In the figure shown the spring is relaxed and mass `m` is attached to the spring. The spring is compressed by `2A` and released at `t = 0`. Mass `m` collides with the wall and loses two third of its kinetic energy and returns. Starting from `t = 0`, find the time taken by it to come back to rest again (instant at which spring is given under maximum compression). Take `sqrt(m)/(k) = (12)/(pi)`

Answer

17

Explanation

The motion starts from position `A` the time taken from `A` to `W_(2) (t_(1)) = (T)/(4) + (T)/(12)`
Before collision the energy of the block just before collision
`K_(i) = (1)/(2)K(2A)^(2) - (1)/(2) KA^(2)` & just after collision `K_(1) = (K_(f))/(3)` (given) `= (1)/(2)KA^(2)`
Now during motion after collision, the energy is again conserved
Hence, `K_(1) + (1)/(2) KA^(2) = (1)/(2)KA^('2)`
`A' =` maximum compresion after collision `rArr = Asqrt(2)`
ie. Now motion has amlitude `Asqrt(2)`
Now time taken by block from
`W_(2)` to positon `B = (T)/(4) + (T)/(8)`
`:.` total time taken `= t_(1) + t_(2) = (T)/(4) + (T)/(12) + (T)/(4) + (T)/(8) = (17)/(24)T = (17pi)/(12)sqrt((m)/(K)) = 17sec {sqrt((m)/(k)) = (12)/(pi)}`