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

Options

A. `(1)/(6)pi sqrt((k)/(m))`
B. `sqrt((k)/(m))`
C. `(2pi)/(3) sqrt((m)/(k))`
D. `(pi)/(4) sqrt((k)/(m))`

Answer

C

Explanation

The total time A to C

`I_(AC) = t_(AB) + t_(BC)`
`= (T//4) +t_(BC)`
where `T = ` time period of oscilation of spring mass system
`t_(BC) =` can obtained from ,`BC = AB sin (2pi//T)t_(BC)`
Putting `(BC)/(AB) = (1)/(2)` we obtain `t_(BC) = (T)/(12)`
`rArr t_(AC) = (T)/(4) + (T)/(12) = (2pi)/(3) sqrt((m)/(k))`.