1. Given data

• Time period: $T = 2\,\text{s}$
• Amplitude: $A = 1\,\text{cm}$
• Total time: $t = 12.5\,\text{s}$

We need to find:

$\frac{D}{d}$

where:

• $D$ = total distance travelled
• $d$ = magnitude of displacement in $12.5\,\text{s}$

2. Find number of oscillations in 12.5 s

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

So the particle completes:

• $6$ full oscillations, and
• an additional $0.25T = \frac{T}{4}$

3. Total distance travelled in one complete oscillation

In one full SHM cycle, the particle goes from one extreme to the other and back, so total distance is:

$4A$

Since $A = 1\,\text{cm}$,

$\text{distance in 1 cycle} = 4\times 1 = 4\,\text{cm}$

Thus, in $6$ full cycles:

$D_1 = 6\times 4 = 24\,\text{cm}$

4. Distance travelled in the remaining }\frac{T}{4}\text{

In SHM, during a quarter period, the particle covers distance equal to amplitude $A$ (whether from mean to extreme or extreme to mean).

Hence,

$D_2 = A = 1\,\text{cm}$

So total distance is:

$D = D_1 + D_2 = 24 + 1 = 25\,\text{cm}$

5. Displacement after 12.5 s

After $6$ complete oscillations, the particle returns to its initial position.

Then in the extra $\frac{T}{4}$, the particle moves by a displacement of magnitude $A$ from its starting point of that quarter-cycle.

Therefore,

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

6. Compute the ratio

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

7. Option check

• A: $10$ ❌
• B: $\frac{16}{5}$ ❌
• C: $25$ ✅
• D: $\frac{15}{4}$ ❌

Hence, the correct answer is:

$\boxed{25}$