In the following arrangements, block is slightly displaced vertically down from its equilibrium position and released. Find time period of vertical oscillations. Assume the pulley to be light.

Explanation

If m is equilibrium tension in string must be mg and spring is stretched by h so that `mg = kh`. If the displace the block downward by a distance A and relesed, it starts executing SHM with amplitude A. During its ocillation we consdier the block at a displacement x below the equilibrium position, if it is moving at a speed b at this position, the pulley will be rotating at an angular speed `omega` given as `omega = v/r`

Thus at this position the total energy of oscillating system is `E_(T) = 1/2mv^(2) + 1/2Iomega^(2) + (1)/(2)k(x+h)^(2) - mgx`
Differentiating with respect to time,
we get `(dE_(T))/(dt) = 1/2m(2v(dv)/(dt)) + 1/2I((1)/(r^(2))) (2v(dv)/(dt)) + 1/2k[2(x+h)(dx)/(dt)] - mg(dx/dt) = 0`
or `mva + (I)/(r^(2)) va + k(x+h)v - mgv = 0` or `a+ ((k)/(m+(I)/(r^(2)))) x = 0` [as `mg = kg`]
comparing above equation with standard differentional equation of SHM we get `omega = sqrt(((k)/(m+(I)/(r^(2)))))`
Thus time period of oscillation is `T = (2pi)/(omega) = 2pisqrt((m+I/(r^(2)))/(k))`