The amplitude of a particle executing SHM is 4 cm. At the mean position, the speed of the particle is 16 `cms^(-1)`. The distance of the particle from the mean position at which the speed of the particle becomes `8 sqrt(3) cms^(-1)` will be

Options

A. `2sqrt3` cm
B. `sqrt3` cm
C. 1 cm
D. 2 cm

Answer

D

Explanation

At mean position velocity is maximum
i.e., `v_"max" = omega a rArr omega = v_"max"/a =16/4=4`
`v=omega sqrt(a^2-y^2) rArr 8sqrt3 = 4sqrt(4^2-y^2)`
`rArr 192=16(16-y^2) rArr 12=16-y^2 rArr y=2 cm `