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

Answer

D

Explanation

At mean position, velocity is maximum
`V_(max)=omega_(a) and omega = (V_(max))/(a)=(16)/(4) = 4`
`therefore" "v= omega sqrt(a^(2)-y^(2))rArr 8 sqrt(3) = 4 sqrt(4^(2)-y^(2))`
`192 = 16(16 - y^(2))`
or `12 = 16 - y^(2) rArr y = 2 cm`