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 $\frac{\mathrm{D}}{\mathrm{d}}$ is

<p>$x(t) = \cos(\pi t)$</p> <p>The particle executes simple harmonic motion (SHM) with amplitude $A = 1 \text{ cm}$ and period $T = 2 \text{ s}$. In one complete cycle (2 s), the motion is as follows:</p> <p><p>It starts at $x = 1 \text{ cm}$.</p></p> <p><p>It moves to $x = -1 \text{ cm}$ (covering a distance of $2 \text{ cm}$).</p></p> <p><p>It returns to $x = 1 \text{ cm}$ (covering another $2 \text{ cm}$).</p></p> <p>Thus, the total distance traveled in one complete cycle is:</p> <p>$D_{\text{cycle}} = 2 + 2 = 4 \text{ cm}.$</p> <p>In 12.5 s, the number of cycles is:</p> <p>$\frac{12.5}{2} = 6.25 \text{ cycles}.$</p> <p>For the 6 complete cycles (12 s), the distance traveled is:</p> <p>$D_{\text{complete}} = 6 \times 4 = 24 \text{ cm}.$</p> <p>For the remaining 0.5 s, determine the displacement. At $t = 12 \text{ s}$, the particle is at:</p> <p>$x(12) = \cos(12\pi) = 1 \text{ cm}.$</p> <p>After an extra 0.5 s (i.e., at $t = 12.5 \text{ s}$), the position becomes:</p> <p>$$ x(12.5) = \cos(12.5\pi) = \cos\left(12\pi + \frac{\pi}{2}\right) = \cos\frac{\pi}{2} = 0 \text{ cm}. $$</p> <p>The distance covered in this interval is:</p> <p>$D_{\text{extra}} = |1 - 0| = 1 \text{ cm}.$</p> <p>So, the total distance traveled is:</p> <p>$D = D_{\text{complete}} + D_{\text{extra}} = 24 + 1 = 25 \text{ cm}.$</p> <p>The net displacement, which is the difference between the final and initial positions, is calculated as follows:</p> <p><p>Initial position at $t = 0$:</p> <p>$x(0) = \cos 0 = 1 \text{ cm},$</p></p> <p><p>Final position at $t = 12.5 \text{ s}$:</p> <p>$x(12.5) = 0 \text{ cm}.$</p></p> <p>Thus, the displacement is:</p> <p>$d = |0 - 1| = 1 \text{ cm}.$</p> <p>The ratio of the total distance traveled to the displacement is:</p> <p>$\frac{D}{d} = \frac{25}{1} = 25.$</p> <p>Therefore, the correct answer is $25$.</p>