Figure shows a system consisting of a massless pulley, a spring of force constant k and a block of mass m. If the block is slightly displaced vertically down from its equilibrium position and then released, the period of its vertical oscillation is

Let us assume that in equilibrium condition spring is x0 elongate from its natural lengthIn equilibrium T0 = mg and kx0 = 2T0⇒kx0 = 2mg-------------(1)If the mass m moves down a distance x from its equilibrium position then pulley will move down by x2. So the extra force in spring will be kx2. From figureFnet= mg-T = mg-k2(x0+x2)Fnet = mg-kx02-kx4from eq.(1), we getFnet = -kx4-----------------(2)Now compare eq.(2) with F = -KSHMxthen KSHM = K4⇒ T = 2πmKSHM = 2π4mK = 4πmK