Consider the situation as shown. The pulley has a moment of inertia I, and radius r, the block is in equilibrium. Find the time period of mass m. Assume no slipping.

Explanation

Let at an instant extension of spring is x (after equilibrium), speed of block is v and angular speed of pulley `omega`.

In equilibrium
`mg=kx_0, x_0`: initial extension of spring
The energy of the system
`E=1/2mv^2+1/2Iomega^2-mgx+1/2k(x+x_0)^2`
`=1/2mv^2+1/2I(v/r)^2-mgx+1/2k(x+x_0)^2`
`=1/2mv^2(1+(I)/(mr^2))-mgx+1/2k(x+x_0)^2`
Since total energy is constant
`(dE)/(dt)=0`
`0=1/2m*2v(dv)/(dt)(1+(I)/(mr^2))-mg((dx)/(dt))`
`+1/2k*2(x_0+x)(dx)/(dt)`
`=mva(1+(I)/(mr^2))-mgv+kx_0v+kxv`
`=ma+I/r^2a+kx`
`(m+I/r^2)a=-kx`
`a=-((k)/(m+I//r^2))x=-omega^2x`
`T=2pisqrt((m+I//r^2)/(k))`