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A particle executed S.H.M. starting from...

A particle executed S.H.M. starting from its mean position at `t=0`, If its velocity is `sqrt3bomega`, when it is at a distance b from the mean positoin, when `omega=(2pi)/(T)`, the time taken by the particle to move from b to the extreme position on the same side is

A

`(5pi)/(6omega)`

B

`(pi)/(3omega)`

C

`(pi)/(2omega)`

D

`(pi)/(4omega)`

Text Solution

Verified by Experts

The correct Answer is:
B
`v^2=omega^2(A^2-b^2)=3omega^2b^2impliesA^2=4b^2`
`b=(A)/(2)=Asinomegatimpliest=(pi)/(6omega)`
Required time `t_1=(T)/(4)-t=(2pi)/(4omega)-(pi)/(6omega)=(pi)/(3omega)`
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Knowledge Check

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    A
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