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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
When the particle crosses `D` , its speed is half the maximum speed.

`:. V = (v_(max))/(R ) sqrt(R^(2) - x^(2))`
`or (v_(max))/(2) = (v_(max))/(R)sqrt(R^(2) - x^(2))` or `x = sqrt(3)/(2)R`
`:.` Distance `BD = 2x = sqrt(3)R`
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