Question

An ideal gas is enclosed in a vertical cylindrical container and supports a freely moving piston of mass M . The piston and the cylinder have equal cross-sectional area A. Atmospheric pressure is$p_0$ and when the piston is in equilibrium, the volume of the gas is V . The piston is now displaced slightly from its equilibrium position. Assuming that the system is completely isolated from its surroundings, show that the piston executes SHM and find the frequency of oscillation.

Answer 1

Answer:

$\mathsf{\frac{1}{2 \pi}\sqrt{\frac{P_{0}A^{2}+MgA}{MV_{0}}}}$

Explanation:

Initially, when the piston of mass M is in equilibrium.

$\mathsf{P = P_{0} + \frac{Mg}{A}}$          ........(i)

Later, when the piston is displayed down slightly from the equilibrium position, a small decrease in volume of gas occurs.

∴ Decrease in volume = dV = -Ax       ....(ii)

As the system is completely isolated from surroundings, the process is adiabatic in nature.

$\mathsf{\therefore PV^{\gamma}= Constant }$

$\mathsf{or \ \gamma PV^{\gamma - 1}dV+V^{\gamma}dP=0}$

$\mathsf{\therefore dP=-\frac{\gamma P}{V}dV}$

$\textsf{Put P and dV from (i) and (ii)}$

$\mathsf{or \ dP = \gamma \Bigg[\frac {\frac{P_{0}+Mg}{A}}{V_{0}} \Bigg]Ax,}$

$\mathsf{where \ V = V_{0} = \ volume \ of \ gas.}$

$\mathsf{\therefore Restoring \ force=-(dP)A}$

$\mathsf{or \ M \times acceleration = \gamma \Bigg[ \frac{P_{0}A+MgA}{MV_{0}} \Bigg]Ax\times A}$

$\mathsf{or \ acceleration = -\gamma \Bigg[ \frac{P_{0}A^{2} + MgA}{MV_{0}} \Bigg]x}$

$\mathsf{\therefore acceleration = \ \propto - \ x}$

$\textsf{Hence, the motion of piston is simple harmonic.}$

$\mathsf{\therefore \omega^{2} = -\frac{acceleration}{x}}$

$\mathsf{or \ (2 \pi f)^{2} = \frac{\gamma (P_{O}A^{2}+MgA)}{MV_{0}}}\\\mathsf{or \ f = \frac{1}{2 \pi}\sqrt{\frac{(P_{0}A^{2}+MgA)}{MV_{0}}}}$

$\mathsf{or \ f = \frac{A}{2 \pi}\sqrt{\frac{\gamma P_{0}}{MV_{0}}} \ \ , \ \ when \ MgA \ll P_{0}A^{2}}$