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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` |
Answer» Correct Answer - 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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