JEE Main 2025 Jan 24 Shift 1, Physics Q13: Simple Harmonic Motion And Oscillation

In one full period $T=2$ s, an SHM particle starting from an extremum returns to the same point, covering a total distance of $4A$ but with zero net displacement.

Number of complete periods in 12.5 s: $\frac{12.5}{2}=6.25$, i.e. 6 full periods plus an additional 0.5 s, which is a quarter period $T/4$.

Distance:

- 6 full periods contribute $6\times 4A=24A$.

- The final quarter period takes the particle from an extremum to the mean position (or vice versa), covering an additional $A$.

Total $D=24A+A=25A=25\times 1=25$ cm.

Displacement: After 6 full periods the particle is back at its starting point (an extremum). The next $T/4$ takes it to the mean position, so the net displacement from the start is $A=1$ cm.

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

$\therefore$ Option (2).