An ideal gas enclosed in a cylindrical container supports a freely moving piston of mass `M`. The piston and the 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

Answer

`00.50`

Explanation

At equilibrium,
`P _(atm) A + MG = P _(0) A ...(i)`
When piston is pushed down a distance x
`( p _(0) + dP ) - ( P _(atm) A + Mg) = M ( d ^(2)x)/(dt ^(2))...(ii)`
For an adiabati system,
`PV ^(gamma)=` constant
`implies V ^( gamma) dP + gamma PV ^( gamma -1) dV = 0`
`implies dP = (- gamma P dV)/( v) = ( - gamma P _(0) (Ax))/( V _(0)) .....(iii)`
From (i),(ii)/& (iii)
`M (d ^(2) x )/(dt ^(2)) = ( - gamma p _(0) A ^(2))/( V _(0)) x `
`implies (d ^(2)x )/( dt ^(2)) = ( - gamma p _(0) A ^(2))/( V _(0) M) x`
`implies omega = sqrt (( gamma P _(0) A ^(2))/( V _(0) M )) and v = (1)/(2pi) sqrt (( gamma P _() A ^(2))/( V _(0) M ))`
`implies N = 2 , q = 1`
Now `(q)/(N) = (1)/(2)= 0. 50`