An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. …

Q: 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 V0 and its pressure is Po. 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

(a) $\displaystyle \frac{1}{2\pi} \frac{A \gamma P_0}{ V_0 M} $

(b) $ \displaystyle \frac{1}{2\pi} \frac{V_0 M P_0}{ A^2 \gamma} $

(c) $ \displaystyle \frac{1}{2\pi} \sqrt{\frac{A ^2 \gamma P_0}{M V_0 }} $

(d) $ \displaystyle \frac{1}{2\pi} \sqrt{\frac{M V_0}{ A \gamma P_0 }} $

Ans: (c)

Sol: For sudden change in pressure and volume of the gas ;

$ \displaystyle PV^\gamma = constant $

Differentiating ,

$ \displaystyle P \gamma V^{\gamma -1}dV + V^\gamma dP = 0 $

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

Net force on piston is

$ \displaystyle F = dP \times A $

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

Let the piston be moved inwards through a small distance x, then decreases in the volume of the gas,
dV = Ax

$ \displaystyle F = -\frac{\gamma P_0}{V_0} \times Ax \times A $

$ \displaystyle F = -\frac{\gamma P_0 A^2}{V_0} x $

$\displaystyle M \omega^2 = \frac{\gamma P_0 A^2}{V_0} $ (Where M = mass of piston)

$ \displaystyle \omega^2 = \frac{\gamma P_0 A^2}{M V_0} $

$ \displaystyle \omega = \sqrt{\frac{\gamma P_0 A^2}{M V_0}} $

Frequency $ \displaystyle \nu = \frac{2\pi}{\omega} $