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

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Answer 1

Total distance covered by the particle $= 4 A$ .

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, $A / 2 = A s i n w t, w = 2 p i / T$ ,

substitute to get $t = T / 12 .$

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

Answer 2

Total distance covered by the particle $= 4 A$ .

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, $A / 2 = A s i n w t, w = 2 p i / T$ ,

substitute to get $t = T / 12 .$

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