A small block of mass m is attached to free end of a spring of force cosntant k , whose other end is fixed to a rigid support . The block is lying on a frictionless horizontal plane and it starts movi0ng away from fixed end of the spring a\t t= 0 with intial velocity u When the speed of the block is `0.6mu` it collides with a fixed heavy mass elastically. After the collision with a fixed heavy mass elastically . Aftere the collision the time at which the block passes through its equilibrium position for the first time is close to

Options

A. `pisqrt(m)/(K)`
B. `1.86sqrt(m)/(K)`
C. `2sqrt(m)/(K)`
D. `2pisqrt(m)/(K)`

Answer

C

Explanation

The given situation is

The collision of bloc with the heavy mass is leastic, So, the block will rebound with same speed u to equilibrium after collision.
In S.H.M. `x = A sin omegat`
`v = A omega cos omegat`
`cos omegat = (3)/(5)`
`cos omegat = (3)/(5)`
`omegat = 0.93`
`t = 0.93 sqrt((m)/(K))`
This is the time taken by th blcok to reach the heavy mass.
`therefore` Time of passing equilibrium position for the first time is
`t_(1) = 2 xx 0.93 sqrt((m)/(K))`
`~~ 1.86sqrt((m)/(K))`