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 an equal cross-sectional area $A$ . Atmospheric pressure is $p_{0}$ and when the piston is in equilibrium, the volume of the gas is $V_{0}$ . The piston is now displaced slightly from its equilibrium position. Assuming that the system, is completely isolated from, its surroundings, what is the frequency of oscillation.

• A. $f=\frac{1}{2 \pi }\sqrt{\frac{\gamma \left(p_{0} A^{2} + M g A\right)}{V_{0} M}}$
• B. $f=\frac{1}{2 \pi }\sqrt{\frac{1}{\gamma } \frac{\left(p_{0} A^{2} + M g A\right)}{V_{0} M}}$
• C. $f=\frac{1}{2 \pi }\sqrt{\frac{\left(p_{0} A^{2} + M g A\right)}{V_{0} M}}$
• D. $f=\frac{1}{2 \pi }\sqrt{\frac{A \left(p_{0} A + M g A\right)}{V_{0} M}}$

Answer 1

Answer: (A)

In equilibrium, pressure inside the cylinder = pressure just outside it or $p = p _{0}+\frac{ Mg }{ A }$
When piston is displaced slightly by an amount $x$, change in volume, $dV =- Ax$
Since, the cylinder is isolated from the surroundings, nature of the process is adiabatic. In adiabatic process,
$\frac{ dp }{ dV }=-\gamma \frac{ p }{ V }$

or increase in pressure inside the cylinder,
$dp =-\frac{\gamma p }{ V } dV =\gamma \frac{ p _{0}+\frac{ Mg }{ A }}{ V _{0}} Ax$
This increase in pressure when multiplied with area of cross-section A will give a net upward force (or the restoring force). Hence,
$F =-\text{dpA}=-\gamma \frac{ p _{0} A ^{2}+ MgA }{ V _{0}} x$
Or $a =\frac{ F }{ M }=-\gamma \frac{ p _{0} A ^{2}+ MgA }{ V _{0} M } x$
Since, $a \propto-x$, motion of the piston is simple harmonic in nature. Frequency of this oscillation is given by
$f =\frac{1}{2 \pi} \sqrt{\frac{ a }{ x }}=\frac{1}{2 \pi} \sqrt{\frac{\gamma p _{0} A ^{2}+ MgA }{ V _{0} M }}$