Question

A cylindrical 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 slightly 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

Answer: (A)

$T=2\pi \sqrt{\frac{Mh}{PA}}$

Let the piston be displaced through distance $x$ towards left then volume decrease and pressure increase. If the process is take place gradually.

So, it is isothermal

$\therefore P_1{V}_1=P_2{V}_2$

$PV=(P+\Delta P)(V-\Delta V)$

$\Delta P$ is increase in pressure, $\Delta V$ is decrease in volume

$PV=PV+\Delta PV-P\Delta V-\Delta P\Delta V$ (Very small value) so neglect

$\Delta PV-P\Delta V=0$

$\Delta P\left(\Delta h\right)=P\left(\Delta x\right)\Rightarrow \Delta P_2\frac{P.x}{h}$

This excess pressure is responsible for providing the resting force $\left(F\right)$ to the $P$ is ton of mass of $M$

$F=\Delta P.A=\frac{PAx}{h}$ --------------(1)

$\left(F\right)=kx$ --------------(2)

On comparing equation (1) and (2)

$\Rightarrow$ K = $M{w}^2$ $=$ $\frac{PA}{h}$

$w=\sqrt{\frac{PA}{Mh}}$

$\Rightarrow T=2\pi \sqrt{\frac{Mh}{PA}}$