Question 139 of 194

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass $\text{M}$. The piston and the cylinder have equal cross-sectional area $\text{A}$. When the piston is in equilibrium, the volume of the gas is ${V}_{0}$ and its pressure is ${M}_{0}$. 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
[Assume the system is in space.]

A$\dfrac{1}{2\pi }\sqrt{\dfrac{{\text{A}}^{2}\gamma {\text{P}}_{0}}{{\text{MV}}_{0}}}$.B$\dfrac{1}{2\pi }\sqrt{\dfrac{{\text{MV}}_{0}}{\text{A}\gamma {\text{P}}_{0}}}$.C$\dfrac{1}{2\pi }\dfrac{\text{A}\gamma {\text{P}}_{0}}{{\text{V}}_{0}\text{M}}$D$\dfrac{1}{2\pi }\dfrac{{\text{V}}_{0}{\text{MP}}_{0}}{{\text{A}}^{2}\gamma }$.

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Answer: (A) $\dfrac{1}{2\pi }\sqrt{\dfrac{{\text{A}}^{2}\gamma {\text{P}}_{0}}{{\text{MV}}_{0}}}$.

For an adiabatic process,

${\text{Pv}}^{\gamma }=\text{constant}$

Differentiating with respect to $v$,

$\dfrac{\text{dP}}{\text{dv}}\left({\text{v}}^{\gamma }\right)+P\gamma {\text{v}}^{\gamma -1}=0$

$\text{dP}=\dfrac{-\gamma P}{\text{v}}\text{dv}=\dfrac{-\gamma P}{\text{v}}\text{Adx}$



Thus, the acceleration, $\text{a}=\dfrac{\text{dF}}{\text{M}}=\dfrac{\gamma {\text{PA}}^{2}}{\text{MV}}\text{dx}$

Comparing with , $\text{a}=-{\omega }^{2}\text{dx}$, we get,

$\omega =\sqrt{\dfrac{\gamma {P}_{0}{A}^{2}}{M{V}_{0}}}=2\pi f$

$f=\dfrac{1}{2\pi }\sqrt{\dfrac{\gamma {P}_{0}{A}^{2}}{M{V}_{0}}}$