The masses in figure slide on a frictionless table. `m_(1)` but not `m_(2)`, is fastened to the spring. If now `m_(1)` and `m_(2)` are pushed to the left , so that the spring is compressed a distance `d`, what will be the amplitude of the oscillation of `m_(1)` after the spring system is released ?

Answer

A, B, D

Explanation

While returning to equilibrium position,
`(1)/(2)kd^(2) = (1)/(2)(m_(1) + m_(2))v^(2)`
`:. v = (sqrt((k_(1))/(m_(1) + m_(2))))d`
Now, after mean position `m_(2)` is detached from `m_(1)` and keeps on moving with this constant velocity `v` towards right. Block `m_(1)` starts SHM with spring and this `v` becomes its maximum velocity at mean position.
`:. v = omega A`
`:. (sqrt((k)/(m_(1) + m_(2))))d = (sqrt((k)/(m_(1))))A`
`:. A= (sqrt((m_(1))/(m_(1) +m_(2))))d`.