Correct Answer - A
`x = (2 xx 10^(-2))cos pi t`
Here, `a = 2 xx 1-^(-2)m = 2cm`
At t = 0, x = 2 cm, i.e. The object is at positive extreme, so to acquire maximum speed, i.e.to reach mean position it takes `(1)/(4)`th of time period.
`therefore` Required time `= (T)/(4)," "["where", omega = (2pi)/(T)=pi]`
`rArr" "T = 2s`
So, required time `= (T)/(4)=(2)/(4)=0.5s`