Correct Answer - A::B::C
`A = 4cm`
`omega = ((pi)/(2))rad//s`
`:. T = (2pi)/(T) = 4sec`
The given equation is `x = Acos omega t`
Therefore, the particle starts from `x = + A`
The given time `t = 3 sec` is `(3T)/(4)`. So, the particle moves from `+ A` to `-A` and then from `- A` to `O`.
So, distance travelled in the given time interval is `3A` or `12cm`.