Question

A cylindrical conducting piston of mass $M$ slides smoothly inside a long cylinder closed at one end, enclosing a certain mass of a 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 (process is isothermal)

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

Answer 1

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

Extra force = $(P-P_0)A=Ma$
$P_1{V}_1=P_2{V}_2$
$P_{0}hA=P(h-x)A$
$P=\frac{P_{0}h}{h-x}$
So acceleration = $-(\frac{P_{0}h}{h-x}-P_0)\frac{A}{M}$

$a=-\frac{P_{0}Ax}{M(h-x)}$
For small $x$,
$a=-\frac{P_{0}Ax}{Mh}$
$-{\omega }^{2}x=-\frac{P_{0}Ax}{Mh}$
$\omega =\sqrt{\frac{Mh}{P_{0}A}}$

$T=2\pi \sqrt{\frac{P_{0}A}{Mh}}$