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
Answers
Answered by
252
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..
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..
Similar questions