A partical excutes SHM with an angular frequency `omega = 4 pi rad//s`. If it is at its extereme position initially, then find the transition when it is at a distance `sqrt2//2` times its amplitude from the mean position.

Explanation

Let a be the amplitude and `phi` the phase constant for the ascillation.
The displacement equation can be written as
`x = a sin (4 pi t + phi) [:' omega = 4 pi rad// s]`
Since at `t = 0: x = + a`, therefore
4 pi (0) + phi = (pi)/(2) implies phi = (pi)/(2)`
x = a sin (4 pi t + pi//2) = a cos (4 pi t)`
Now let `t_(n)` be the instant when the partical's displacement is `sqrt2a//2`. nth
Evidently `+- (sqrt3a)/(2) = a cos (4 pi t_(n))`
` implies 4 pi t_(n) = n pi +- (pi)/(6) implies r_(n) = (n^(2))/(4) +- (1)/(24)`
Taking only the positive value for `t`. ,
`t_(1) = (1)/(24) s, t_(2) = (5)/(24) s,t_(3) = (7)/(24) s,t_(4) = (11)/(24) s,t_(5) = (13)/(24) s`. etc.