QJEEM25Phabc6: JEE Main 2025 Physics - Class 11 JEEM25Phabc6 Solution: A particle is executing simple harmonic motion with time period …

$x(t) = \cos(\pi t)$

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:

It starts at $x = 1 \text{ cm}$ .

It moves to $x = -1 \text{ cm}$ (covering a distance of $2 \text{ cm}$ ).

It returns to $x = 1 \text{ cm}$ (covering another $2 \text{ cm}$ ).

Thus, the total distance traveled in one complete cycle is:

$D_{\text{cycle}} = 2 + 2 = 4 \text{ cm}.$

In 12.5 s, the number of cycles is:

$\frac{12.5}{2} = 6.25 \text{ cycles}.$

For the 6 complete cycles (12 s), the distance traveled is:

$D_{\text{complete}} = 6 \times 4 = 24 \text{ cm}.$

For the remaining 0.5 s, determine the displacement. At $t = 12 \text{ s}$ , the particle is at:

$x(12) = \cos(12\pi) = 1 \text{ cm}.$

After an extra 0.5 s (i.e., at $t = 12.5 \text{ s}$ ), the position becomes:

$x(12.5) = \cos(12.5\pi) = \cos\left(12\pi + \frac{\pi}{2}\right) = \cos\frac{\pi}{2} = 0 \text{ cm}.$

The distance covered in this interval is:

$D_{\text{extra}} = |1 - 0| = 1 \text{ cm}.$

So, the total distance traveled is:

$D = D_{\text{complete}} + D_{\text{extra}} = 24 + 1 = 25 \text{ cm}.$

The net displacement, which is the difference between the final and initial positions, is calculated as follows:

Initial position at $t = 0$ :

$x(0) = \cos 0 = 1 \text{ cm},$

Final position at $t = 12.5 \text{ s}$ :

$x(12.5) = 0 \text{ cm}.$

Thus, the displacement is:

$d = |0 - 1| = 1 \text{ cm}.$

The ratio of the total distance traveled to the displacement is:

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

Therefore, the correct answer is $25$ .