1.

A particle is executing shm with time period t starting from mean position the time taken by it to complete 5/8 th oscillations is

Answer»

U can get the answer as....

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 completed 1/2 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 TYM taken = tym to complete
previous ONE half(2A) + tyn taken to complete

A/2 = t/2 + t/12 = 7T/12.

hope it will help you..


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