A particle starts simple harmonic motion from the mean position. If It's amplitude is A and time period T, then at a certain instant, its speed is half of its maximum speed at this instant, the displacement is

Options

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

Answer

B

Explanation

`v=omegasqrt(A^(2)-x^(2))`
`therefore (A^(2))/(4)=A^(2)-x^(2) therefore (sqrt(3))/(2)A=x`