The string, spring and the pulley shown in figure are light. Find the time period of the mass `m`.

Explanation

Force Method
Let in equilibrium position of the block, extension in spring is `x_(0)`.
`:. Kx_(0) -1`
Now if we displace the block by `x` in the downward direction, net force on the block towards mean position is
`F = k(x + x_(0)) - mg, = kx` using `(1)`
Hence the net force is acting towards mean position and is also propotional to `x`. So, the particle will perform `S.H.M.` and its time period would be
`T = 2pi sqrt((m)/(k))`
(b) Energy Method
Let gravitational potential energy is to be zero at the level of the block when spring is in its natural length.
Now at a distance `x` below that level, let speed of the block be `v`.
Since total mechcanical energy is conserved in `S.H.M.`
`:. -mgx + (1)/(2)kx^(2) + (1)/(2)mv^(2) = constant`
Differentiating `w.r.t.` time, we get
`-mgv + kxv + mva = 0`
where `a` is accleration.
`:. F = ma = -kx + mg` or `F = -k(x - (mg)/(k))`
This shows that for the motion, force constant is `k` and equilibrium position is `x = (mg)/(k)`.
So, the particle will perform `S.H.M` and its timne period would be `T = 2pisqrt((m)/(k))`