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The time period of a particle executing ...

The time period of a particle executing `SHM` is `T`. After a time `T//6` after it passes its mean position, then at`t = 0` its :

A

displacement will be one-half of its amplitude

B

velocity will be one-half of its maximum velocity

C

kinetic energy `= 1//3` (potential energy)

D

acceleration will be `(sqrt(3))/(2)` times of its maximum acceleration

Text Solution

Verified by Experts

The correct Answer is:
B,C,D
Let `x = Asinomegat` where `omega = (2pi)/(T)`
For `(A) = x = A sin ((2pi)/T) (T/6) = Asin((pi)/3) = (sqrt(3))/(2)`
For (B) : `v = (dx)/(dt) = Aomegacosomegat` At `t = (T)/(6), v = Aomegacos((2pi)/T) (T/6) = Aomegacos"(pi)/(3) = (A)/(2)`
For (C) : `KE = (1)/(2)m ((v_("max"))/(2))^(2) = 1/4(KE)_("max") = 1/4(TE)` & `PE = TE - KE = 3/4 TE rArr KE = 1/3 (PE)`
For (D) : Acceleration `a = (dv)/(dt) = - Aomega^(2) sin omegat = - omega^(2) x`
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