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An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and the cylinder have cross sectional area A. When the piston is in equilibrium, the volume of the gas is V₀ and its pressure is P₀ . The piston is slightly displaced from the equilibrium position and released.Assuming that the system is completely isolated from its surrounding,the piston executes a simple harmonic motion with frequency:

Option 1)
$\frac{1}{2\pi }\sqrt{\frac{MV_{0}}{A\gamma P_{0}}}$

Option 2)
$\frac{1}{2\pi }\, {\frac{A\gamma P_{0}}{V_{0}M}}$

Option 3)
$\frac{1}{2\pi }\frac{V_{0}MP_{0}}{A^{2}\gamma }$

Option 4)
$\frac{1}{2\pi }\sqrt{\frac{A^{2}\gamma P_{0}}{MV_{0}}}$

Answers (1)

best_answer

As we have learned

Relation between slope of isothermal and adiabatic process -

$\frac{dP}{dV}_{isothermal}= -\frac{P}{V}$

$\frac{dP}{dV}_{adiabatic}= -\gamma \frac{P}{V}$

- wherein

$Slope \, of \, adiabatic\, process = \gamma\, \times Slope \, of \, isothermal\, process$

At equilibrium

$V = V_0 \; \; \; \; P = P_0$

Let the atmospheric pressure is pa

$(P_0 - Pa )A =Mg ...........(1)$

Let piston is displaced slightly into the cylinder then pressure become $P_0 + dP$

$\Rightarrow Net \: \: \: force = Mg - (P_0 + dP )A +P_a A \\ = (Mg + P_a )A - P_0A - dP A$

$F = -dP A ...........(2)$

Since cylinder is isolated , then the process is adiabatic $\Rightarrow PV^r = constant$

$\Rightarrow dP = V^\gamma + pr v^{\gamma -1}dv = 0 \\ or\: \: dP = \frac{-\gamma P}{V }dV$

$\therefore F = - \frac{\gamma P_0}{V_0}A dV$

$dV + A dx \\ F = \frac{- P_0 \gamma A}{V_0 }Adx = \frac{- \gamma A^2P}{V_0 }dx$

Acceleration = F/M =

$= \frac{- \gamma A^2P_0}{MV_0 }dx$

This equation is of SHM with

$w^2 = \left ( \frac{\gamma P_0A^2}{MV_0 } \right )$

$T = 2 \pi \sqrt{\frac{MV_0}{\gamma P_0A^2 }}$

$= 1/T = \frac{1}{2\pi}\sqrt{\frac{\gamma P_0A^2}{MV_0}}$

Option 1)

$\frac{1}{2\pi }\sqrt{\frac{MV_{0}}{A\gamma P_{0}}}$

Option 2)

$\frac{1}{2\pi }\, {\frac{A\gamma P_{0}}{V_{0}M}}$

Option 3)

$\frac{1}{2\pi }\frac{V_{0}MP_{0}}{A^{2}\gamma }$

Option 4)

$\frac{1}{2\pi }\sqrt{\frac{A^{2}\gamma P_{0}}{MV_{0}}}$

Posted by

Avinash

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