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 is equilibrium position and then released, the period of its vertical oscillation is

Explanation

Case (A)
As the pully is fixed and string in inexrensible, if mass m is displaced by y the spring will stretch by y, and as there is no mass between string and spring (as pully is massless) `F = T = ky` i.e., restoring force is linear and so motion of mass m will be linear simple harmonic with frequnecy
`n_(A) = (1)/(2pi) sqrt((k)/(m)) = 1/(2pi) sqrt((4000)/(1)) ~~ 10 Hz`

Case (B) :
The pulley is mobale and string inetensible, so if mass m moves down a distance y, the pulley will move down by `(y//2)`.
Now as pulley is massless `F = 2T, rArr T = f//2 = (k//4y)`.
So the force in the spring `F = k(y//2)`
So, `n_(B) = (1)/(2pi)sqrt((k)/(m)) = (1)/(2pi) sqrt((k)/(4m)) = (n_(A))/(2) = 5 Hz`

Case (C) :
In this sitution if the mass m moves by y the pulley will also move by y and so the string will stretch by `2y` (as string is inextensible) and so `T' = F = 2ky`.Now as pulley is massless so `T = F + T' = 4ky`, i.e., the restoring force on the mass m
`T = 4ky = k' = 4k`
so `n_(C) = (1)/(2pi)sqrt((k')/(m)) = (1)/(2pi)sqrt((4k)/(m)) = 2n_(A) = 20 Hz`