InterviewSolution
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InterviewSolution
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A body executes `SHM`, such that its velocity at the mean position is `1ms^(-1)` and acceleration at exterme position is `1.57 ms^(-2)`. Calculate the amplitude and the time period of oscillation. |
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Answer» `(a_(max))/(v_(max)) = (A•^(2))/(A•) = (1.57)/(1) rArr• = 1.57 rad` `:.` Time period `T = (2•)/(1.57) = (2(3.14))/(1.57) = 4s`. but `A• = 1`, i.e., `A (1.57) = 1 or A = (1)/(1.57)` `:.` Amplitude `A = 0.637m`. |
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