A particle is executing SHM with time period 'T.Starting from mean position,time taken by it to complete 5/8 oscilllations,is..........

Pls explain..........

Hey vishwajit,

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.