Question

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

Answer 1

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 time taken = time to complete previous one half(2A) + time taken to complete$A/2=t/2+t/12=7T/12.$