A particle is executing simple harmonic motion with an angular at a `4cm`. At the mean position the velocity of the particle is `10cm//sec` .The particle from the mean position when its speed becomes `2cm//s` is

Options

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

Answer

C

Explanation

`v = omega sqrt(a^(2) - y^(2))`
At `y = 0 , v_(max) = 10 cm//s`
As `a = 4cm, v_(max) = omega` a hence omega `= 2.5 rad//s`
Hence `5 = 2.5 sqrt(16 - y^(2))`
or ` y = 2sqrt(3) cm`