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A particle performs SHM of time period T...

A particle performs `SHM` of time period `T` , along a straight line. Find the minimum time interval to go from position `A` to position `B`. At `A` both potential energy and kinetic energy are same and at `B` the speed is half of the maximum speed.

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The correct Answer is:
`(T)/(24)`
`x_(A)=(A)/(sqrt(2))`
and for `x_(B),(omegaA)/(2)=omegasqrt(A^(2)-x_(B)^(2))`
or `x_(B)=(sqrt(3))/(2)A`
or `omega t_(A)=(pi)/(4)` and `omega t_(B)=(pi)/(3)`
or `omega (t_(B)-t_(A))=(pi)/(3)-(pi)/(4)`
or`(pi)/(3)-(pi)/(4)=(2pi)/(T)t`
or `t=(T)/(2pi)xx(pi)/(12)=(T)/(24)`
Ans. `(T)/(24)`
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