A block of mass m = 1 kg hangs by means of a string which goes over a pulley of mass M = 4m and moment of inertia I=14MR2, as shown in the diagram. The string does not slip relative to the pulley. If the force constant of the spring K = 900 N/m, then find the angular frequency of small oscillations (in rad/s).

At the time of equilibrium : Kx0=2mg+Mg        …(i)If the block is depressed by x, the pulley owing to the constraint is depressed by x2. If the acceleration of the block is a, then the acceleration of the pulley should be a2.Writing equation of motionFor block : T−mg=ma           …(ii)For pulley : Kx0+x2−T−T′−Mg=Ma2               …(iii)The acceleration of point 'P' should be zero as it is directly connected to a fixed point.Hence, αR=a2⇒α=a2R            …(iv)Now writing torque equation for pulley : T′R−TR=Iα=I⋅a2R or T′−T=Ia2R2                …(v)Using above equations, we get a=Kx4m+M+IR2As the acceleration vector and displacement vectors both are opposite to each other,hence above equation can be written in vector form as,a→=K4m+M+IR2⋅x→Comparing with equation : a→=−ω2x→;we get ω2=K4m+M+IR2 or ω=K4m+M+IR2⇒ω=9004×1+4×1+4mR24R2=10 rad/s