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A particle performs simple harmonic mition with amplitude A. Its speed is trebled at the instant that it is at a destance `(2A)/3` from equilibrium position. The new amplitude of the motion is:A. `(A)/(3)sqrt(41)`B. `3A`C. `Asqrt(3)`D. `(7A)/(3)` |
Answer» Correct Answer - D Applying conservation of energy `(1)/(2) m omega^(2) ((2A)/(3))^(2) + 9 xx (1)/(2) m omega^(2) [A^(2) -((2A)/(3))^(2)]` `= (1)/(2) m omega^(2) A_(1)^(2)` `rArr A_(1) = (7A)/(3)` |
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