Block of mass `m_(2)` is in equilibrium and at rest. The mass `m_(1)` moving with velocity u vertically downwards collides with `m_(2)` and sticks to it. Find the energy of oscillation.

Explanation

At equilibrium position `m_(2)g = kx_(0)`
`implies x_(0) = (m_(2)g)/(K)`
After collision m2 sticks to `m_(1)`.
`:. " "` By momentum conservation.
`m_(1)u = (m_(1)+m_(2))v`
`v= (m_(1)u)/(m_(1)+m_(2))`
Now both the blocks are executing S.H.M. which can be interpreted as follows:
Now , we know that `v^(2) = omega^(2) (A^(2)-x^(2)) " " cdots (1)`
`omega^(2) = (k)/(m_(1)+m_(2))`
`= > x = (m_(1)g)/(k) `
Put the values of v, `omega^(2)` & x in eq. (1)
`((m_(1)u)/(m_(1)+m_(2)))^(2)=((k)/(m_(1)+m_(2)))[A^(2)-((m_(1)g)/(k))^(2)]`

`implies kA^(2)= [((m_(1)^(2)u^(2))/(m_(1)+m_(2)))+((m_(1)g)/(k))^(2)]`
`implies ` Energy of oscillation =` (1)/(2)kA^(2)`
`=(1)/(2)[((m_(1)^(2)u^(2))/(m_(1)+m_(2)))+((m_(1)^(2)g^(2))/(k))]`