A particle is executing simple harmonic motion with time period $$2\,s$$ and amplitude $$1\,cm.$$ If $$D$$ and $$d$$ are the total distance and displacement covered by the particle in  $$12.5\,s,$$ then  $$\frac{D}{d}$$ is:

A particle executes SHM with time period $$T = 2$$ s and amplitude $$A = 1$$ cm. We need the ratio $$D/d$$ after 12.5 s, where $$D$$ = total distance and $$d$$ = displacement.

Since $$12.5 \text{ s} = 6 \times 2 + 0.5 = 6T + T/4$$, the particle completes 6 full periods and an additional $$T/4$$.

In each complete period the particle traverses a distance of $$4A$$, so in 6 complete periods the distance is $$6 \times 4A = 24A$$. In the additional $$T/4$$ the particle moves from the mean position to the extreme, covering a distance $$A$$. Therefore $$D = 24A + A = 25A = 25 \text{ cm}$$.

After 6 complete periods the displacement is zero, and during the additional $$T/4$$ the particle reaches the extreme, so $$d = A = 1 \text{ cm}$$.

Therefore $$\frac{D}{d} = \frac{25}{1} = 25$$, which corresponds to Option (4): 25.