JEE Mains · Physics · STD 11 - Oscillations
- JEE Mains 2013
- Hard
- MCQ
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
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Answer & Solution
Correct Answer
(C) \(\;\frac{1}{{2\pi }}\sqrt {\frac{{{A^2}\gamma {P_o}}}{{M{V_0}}}} \)
Step-by-step Solution
Detailed explanation
\({\frac{M g}{A}=P_{0}}\) \( {P_{0} V_{0}^{\gamma}=P V^{\gamma}}\)
\({\mathrm{Mg}=\mathrm{P}_{0} \mathrm{A}}{\ldots(1)}\) \( {P_{0} A x_{0}^{\gamma}=P A\left(x_{0}-x\right)^{\gamma}}\)
\(P=\frac{P_{0} x_{0}^{\gamma}}{\left(x_{0}-x\right)^{\gamma}}\)
Let piston is displaced by distance \(x\)
\(M g-\left(\frac{P_{0} x_{0}^{\gamma}}{\left(x_{0}-x\right)^{\gamma}}\right) A=F_{\text {restoring }}\)
\(P_{0} A\left(1-\frac{x_{0}^{\gamma}}{\left(x_{0}-x\right)^{\gamma}}\right)=F_{\text {restoring }} \quad\left[x_{0}-x \approx x_{0}\right]\)
\(F=-\frac{\gamma P_{0} A x}{x_{0}}\)
Frequency with which piston executes \(SHM.\)
\(f=\frac{1}{2 \pi} \sqrt{\frac{\gamma P_{0} A}{x_{0} M}}=\frac{1}{2 \pi} \sqrt{\frac{\gamma P_{0} A^{2}}{M V_{0}}}\)
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