The time period of oscillation of a particle that executes `SHM` is `1.2s`. The time starting from mean position at which its velocity will be half of its velocity at mean position is

Options

A. `2 A`
B. `(sqrt(3))/(2)xxA`
C. A
D. `(A)/(2)`

Answer

B

Explanation

`V=(V_(m))/(2)`
`omegasqrt(A^(2)-x^(2))=(Aomega)/(2)`
`A^(2)-x^(2)=(A^(2))/(4)`
`A^(2)-(A^(2))/(4)=x^(2)`
`x=sqrt(3)/(2)A`.