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  • Help me please, 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 position is disturbed from its equillibrium position, its

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 position is disturbed from its equillibrium position, its oscillates simple harmonically. The period of the oscillation will be

 

  • Option 1)

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

  • Option 2)

    T = 2\pi \sqrt{[\frac{MA}{Ph}]}

  • Option 3)

    T = 2\pi \sqrt{[\frac{M}{PAh}]}

  • Option 4)

    T = 2\pi \sqrt{MPhA}

 

Answers (1)

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 is decrease in 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 P V -P\Delta V - \Delta P \Delta V\\*\Rightarrow \Delta PV - P\Delta V = 0 \;\;\;\;(neglecting \; \Delta P \Delta V)\\* \Rightarrow \Delta P (Ah) = P(Ax) \Rightarrow \Delta P = \frac{P\cdot x}{h}

This execess pressure  is resonsible for providing the restoring force (F) to the piston of mass M.

Hence F = \Delta P \cdot A = \frac{PAx}{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}}

 

Time period of oscillation of simple pendulum -

T=2\pi \sqrt{\frac{l}{g}}

- wherein

l = length of pendulum 

g = acceleration due to gravity.

 

 


Option 1)

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

This is correct.

Option 2)

T = 2\pi \sqrt{[\frac{MA}{Ph}]}

This is incorrect.

Option 3)

T = 2\pi \sqrt{[\frac{M}{PAh}]}

This is incorrect.

Option 4)

T = 2\pi \sqrt{MPhA}

This is incorrect.

Posted by

Vakul

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