The maximum speed of a particle executing S.H.M. is 1 m/s and its maximum acceleration is`1.57 m// sec^(2)` . The time period of the particle will be

Options

A. 0.25 `s^(-1)`
B. 2 `s^(-1)`
C. 1.57 `s^(-1)`
D. 2.57 `s^(-1)`

Answer

A

Explanation

`v_(0) = r" "omega = 1 and a_(0) = omega^(2)r = 1.57 ms^(-2)`
`omega = omega^(2)r//r omega = 1.57//1 or 2 pi v = 1.57`
`v = (1.57)/(2 xx 3.14)=(1)/(4)=0.25 s^(-1)`