An ideal gas is enclosed in a vertical cylindrical container and supports a freely moving piston of mass M . The piston and the cylinder have an equal cross-sectional area A . Atmospheric pressure is p0 and when the piston is in equilibrium, the volume of the gas is V0 . The piston is now displaced slightly from its equilibrium position. Assuming that the system, is completely isolated from, its surroundings, what is the frequency of oscillation.

Question

An ideal gas is enclosed in a vertical cylindrical container and supports a freely moving piston of mass M . The piston and the cylinder have an equal cross-sectional area A . Atmospheric pressure is p0 and when the piston is in equilibrium, the volume of the gas is V0 . The piston is now displaced slightly from its equilibrium position. Assuming that the system, is completely isolated from, its surroundings, what is the frequency of oscillation. (A) f=(1/2 π )√(&gamma; (p0 A2 + M g A)/V0 M) (B) f=(1/2 π )√(1/&gamma; ) ((p0 A2 + M g A)/V0 M) (C) f=(1/2 π )√((p0 A2 + M g A)/V0 M) (D) f=(1/2 π )√(A (p0 A + M g A)/V0 M)

Tardigrade Admin · Accepted Answer

In equilibrium, pressure inside the cylinder = pressure just outside it or

p = p_{0} + \frac{M g}{A}

When piston is displaced slightly by an amount

x

, change in volume,

d V = - A x

Since, the cylinder is isolated from the surroundings, nature of the process is adiabatic. In adiabatic process,

\frac{d p}{d V} = - γ \frac{p}{V}

or increase in pressure inside the cylinder,

d p = - \frac{γ p}{V} d V = γ \frac{p _{0} + \frac{M g}{A}}{V _{0}} A x

This increase in pressure when multiplied with area of cross-section A will give a net upward force (or the restoring force). Hence,

F = - dpA = - γ \frac{p _{0} A ^{2} + M g A}{V _{0}} x

Or

a = \frac{F}{M} = - γ \frac{p _{0} A ^{2} + M g A}{V _{0} M} x

Since,

a \propto - x

, motion of the piston is simple harmonic in nature. Frequency of this oscillation is given by

f = \frac{1}{2 π} \frac{a}{x} = \frac{1}{2 π} \frac{γ p _{0} A ^{2} + M g A}{V _{0} M}

$f = \frac{1}{2 π} \frac{γ ( p _{0} A ^{2} + M g A )}{V _{0} M}$

$f = \frac{1}{2 π} \frac{1}{γ} \frac{( p _{0} A ^{2} + M g A )}{V _{0} M}$

$f = \frac{1}{2 π} \frac{( p _{0} A ^{2} + M g A )}{V _{0} M}$

$f = \frac{1}{2 π} \frac{A ( p _{0} A + M g A )}{V _{0} M}$

f=2π1​V0​Mγ(p0​A2+MgA)​​

f=2π1​γ1​V0​M(p0​A2+MgA)​​

f=2π1​V0​M(p0​A2+MgA)​​

f=2π1​V0​MA(p0​A+MgA)​​

$f = \frac{1}{2 π} \frac{γ ( p _{0} A ^{2} + M g A )}{V _{0} M}$

$f = \frac{1}{2 π} \frac{1}{γ} \frac{( p _{0} A ^{2} + M g A )}{V _{0} M}$

$f = \frac{1}{2 π} \frac{( p _{0} A ^{2} + M g A )}{V _{0} M}$

$f = \frac{1}{2 π} \frac{A ( p _{0} A + M g A )}{V _{0} M}$