A body performs simple harmonic oscillations along the straight line `ABCDE` with `C` as the midpoint of `AE`. Its kinetic energies at `B` and `D` are each one fourth of its maximum value. If `AE = 2R`, the distance between `B` and `D` is

Options

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

Answer

C

Explanation

A and E are extreme positions and C is the mean position
`K_(D) = (1)/(4) K_(max)`
`(1)/(2) mv^(2) = (1)/(4).(1)/(2) m (A omega)^(2)`
`rArr v = (1)/(2)A omega`
`rArr omega sqrt((A^(2) - x^(2))) = (A omega)/(2) rArr x = (sqrt(3)A)/(2)`
distance between B and D `rArr 2x = sqrt(3) A`