Figure shown a spring block system hanging in equilibrium. If a velocity `v_(0)` is imparted to the block in downwards direction . Find the amplitude of SHM of the block and the time after which it will reach a point at half the amplitude of block

Explanation

Initially in equlibrium if block is at a depth h below the nature length of length of spring , then we have
`mg = kh`
If at mean position block is imparted a velocity of block during its oscillation. If its amplitude of oscillation is A , then it is given as
`v_(0) = A omega [where omega = sqrt((k)/(m))]`
or `A = (v_(0))/(omega) = v_(0) sqrt(m)/(k)`
Now to find the taken by block its half of amplitude point, we SHM as shown in the figure.

Here block p will reach to half of its amplitude when particle `p_(0)` will reach point E shown in the figure at an angular displacement `theta` from mean position relative to point A,
thus time by it is `t = (theta)/(omega) = (sin^(-1) ((1//2))/(sqrt(k//m))) = (pi)/(6) sqrt((m)/(k))`