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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)pisqrt((k)/(m))`

B

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

C

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

D

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

Text Solution

Verified by Experts

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))`
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