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

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 from A to C
`= t_(AC) = t_(AB) + t_(BC) = (T//4) + t_(BC)`
where T = time period of oscillation of spring mass system
`t_(BC)` can be 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))`
Hence (C ) is correct