Figure shows a system consisting of a massles pulley, a spring of force constant `k` and a block of mass `m`. If the block is sligthtly displaced vertically down from its equilibrium and released, find the period of its vertical oscillation in cases (a) and (b).

Explanation

case (a) At point P:F=T

And for the equilibrium of pulley
`2T=F_(s) ` …(ii)
But as due to the shift of point P by y, the spring stretches by `(y//2)`. So,
`F_(s)=k(y//2)` …(iii)
So substituting `F_(s)` from (iii) in (ii) and then T from (ii) in (i) we get
`F=(k//4)y` ..(A)
Case (b) As the tension in massless string and spring will be same,
`T=F'_(s)` ...(i)
For pulley: `F=2F'_(s)`...(ii)

Now if the mass M shifts by y, the spring will stretch by `2y` (as string is inextensible)
`F'_(s)=k(2y)` ...(iii)
So substituting F' from eq, (ii) in (iii)
`F=(4k)y` ..(B)
Force of spring does not change instantaneously.