The amplitude of a executing `SHM` is `4cm` At the mean position the speed of the particle is `16 cm//s` The distance of the particle from the mean position at which the speed the particle becomes `8 sqrt(3)cm//s` will be

Options

A. `2sqrt(3)` cm
B. `sqrt(3)` cm
C. 1 cm
D. 2 cm

Answer

D

Explanation

At mean position velocity is maximum
i.e. `""^(v)"max"=omega aimpliesomega=(v_("max"))/(a)=(16)/(4)=4`
`:. V=omega sqrt(a^(2)-y^(2))implies8sqrt(3)=4sqrt(4^(2)-y^(2))`
`implies192=16(16-y^(2))implies12=16-y^(2)impliesy=2" cm"`.