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 `V_(0)` and 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, the piston executes a simple harmonic motion with frequency:

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))`