InterviewSolution
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InterviewSolution
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A particle performs SHM of amplitude A along a straight line. When it is at distance `sqrt(3)/2` A from mean position, its kinetic energy gets increased by an amount `1/2momega^(2)A^(2)` due to an impulsive force. Then its new amplitude becomes.A. `(sqrt(5))/(2)A`B. `(sqrt(3))/(2)A`C. `sqrt(2)A`D. `sqrt(5)A` |
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Answer» Correct Answer - C at `x = (sqrt(3))/(2)A` `KE = 1/2 momega^(2) (A^(2) - (3)/(4)A^(2)) = 1/8 momega^(2)A^(2)` `KE` is increased by an amount of `1/2momega^(2)A^(2)`. Let now amplitude be `A_(1)` then total `KE` `KE_(1) = 1/8momega^(2)A^(2) + 1/2 momega^(2)A^(2)` `= 5/8momega^(2)A^(2) = 1/2momega^(2)(A_(1)^(2) - (3)/(4)A^(2)) rArr A_(1) = sqrt(2)A` |
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