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 period of oscillation will be:

A) \[T=2\pi \,\sqrt{\frac{Mh}{PA}}\]           

B) \[T=2\pi \,\sqrt{\frac{Mh}{Ph}}\]

C) \[T=2\pi \,\sqrt{\frac{Mh}{PAh}}\]         

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. AP = increase in pressure and v = decrease in volume, for isothermal compression \[(P+\Delta p)\,(V-v)=PV\] \[(P+\Delta p)\,\,(Ah-Ax)=PV=PAh\] \[PAh-PAx+Ah\,\Delta p-\Delta p\,Ax=PAh\text{ }\]                                     \[\text{ }\!\![\!\!\text{ }\Delta P\,Ax\,\,is\,negligible]\] \[-PAx+\Delta PAh=0\] \[\Delta p=\frac{Px}{h}\] \[\Delta F=\Delta pA\,=\frac{pAx}{h}\] Restoring force \[\Delta F\,=\frac{-PAx}{h}\] \[\Delta F\,=\frac{-PAx}{h}\]      Compare it will \[a={{\omega }^{2}}x\] \[{{\omega }^{2}}=\frac{PA}{Mh}\Rightarrow \omega =\sqrt{\frac{PA}{Mh}}\Rightarrow T=2\pi \sqrt{\frac{Mh}{PA}}\]