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

$\mathrm{A}=1\mathrm{cm}$

Step 1: Number of Complete Oscillations

The number of complete oscillations in time $t$ is:

$$N = \frac{t}{T} = \frac{12.5}{2} = 6.25$$

So the particle completes 6 full oscillations and one-fourth (0.25) of an oscillation.

Step 2: Total Distance $D$ Covered

One full oscillation covers a total distance of $4A = 4 \times 1 = 4$ cm.

In 6 full oscillations, the total distance is:

• $$D_6 = 6 \times 4 = 24 \text{ cm}$$

In the remaining 0.25 oscillation, the particle moves from mean position to amplitude $A$, which adds:

• $$D_{0.25} = A = 1 \text{ cm}$$

Thus, the total distance covered is:

$$D = 24 + 1 = 25 \text{ cm}$$

Step 3: Total Displacement $d$

• Since the motion starts from the mean position and completes 6 full oscillations, it returns to the mean position after 6 oscillations.
• In the remaining 0.25 oscillation, it moves to amplitude $A$ (1 cm).

Thus, the net displacement from the initial position is:

$$d = 1 \text{ cm}$$

Step 4: Find the Ratio $\frac{D}{d}$

$$\frac{D}{d} = \frac{25}{1} = 25$$

Final Answer:

$$\boxed{25}$$