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One end of a spring of force constant k ...

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

A

`(1)/(6) pi sqrt((k)/(m))`

B

`sqrt((k)/(m))`

C

`(2pi)/(3) sqrt((m)/(k))`

D

`(pi)/(4) sqrt((k)/(m))`

Text Solution

Verified by Experts

The correct Answer is:
C
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
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