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A particle executes `SHM` on a straight line. At two positions, its velocities are `u` and `v` whle accelerations are `alpha` and `beta` respectively `[beta gt alpha gt ]`.The distance between these two positions isA. `(u^(2)-v^(2))/(alpha+beta)`B. `(u^(2)+v^(2))/(alpha+beta)`C. `(u^(2)-v^(2))/(alpha-beta)`D. `(u^(2)+v^(2))/(alpha-beta)` |
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Answer» Correct Answer - A Let distance by `x_(1)` when velocity if u and accleration `alpha`. Let distance by `x_(2)` when velocity is v and acceleration `beta`. If `omega` is the angular frequencey then, `alpha=omega^(2)x_(2)` `therefore alpha+beta=omega^(2)(x_(1)+x_(2)),.......(1)` `"Also," u^(2)=omega^(2)A^(2)-omega^(2)x_(1)^(2)` `and v^(2)=omega^(2)A^(2)-omega^(2)x_(2)^(2` `v^(2)-u^(2)=omega^(2)(x_(1)^(2)-x_(2)^(2))` `v^(2)-u^(2)=omega^(2)(x_(1)-x_(2))(x_(1)+x_(2)).....(ii)` By. eq (i) we get `v^(2)-u^(2)=(x_(1)-x_(2))(alpha+beta)` `therefore x_(1)-x_(2)=(u^(2)-v^(2))/(alpha+beta) or x_(2)-x_(1)=(u^(2)-v^(2))/(alpha+beta)` |
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