A uniform cylinder of mass m and radius R is in equilibrium on an inclined by the action of a light spring of stiffness, gravity and reaction force acting on it .If the angle of inclination of the plane is `phi` , find the angular frequency of small oscillation of the cylinder

Explanation

If the CM is pushed (shifted) down by the addition by the additional distance x, the additional elongation of the spring will be `2x` Letx' be the equilibrium strech of spring .
The total elongation of the spring `x'' = (x' + 2x)`

The spring force`= F_(s) = kx' = k (x' + 2x)`
The tarque acting on the cylinder about P by the spring force is `yau_(s) = F_(s) (2 R) = 2k (x' + 2x) R`
The gravitational torque about P is `tau_(s) = - mg R sin phi`
The net tarque about is P is
`tau_("net") = tau_(S) + tau_(g) = 2kx' R + 4 kxR - mg R sin phi`
Since, `2 kx' = sin phi` from Eq. (ii), we have` tau_("net") = 4 kxR` Substituting `x = T theta` , we have the vector equation,
`tau_("net") = 4 kR^(2) phi`
Torque equation: The torque rotetes the cylinder with angular acceleration `alpha` at the angular position `theta`. Appliying Newton's second low of the rotation , we have `alpha = (tau_("net"))/(l_(p))`
were `tau_("net") = - 4 kR^(2) theta`
This gives `alpha = (-4kR^(2))/(l_(p)) theta`
Comparing the above equation with `alpha = - omega^(2) theta`.
we have `omega = sqrt((4 kr^(2))/(l_(p)))`, where `l_(p) = (3 mR^(2))/(2)`
Then, `omega = 2 sqrt((2k)/(3m))`