Figure shows a pulley block system in which a block `A` is hanging one side of pulley and an other side a small bead `B` of mass `m` is welded on pulley. The moment of intertia of pulley is I and the sytem is in equilibrium position when bead is at an angle `alpha` from the vertical. If the system is slightly disturbed from its equlibrium position, find the time period of its oscillations.

Explanation

In equilibrium the net torque on pulley must be ze4ro, thus we have
`MgR = mgRsinalpha`
or `M = msinalpha` [if mass of block A is assumed tc be M]
Now if block is displace dow by distance A and released, it starts oscillating with amplitude A. Now consider Now if block at distance x below the equilibrium position when it is going down at speed v. Fifure shows the corresponding sitution at this instant and the total energy of oscillating system can be written as
`E_(T) = (1)/(2)Mv^(2) + (1)/(2)mv^(2) + 1/2l((v)/(R))^(2) - Mgx + mgR[cos alpha - cos(theta + alpha)]`

Differenting the abvoe equation w.r.t to time, we get,
`(dE_(T))/(dt) = 1/2M(2v(dv)/(dt))+1/2m(2v(dv)/(dt))+1/2(1)/(R^(2))(2v(dv)/(dt)) - Mg"((dx)/(dt))+mgR[+sin(0+alpha)(d0)/(dt)]=0`
`rArr Mva = mva + (1)/(R^(2))va - Mgv + mgR sin(theta+alpha)((v)/(R)) =0 , ["as" (d theta)/(dt) = omega = (v)/(R)]`
`rArr (M + m+(I)/(R^(2))) a - Mg - mg[cosalpha + sinalpha] = 0` [For small angle `theta , sintheta = theta, cos 1]`
`rArr (M + m + 1/(R^(2))) a + mg cosalpha. x/R = 0 ["as" M = m sin alpha "and" theta = (x)/(R)] rArr a = -(mgcosalpha)/(R(M+m+(I)/(R^(2))))x`
Comparing equation with basic differential equation of SHM, we get the angular frequency of SHM of system
as `omega = sqrt((mgcosalpha)/(R(M+m+(I)/(R^(2)))))`