An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and the cylinder have equal cross sectional area A. [2013]

Question

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and the cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of the gas is $V_{0}$ and its pressure is $P_{0}$ . 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 [2013]

Smart Achievers · Accepted Answer

(3)

$\frac{M g}{A} = P_{0}$

$P_{0} V_{0}^{γ} = P V^{γ}$

$M g = P_{0} A . . . (1)$

Let piston is displaced by distance $x$

$P_{0} A x_{0}^{γ} = P A {(x_{0} - x)}^{γ}$

$P = \frac{P_{0} x_{0}^{γ}}{{(x_{0} - x)}^{γ}}$

$M g - (\frac{P_{0} x_{0}^{γ}}{(x_{0} - x)^{γ}}) A = F_{restoring}$

$P_{0} A (1 - \frac{x_{0}^{γ}}{{(x_{0} - x)}^{γ}}) = F_{restoring}$ $[x_{0} - x \approx x_{0}]$

$P_{0} A (1 - \frac{1}{{(1 - \frac{x}{x_{0}})}^{γ}}) = P_{0} A [1 - {(1 - \frac{x}{x_{0}})}^{γ}]$

$= P_{0} A [1 - (1 + \frac{γ x}{x_{0}})]$

$F = - \frac{γ P_{0} A x}{x_{0}}$ $\Rightarrow M ω^{2} x = \frac{γ P_{0} A x}{x_{0}}$ $\Rightarrow ω = \sqrt{\frac{γ P_{0} A}{x_{0} M}}$

$∴$ $Frequency with which piston executes SHM$

$f = \frac{1}{2 π} \sqrt{\frac{γ P_{0} A}{x_{0} M}}$ $= \frac{1}{2 π} \sqrt{\frac{γ P_{0} A^{2}}{M V_{0}}}$

Solution

1 $\frac{1}{2 π} \frac{A γ P_{0}}{V_{0} M}$

2 $\frac{1}{2 π} \frac{V_{0} M P_{0}}{A^{2} γ}$

3 $\frac{1}{2 π} \sqrt{\frac{A^{2} γ P_{0}}{M V_{0}}}$

4 $\frac{1}{2 π} \sqrt{\frac{M V_{0}}{A γ P_{0}}}$

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Solution

1 12πAγP0V0M

2 12πV0MP0A2γ

3 12πA2γP0MV0

4 12πMV0AγP0

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1 $\frac{1}{2 π} \frac{A γ P_{0}}{V_{0} M}$

2 $\frac{1}{2 π} \frac{V_{0} M P_{0}}{A^{2} γ}$

3 $\frac{1}{2 π} \sqrt{\frac{A^{2} γ P_{0}}{M V_{0}}}$

4 $\frac{1}{2 π} \sqrt{\frac{M V_{0}}{A γ P_{0}}}$