Correct Answer - D
FBD off piston at equilibrium
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)
As the system is completely isolated from its surrounding therefore the change is adiabatic.
For an adiabatic process.
`PV^(gamma)=`constant `thereforeV^(gmma)dP+PgammaV^(gamma-1)dV=0`
or `dP=-(gammaPdV)/(V)thereforedP=-(gammaP_(0)(Ax))/(V_(0))` . .. (iii)
Using (i) and (iii) in (ii), we get
`M(d^(2)x)/(dt^(2))=-(gammaP_(0)A^(2))/(V_(0))x` or `(d^(2)x)/(dt^(2))=-(gammaP_(0)A^(2))/(MV_(0))x`
Comparing it with standard equation of SHM,
`(d^(2)x)/(dt^(2))=-omega^(2)x`
we get, `omega^(2)=(gamma)/(2pi)=(1)/(2pi)sqrt((gammaP_(0)A^(2))/(MV_(0)))`
Frequency, `upsilon=(omega)/(2pi)=(1)/(2pi)sqrt((gammaP_(0)A^(2))/(MV_(0)))`