A spring block pendulum is shown in figure . The system is hanging in equilibrium. A bullet of mass `m//2` moving at a speed u bites the block from downwards direction and gets embedded in it. Find the amplitude of oscillation of the block now.

Explanation

If block is in equilibrium , then spring must be at some steretch if it is h, have `mg = kh`.
If a bullet of mass `m//2` gets embedded in the block, due to this inelastic impact its new mass becoms `3m//2` and new mean positionof the block will be , say at a depth `h_(1)` from old mean position, then we must have
`(3)/(2) mg = k(h + h_(1))`
`(3m)/(2) g = k(h + h_(1))`
`h_(1) = (mg)/(2k) (as mg = kh)`

just after impact due to inelastic collition if the velocity of block becomes v, we have, according to momentum conservation.
`(m)/(2) mu = (3 m)/(2) nu or nu = (u)/(3)`
Now the block executes SHM and at `t = 0` block is at a distance `h = (mg)/(2k)` above its mean position and having a velocity `u//3`. If amplitude of oscillation is A, we have
`(u)/(3) = omega sqrt(A^(2) - ((mg)/(2 k))^(2))`
or `(u^(2))/(9) = (2k)/(3m) [A^(2) - ((mg)/(2k))^(2)]`
[as for this spring block system omega `= sqrt((k)/((3m//2))) ]`
`A = sqrt((mu^(2))/(6k) + ((mg)/(2k))^(2))`
Now time taken by partical to reach the topmost point can be obtained by circular motion representation as shown in figure. This figure shows the position of block P and its corresponding circular motion partical `p_(0)` at the`t = 0`, Block P will reach its upper extereme position when partical `P_(0)` will traveres the angle `theta` and reach the topmost point as `P_(0)` moves at constant angular velocity `omega`, it will take a time given as
`t = (theta)/(omega) = (cos^(-1) (h_(1)//A))/sqrt((k)/(3m//2)) = sqrt((3m)/(2k)) cos^(-1) ((3 mg)/(2 kA))`