A partical excutes SHM with an angular an amplitude `8 cm` and a frequency `4 cm` initially, in the possitive direction , determine its displecement equation and the maximum velocity and acceleration.

Explanation

Given `a = 8` cm and `omega = 2 pi n = 20 pi rad//s`
Let the phase constant be `phi`
The displacement equation can be written as
`x = 8 sin (20 pi t + phi)`.
Given at `t = 0: x = 4cm`, therefore
`4 = 8 sin (20pi (0) + phi) implies sin (phi) = (1)/(2) implies phi = (pi)/(6)`
`:. "Displacement equation" x = 8 sin (20pi t + (pi)/(6))`
Differenting the above equation w.r.t. time 't'
`(dx)/(dt) = v = 160pi cos (20pi + (pi)/(6))`
`:. v_(max) = +- 160 picm//s [when cos (20 pi t + pi//6) = +-1]`
Differenting once again w.r.t. time 't'
`(d^(2) x)/(dt^(2)) = f = - 3200 pi^(2) sin (20pi t + (pi)/(6))`
`f_(max) = +- 3200 pi^(2)`