JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Elasticity & SHM

With period $T=2$ s and total time $t=12.5$ s, the number of complete oscillations is
$n=\frac{t}{T}=\frac{12.5}{2}=6.25$

The particle completes 6 full oscillations and then a further quarter oscillation.

In one full oscillation (mean → extreme → mean → opposite extreme → mean) it covers a distance $4A$; in the extra quarter oscillation, starting from the mean position, it covers a further $A$ to reach an extreme.

Total distance: $D=6(4A)+A=25A=25$ cm (with $A=1$ cm)

After 6 complete oscillations the particle is back at the mean position; the additional quarter oscillation carries it to an extreme, so its net displacement from the mean is
$d=A=1$ cm

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