An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass $\mathit{\text{M}}$. The piston and the cylinder have equal cross-sectional area $\mathit{\text{A}}$. 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.]

• Correct answer
  A

  $\frac{1}{2\mathrm{\pi }}\sqrt{\frac{{\mathit{\text{A}}}^2\mathrm{\gamma }{\mathit{\text{P}}}_0}{{\mathit{\text{MV}}}_0}}$.
• B

  $\frac{1}{2\mathrm{\pi }}\sqrt{\frac{{\mathit{\text{MV}}}_0}{\mathit{\text{A}}\mathrm{\gamma }{\mathit{\text{P}}}_0}}$.
• C

  $\frac{1}{2\mathrm{\pi }}\frac{\mathit{\text{A}}\mathrm{\gamma }{\mathit{\text{P}}}_0}{{\mathit{\text{V}}}_0\mathit{\text{M}}}$
• D

  $\frac{1}{2\mathrm{\pi }}\frac{{\mathit{\text{V}}}_0{\mathit{\text{MP}}}_0}{{\mathit{\text{A}}}^2\mathrm{\gamma }}$.

Solution

For an adiabatic process,

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

Differentiating with respect to $v$,

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

$\mathit{\text{dP}}=\frac{-\mathrm{\gamma P}}{\mathit{\text{v}}}\mathit{\text{dv}}=\frac{-\mathrm{\gamma P}}{\mathit{\text{v}}}\mathit{\text{Adx}}$



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

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

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


$f=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{\mathrm{\gamma }{P}_0{A}^2}{M{V}_0}}$

Hence, the correct answer is option: 'A'.