A particle executing SHM with time period of `2 s`: Find the time taken by it to move from one amplitude to half the amplitude position.

Explanation

Let `a` be the amplitude of oscillation then evidently
`omega = (2 pi)/(T) = pi rad//s`
The displlacement equation can betwritten as `x = a sin (pit + phi)` Let, for simplicity, the time be reckoned from the instant when the partical was at its exterme position.
Thus, at `t = 0: x = a`. Therefore .
`alpha = a sin (phi) implies phi = (pi)/(2)`
Let at `t = t_(1), x = a//2`,
`:. (a)/(2) = a sin (pi t_(1) = (pi)/(2))`
`implies (pi t_(1) = (pi)/(2)) = (5 pi)/(6) implies pi t_(1) = (pi)/(3) implies t_(1) = (1)/(3) s`