An ideal gas enclosed in a vertical cylindric | vedclass

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(C) At equilibrium, the pressure of the gas $P_0$ balances the weight of the piston: $P_0 A = Mg$.Since the system is isolated, the process is adiabat

Vedclass · Accepted Answer

$\frac{1}{2\pi}\sqrt{\frac{A^2\gamma P_0}{MV_0}}$

Vedclass · Answer

(C) At equilibrium, the pressure of the gas $P_0$ balances the weight of the piston: $P_0 A = Mg$.Since the system is isolated, the process is adiabatic: $P V^{\gamma} = 	ext{constant}$.Let the piston be displaced downwards by a small distance $x$. The new volume is $V = A(x_0 + x)$, where $V_0 = Ax_0$.The new pressure $P$ is given by $P = P_0 \left(\frac{V_0}{V}ight)^{\gamma} = P_0 \left(\frac{Ax_0}{A(x_0 + x)}ight)^{\gamma} = P_0 \left(1 + \frac{x}{x_0}ight)^{-\gamma}$.Using binomial approximation for small $x$: $P \approx P_0 \left(1 - \frac{\gamma x}{x_0}ight)$.The net restoring force on the piston is $F = (P_0 - P)A = P_0 A \left(1 - (1 - \frac{\gamma x}{x_0})ight) = \frac{P_0 A \gamma x}{x_0}$.Since $V_0 = Ax_0$, we have $x_0 = V_0/A$, so $F = \frac{P_0 A^2 \gamma}{V_0} x$.Comparing with $F = kx$, the effective spring constant is $k = \frac{\gamma P_0 A^2}{V_0}$.The frequency of $SHM$ is $f = \frac{1}{2\pi} \sqrt{\frac{k}{M}} = \frac{1}{2\pi} \sqrt{\frac{\gamma P_0 A^2}{M V_0}}$.

Similar Questions

$A$ body performs linear $S$.$H$.$M$. with amplitude $a$. When it is at a distance $\frac{a}{3}$ from the extreme position,the magnitude of velocity is $\frac{1}{3}$ times the magnitude of acceleration. The period of $S$.$H$.$M$. is:

Obtain the ratio of maximum acceleration and maximum velocity of a $SHM$ particle.

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