A cylindrical piston of mass M slides smoothl | vedclass

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Question

(A) Let the piston be displaced through a small distance $x$ towards the left. The volume of the gas decreases,and its pressure increases. Let $\Delta

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

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(A) Let the piston be displaced through a small distance $x$ towards the left. The volume of the gas decreases,and its pressure increases. Let $\Delta P$ be the increase in pressure and $\Delta V$ be the decrease in volume. Assuming the process is isothermal,we have $P_1V_1 = P_2V_2$.$PV = (P + \Delta P)(V - \Delta V)$$PV = PV - P\Delta V + \Delta P V - \Delta P \Delta V$Neglecting the small term $\Delta P \Delta V$,we get $P \Delta V = \Delta P V$.Since $\Delta V = A x$ and $V = A h$,we have $P(Ax) = \Delta P(Ah)$.$\Delta P = \frac{Px}{h}$.This excess pressure creates a restoring force $F$ on the piston of mass $M$:$F = -(\Delta P)A = -\left(\frac{PA}{h}\right)x$.Comparing this with the equation for simple harmonic motion $F = -kx$,we identify the force constant $k = \frac{PA}{h}$.The angular frequency is $\omega = \sqrt{\frac{k}{M}} = \sqrt{\frac{PA}{Mh}}$.The time period $T$ is given by $T = \frac{2\pi}{\omega} = 2\pi \sqrt{\frac{Mh}{PA}}$.

$A$ 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

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