Question

A cylindrical piston of mass $M$ slide 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. Its period is:

• A. $T=2\pi \sqrt{\left(\frac{Mh}{PA}\right)}$
• B. $T=2\pi \sqrt{\left(\frac{MA}{Ph}\right)}$
• C. $T=2\pi \sqrt{\left(\frac{M}{PAh}\right)}$
• D. $T=2\pi \sqrt{MPhA}$

Answer 1

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

Let the piston be displaced through distance $x$ towards left, then volume decrease, pressure increases. If $△P$ is increase is pressure and $△V$ is decreases in volume, then considering the process to take place gradually (i.e. isothermal)
$P_1{V}_1-P_2{V}_2$
$\Rightarrow PV=(P+△P)(V-△V)$
$\Rightarrow PV=PV+△PV-P△V-△P△V$
$\Rightarrow △PV-P△V=0\left(neglecting△P△V\right)$
$△P\left(Ah\right)=P\left(Ax\right)\Rightarrow △P=\frac{P.x}{h}$
This excess pressure is responsible for providing the restoring force $\left(F\right)$ to the piston of mass $M$.
Hence $F=△P.A=\frac{P.Ax}{h}$
Comparing it with $|F|=kx\Rightarrow k=M{\omega }^2=\frac{PA}{h}$
$\Rightarrow \omega =\sqrt{\frac{PA}{Mh}}\Rightarrow T=2\pi \sqrt{\frac{Mh}{PA}}$