A particle executes simple harmonic motion with an amplitude of 4 cm . At the mean position the velocity of the particle is 10 cm/ s . The distance of the particle from the mean position when its speed becomes 5 cm/s is

Options

A. `sqrt(3)` cm
B. `sqrt(5)` cm
C. `2 sqrt(3)` cm
D. `2 sqrt(5)` cm

Answer

C

Explanation

`V_(max) = a omega`
`rArr" "omega = (V_(max))/(a) = (10)/(4) rads^(-1)`
Now, `v = omega sqrt(a^(2) - y^(2))`
`rArr" "v^(2) = omega^(2) (a^(2)-y^(2))`
`rArr" "y^(2) = a^(2) - (v^(2))/(omega^(2))`
`rArr" "y = sqrt(a^(2) - (v^(2))/(omega^(2)))`
`= sqrt(4^(2) - (5^(2))/((10//4)^(2)))=2 sqrt(3)` cm