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Paragraph : A particle vibrates in S.H.M. along a straight line. Its velocity is 4 cm/s when its displacement is 3 cm and velocity is 3 cm/s when displacement is 4 cm. A particle executes simple harmonic motion between x=-A and x=+A. The time taken for it to go from 0 to A/2 is T_(1) and go from A/2 to A is T_(2). Then : |
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Answer» `T_(1)lt T_(2)` `(A)/(2)=A sin OMEGA T_(1)""implies""omega T_(1)=(PI)/(6)""implies""(2pi T_(1))/(T)=(pi)/(6)` `T_(1)=(T)/(12)` `:.` Now `T_(1)+T_(2)=(T)/(4)impliesT_(2)=(T)/(4)-(T)/(12)=(T)/(6)` CLEARLY `T_(1) lt T_(2)`. Correct choice is (a). |
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