The dispalcement of an object attached to a spring and excuting simple harmonic motion is given by `x = 2 xx 10^(-2) cospit` metres. The time at which at maximum speed first occurs is :

Options

A. 0.5 s
B. 0.75 s
C. 0.125 s
D. 0.25 s

Answer

A

Explanation

`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`