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

Answer

D

Explanation

At the mean position, the speed will be maximum,
`V_(max)=16 cms^(-1)=a omega = 4 omega`
So, `omega = (16)/(4) = 4 rads^(-1)`
and `v = omega sqrt(A^(2)-y^(2))`
`8 sqrt(3) = 4 sqrt(A^(2)-y^(2)) or A^(2) - y^(2) = 12`
or `y^(2) = A^(2) - 12 = (4^(2)) - 12 = 4 or y = 2` cm