The maximum speed of a particle performing a linear S.H.M. is 0.16 m/s and the maximum acceleration is 0.64 `m//s^(2)`. The period of S.H.M. is

Options

A. 1 second
B. 2 second
C. 1.57 second
D. 3.57 second

Answer

C

Explanation

`(a_("max"))/(v_("max"))=(a omega^(2))/(a omega)=omega " " therefore omega =(0.64)/(0.16)=4`
` therefore T=(2pi)/(omega)=(2pi)/(4)=(pi)/(2)=(3.14)/(2)=1.57 s`