In the string mass system shown in the figure, the string is compressed by `x_(0) = (mg)/(2k)` from its natural length and block is relased from rest. Find the speed o the block when it passes through P (`mg//4k` distance from mean position)

Explanation

`omega = sqrt((3k)/(m))`
`x = A sin (omega t + phi)`
`v = A omega cos (omega t + phi)`
at `t = 0, x = 0`
`rArr phi = 0`
`v = A omega cos omega t`
`A = mg//2k`
`x = A sin omega t`
`(mg)/(4k) = (mg)/(2k) sin omega t`
`sin omega t = (1)/(2) rArr omega t = (pi)/(6) rArr t = (pi)/(6 omega) =(T)/(12)`
`v = A omega cos omega t`
`v = (mg)/(2k) sqrt((3k)/(m)) cos [sqrt((3k)/(m)) (2pi sqrt((m)/(3k))//12)} = g sqrt((9m)/(16))k`