One end of a spring of force constant k is fixed to a vertical wall and the other to a blcok of mass m resting on a smooth horizontal surface. There is another wall at distance `x_(0)` from the block. The spring is then compressed by `2x_(0)` and released. The time taken to strike the wall is

Explanation

Let the position of block be A when it is compressed by `2x_(0)`. When it is released it will be executing simple harmonic motion. Let the time period be T and amplitude of resulting SHM will be `2x_(0)`.
Time taken by the block from A to B is `t_(1) = (T)/(4)`
Time taken `(t_(2))` from B to C by the block can be calculated. As equation of motion is
`x = A sin omega t rArr x_(0) = 2 x_(0) sin omega t_(2)`

`rArr" "sin omega t_(2) = (1)/(2) rArr omega t_(2) = (pi)/(6)`
`rArr" "t_(2) = (pi.T)/(6.2 pi) = (T)/(12)`
Total time `t = t_(1) + t_(2) = (T)/(3)`
`rArr" "t = (2pi)/(3) sqrt((m)/(k))" "because" "T = 2pi sqrt((m)/(k))`