The string the spring and the puley 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))`