if the displacement of the particle at an instant is given by
`y = r sin (omega t - theta)`
where r is amplitude of oscillation. `omega` is the angular velocity and `-theta` is the initial phase of the particle, then find the particle velocity and particle acceleration.

Answer

`r omega cos (omega t - theta) ; - omega^2 y`

Explanation

`y =r sin (omegat = theta)`
Velocity, `upsilon = (dy)/(dt) =r cos (omega t -theta)(d)/(dt)(omega t - theta) = r cos (omegat - theta)xx omega =r omega cos (omegat - theta)`
Acceleration, `A =(d upsilon)/(dt) = r omega [-sin(omegat - theta)]xxomega=- omega^2 r sin (omegat - theta) =- omega^2 y`