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:

Answer

`1/2pi sqrt(gamma(p_(0)A^(2)+MgA)/V_(0)M)`

Explanation

`((1)/(2pi)sqrt((gamma(p_(0)A^(2)+MgA))/(V_(0)M)))`
In equilibrium pressure inside the cylinder = pressure just outside it
or `P=P_(0)+(Mg)/(A)`
When piston is displaced slightly by an amount x, change in volume,
Since, the cylinder is isolated from the surroundings, process is adiabatic in nature. In adiabatic process,
`(dP)/(dV)=-gamma(P)/(V)`
or increase in pressure inside the cylinder, `dp=-((gammaP))/(V)(dV)=gamma((P_(0)+(Mg)/(A))/V_(0))(Ax)`
This increase in pressure when multiplied with area of cross-section A will give a net upward force (or the restoring force). Hence,
`F=-(dP)A=-gamma((P_(0)A^(2)+MgA)/(V_(0)))x` or `a=(F)/(M)=-gamma((P_(0)A^(2)+MgA)/(V_(0)M))x`
Since, `aprop-x` motion of the piston is simple harmonic in nature. Frequency of this oscillation is given by
`f=(1)/(2pi)sqrt(|(a)/(x)|)=(1)/(2pi)sqrt((gamma(P_(0)A^(2)+MgA))/(V_(0)M))`