left(P_{0}+dpright)A-left(p_{atm} A+Mgright)=M frac{d^{2}x}{dt^{2}} ldotsleft(iiright)$

As the system is completely isolated from its surrounding therefore the change is adiabatic

For an adiabatic process

$PV^{gamma}$ =constant $ therefore V^{gamma} dP+Pgamma V^{gamma-1} dV=0$

or $dP=-frac{gamma PdV}{V} $

$therefore dp=-frac{gamma P_{0} left(Axright)}{V_{0}} ldotsleft(iiiright)$

Using $left(iright)$ and $left(iiiright)$ in $left(iiright)$, we get

$M frac{d^{2} x}{dt^{2}}=-frac{gamma P_{0}A^{2}}{V_{0}}x $

or $ frac{d^{2}x}{dt^{2}}=-frac{gamma P_{0}A^{2}}{MV_{0}}x$

Comparing it with standard equation of $SHM$,

$frac{d^{2}x}{dt^{2}}=-omega^{2}x$

We get,$ omega^{2}=frac{gamma P_{0}A^{2}}{MV_{0}}$ or $omega=sqrt{frac{gamma P_{0} A^{2}}{MV_{0}}}$

Frequency, $upsilon=frac{omega}{2pi}=frac{1}{2pi}sqrt{frac{gamma P _{0}A^{2}}{MV_{0}}}$