FBD of piston at equilibrium⇒Patm​A+mg=P0​A........(1)
FBD of piston when piston is pushed down a distance xPatm​+mg−(P0​+dP)a=mdt2d2x​................(2)Process is adiabatic ⇒PVγ=C⇒−dP=VγPdV​Using 1, 2, 3 me get f=2π1​MV0​A2γP0​​​
