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

Answer

D

Explanation

At mean position velocity is
maximum `i.e.," "v_(max)=omegaA`
`impliesomega=(v_(max))/(A)=(16)/(4)=4`
`thereforev=omegasqrt(A^(2)-y^(2))`
`implies8sqrt(3)=4sqrt(4^(2)-y^(2))`
`impliesy=2cm`