A spring block system is kept inside a lift moving with a constant velocity `v_(0)` as shown in figure. Block is in equilibrium and at rest with respect to lift. The lift is suddenly stoped at time `t = 0`. Taking upward direction as the positive direction, write `x - t` equation of the block.

Answer

A

Explanation

Lift is suddenly stopped but block will not stop instantly. It will have same velocity `v_(0)` at `t = 0`. But this is the velocity at its mean position (as it was at rest w.r.t lift). So, this is also its maximum velocity of SHM.
`:. v_(0) = v_(max) = omega A` `rArr A = v_(0)/(omega)`
(where, `omega = sqrt((K)/(m))`
Now, at `t = 0`, block is at mean position and has a velocity `v_(0)` in upward or positive direction.
Therefore, the `x-t`eqution is
`x = Asin omega t`
where, `A = (v_(0))/(omega)` and `omega = sqrt(K)/(m)`