question_answer

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) \[\frac{1}{2\pi }\,\frac{A\gamma {{P}_{0}}}{{{V}_{0}}M}\]     

B) \[\frac{1}{2\pi }\,\frac{{{V}_{0}}M{{P}_{0}}}{{{A}^{2}}\gamma }\]

C) \[\frac{1}{2\pi }\,\sqrt{\frac{{{A}^{2}}\gamma {{P}_{0}}}{M{{V}_{0}}}}\]    

D) \[\frac{1}{2\pi }\,\sqrt{\frac{M{{V}_{0}}}{A\gamma {{P}_{0}}}}\]

Correct Answer: C

Solution :

\[\frac{Mg}{A}={{P}_{0}}\]

\[Mg={{P}_{0}}A\]       ...(1)

\[{{P}_{0}}{{V}_{0}}^{\gamma }=P{{V}^{\gamma }}\]

\[{{P}_{0}}A{{x}_{0}}^{\gamma }=PA{{({{x}_{0}}-x)}^{\gamma }}\]

\[P=\frac{{{P}_{0}}x_{0}^{\gamma }}{{{({{x}_{0}}-x)}^{\gamma }}}\]

Let piston is displaced by distance .

\[Mg-\left( \frac{{{P}_{0}}x_{0}^{\gamma }}{{{({{x}_{0}}-x)}^{\gamma }}} \right)A={{F}_{restoring}}\] \[{{P}_{0}}A\left( 1-\frac{x_{0}^{\gamma }}{{{({{x}_{0}}-x)}^{\gamma }}} \right)={{F}_{restoring}}\]

\[F=-\frac{\gamma {{P}_{0}}Ax}{{{x}_{0}}}\]

\[\therefore \]  Frequency with which piston executes SHM.

\[f=\frac{1}{2\pi }\sqrt{\frac{\gamma {{P}_{0}}A}{{{x}_{0}}M}}=\frac{1}{2\pi }\sqrt{\frac{\gamma {{P}_{0}}{{A}^{2}}}{M{{V}_{0}}}}\]