A particle is executing simple harmonic motion with time period 2 s and amplitude 1 cm . If D and d are the total distan…

The period T is 2 seconds, so the number of complete cycles in 12.5 seconds is:

\(n=\frac{12.5}{2}=6.25\text{ cycles. }\)

In one full cycle, the particle moves a distance of 4 A (since it goes from +A to -A and back to +A).
So, in 6.25 cycles, the total distance covered is:

\(D=6.25\times 4A=6.25\times 4\times 1=25cm\)

Since after 6 full cycles, the particle returns to its starting position, it only moves A in the last incomplete cycle.
Therefore, the displacement is d =1 cm.

\(\frac{D}{d}=\frac{25}{1}=25\)