A solid uniform cylinder of mass `M` performs small oscillations in horizontal plane if slightly displaced from its mean position shown in figure. If it is given that initially springs are in natural lengths and cylinder does not slip on ground during oscillaitons due to frications between ground and cylinder. Force constant of each spring is `k`.
Find time period of these oscillation.

Explanation

In the sitution given in problem, the cylinder is in its equilibrium position, when position when springs are unstrained. When it slightly rolled and released, it starts excuting SHM and due to friction, the cylinder is in rollting motion. Now during oscillations we consider the cylinder when it is at a distance x from the mean position and moving with a speed v as shown in figure. As cylinder in pure rolling, its angular speed of rotation can be given as

As centre of cylinder is at a distance x from the initial position, the springs which are connected at a point on the oscillating system can be given as
`E_(T) = (1)/(2)mv^(2)+(1)/(2)(1/2MR^(2))omega^(2) + 1/2k(2x)^(2) xx 2`
Differentiating with respect to time, we get
`(dE_(T))/(dt) = (1)/(2)M(2v(dv)/(dt)) + 1/2(1/2MR^(2)) (1)/(R^(2)) (2v(dv)/(dt)) + 4k (2x(dx)/(dt)) = 0`
`rArr Mva + 1/(2)Mva + 8kxv = 0 rArr a = - (16)/(3) (k)/(M) x`
Comparing equation with basic differntial equation of SHM.
We get, the angular frequency of SHM as `omega = sqrt((16k)/(3M))`
Thus time period these oscillation is `T = (2pi)/(omega) = 2pisqrt((3M)/(16k)) = (pi)/(2)sqrt((3M)/(k))`