An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and cylinder have equal 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?

Let the piston be displaced by small amount X and the increase in pressure be Δp.Initial pressure P1=P0+MgA Volume V1=V0

For adiabatic process, PVγ = constant

(P0+MgA)V0γ=(P0+MgA+Δp)(V0-Ax)γ

Since x is small, we can use binomial approximation:

(1+x)n≈1+nx for small x

(P0+MgA)V0γ=(P0+MgA)(1-Ax/V0)γ(1+Δp/(P0+MgA))

(P0+MgA)V0γ=(P0+MgA)(1-γAx/V0)(1+Δp/(P0+MgA))

Ignoring higher order terms, we have:

1=1-γAx/V0+Δp/(P0+MgA)

Δp=γ(P0+MgA)x/V0A

Force = pressure × area = AΔp = γ(P0+MgA)x/V0

The restoring force is proportional to the displacement x, hence it is SHM.

Restoring force F = - γ(P0+MgA)x/V0

Comparing with F = -kx, we have k= γ(P0+MgA)/V0

Frequency ω=√(k/m)=√(γ(P0+MgA)/mV0)

ω=1/2π√(γ(P0+MgA)/MV0)=1/2π√(Aγ(P0+MgA)MV0)

Therefore, the frequency is 1/2π√(Aγ(P0+MgA)MV0)