question_answer

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) \[T=2\pi \sqrt{\left( \frac{Mh}{PA} \right)}\] 

B) \[T=2\pi \sqrt{\left( \frac{Mh}{Ph} \right)}\]

C) \[T=2\pi \sqrt{\left( \frac{M}{PAh} \right)}\]

D) \[T=2\pi \sqrt{MPhA}\]  

Correct Answer: A

Solution :

[a] Let the piston be displaced through distance x towards left, then volume decreases, pressure increases. If \[\Delta P\]is increase in pressure and \[\Delta V\] volume, then considering the process to take place gradually (i.e. isothermal) \[{{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}\] \[\Rightarrow PV=(P+\Delta P)(V-\Delta V)\] \[\Rightarrow PV=PV+\Delta PV-P\Delta V-\Delta P\Delta V\] \[\Rightarrow \Delta P.V-P.\Delta V=0\](neglecting\[\Delta P.\Delta V\]) \[\Delta P(Ah)=P(Ax)\Rightarrow \Delta P=\frac{P.x}{h}\] This excess pressure is responsible for providing the restoring force (F) to the piston of mass M. Hence \[F=\Delta P.A=\frac{PAx}{h}\] Comparing it with \[\left| F \right|=kx\Rightarrow M{{\omega }^{2}}=\frac{PA}{h}\] \[\Rightarrow \omega =\sqrt{\frac{PA}{Mh}}\Rightarrow T=2\pi \sqrt{\frac{Mh}{PA}}\]