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Question

A particle is executing SHM with time period T. Starting from mean position, time taken by it to complete 5/8 oscillations, is

Solution
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Total distance covered by the particle=4A.

we divide this whole path in 8 intervals of A/2.

so, 5/8 oscillations means, it has already completed1/2 oscillation(i.e. total dist. = 2A) and is half way to the other side i.e. A/2.

so,A/2=Asinwt,w=2pi/T ,

substitute to get t=T/12.

now, total time taken = time to complete previous one half(2A) + time taken to completeA/2=t/2+t/12=7T/12.

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