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 - section area `A`. When the piston is in equilibrium, the volume of the gas is `(V_0)` and the 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, Show that the piston executes simple harmonic motion and find the frequency of oscillations.

Options

A. `1/(2pi) (v_(0)Mp_(0))/(A^(2)gamma)`
B. `1/(2pi) sqrt((A^(2)gammaP_(0))/(MV_(0)))`
C. `1/(2pi) sqrt( (MV_(0))/(AgammaP_(0)))`
D. `1/(2pi) (AgammaP_(0))/(V_(0)M)`

Answer

B

Explanation

From FBD of piston at equilibrium
`P_(atm)A+Mg=P_(0)A ` … (i)
From FBD of piston when piston is pushed down a distance x
`(P_(0)+dp)A-(P_(atm)A+Mg)=M(d^(2)x)/(dt^(2))`…..(ii)
The system is completely isolated from its surrounding hence the change is adiabatic.
For an adiabatic process,`PV^(gamma)`= constant
or `dp=(gammapdV)/V`
but dV=Ax
`therfore dp=-(gammap_(0)(Ax))/v_(0)`
Using (i) and (iii) in (ii), we get `M(d^(2)x)/(dt^(2))=-(gammaP_(0)A^(2))/(V_(0))xor(d^(2)x)/(dt^(2))=-(gammaP_(0)A^(2))/(M V_(0))`