Q 11-13-049JEE MainJEE Main 2025 (24 Jan, Shift 1)Medium
A particle is executing simple harmonic motion with time period $2\ \text{s}$ and amplitude $1\ \text{cm}$. If $D$ and $d$ are the total distance and displacement covered by the particle in $12.5\ \text{s}$, then $\dfrac{D}{d}$ is
Answer: (D) $25$
$12.5\ \text{s} = 6.25\,T$. Take the particle to start from the mean position (the standard reading).
- In each full period it covers $4A = 4\ \text{cm}$: $6\times4 = 24\ \text{cm}$.
- In the extra quarter period it moves from the mean position to an extreme: $1\ \text{cm}$.
$D = 25\ \text{cm}$. After six full periods plus a quarter it is at the extreme, so $d = A = 1\ \text{cm}$.
$$\frac{D}{d} = 25$$
Solution by Sreeraj P, M.Sc Physics