Question

A cylindrical piston of mass $M$ slides smoothly inside a long cylinder closed at one end, enclosing a certain mass of gas.
The cylinder is kept with its axis horizontal. If the piston is disturbed from its equilibrium position, it oscillates simple harmonically. The period of oscillation will be

• A. $T=2\pi \sqrt{\frac{Mh}{PA}}$
• B. $T=2\pi \sqrt{\frac{MA}{Ph}}$
• C. $T=2\pi \sqrt{\frac{M}{PAh}}$
• D. $T=2\pi \sqrt{MPHA}$

Answer 1

The correct option is A $T=2\pi \sqrt{\frac{Mh}{PA}}$


Let $V(=Ah)$ be the initial volume of the cylinder. If we compress the gas by displacing the piston through $x$, restoring force acting on the piston due to the gas will be
$F=-P\left(\frac{\Delta V}{V}\right).A$
where, $\Delta V(=Ax)$ is the change in volume
$\Rightarrow F=-\frac{P\left(Ax\right)}{\left(Ah\right)}.A$
$\Rightarrow Ma=-\frac{PA}{h}x$
where, $a$ is the acceleration of piston.
$\Rightarrow a=-\frac{PA}{Mh}x$
Comparing with $a=-{\omega }^{2}x$, we get
${\omega }^2=\frac{PA}{Mh}$
$\Rightarrow \omega =\sqrt{\frac{PA}{Mh}}$
$\therefore T=2\pi \sqrt{\frac{Mh}{PA}}$.
Hence, option $\left(a\right)$ is the correct answer.