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A body performs simple harmonic oscillat...

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

A

`sqrt(3)/(2)R`

B

`( R)/sqrt(2)`

C

`sqrt(3) R`

D

`sqrt(2) R`

Text Solution

Verified by Experts

The correct Answer is:
C
`(1)/(2)k(A^(2) - x^(2)) = (1)/(4)((1)/(2)kA^(2))`
`:. x = sqrt(3)/(2) A`
`:. CD = CB = sqrt(3)/(2)R`
or `BD = 2 (CD)`
or `2 (CB) = sqrt(3)R`
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