JEE Physics 2025 Question with Solution

Given: Time period $T = 2 \, \text{s}$, amplitude $A = 1 \, \text{cm}$, and total time $t = 12.5 \, \text{s}$.

Find: The ratio $\frac{D}{d}$, where $D$ is the total distance and $d$ is the displacement.

First calculate the number of oscillations:

In SHM, the total distance covered in one complete oscillation is:

So, in $6$ complete oscillations and an additional $0.25$ oscillation, the total distance is:

After $6$ full oscillations, the particle returns to its starting position, so the displacement for that part is zero. During the extra $0.25$ oscillation, the particle moves from the equilibrium position to the extreme position, so:

Therefore,

The correct option is B.

The solution states that in SHM, the total distance traveled is the sum of distances moved in each cycle, while displacement is the straight-line change between initial and final positions.

Given: $T = 2 \, \text{s}$, $A = 1 \, \text{cm}$, and $t = 12.5 \, \text{s}$.

Find: The ratio $\frac{D}{d}$.

The number of cycles completed in $12.5 \, \text{s}$ is:

A particle in SHM travels from one extreme to the other and back in one full cycle, so the total path length in one cycle is:

Hence the total distance in $6.25$ cycles is:

For displacement, $6$ full cycles contribute zero net displacement because the particle returns to the initial position. The remaining quarter cycle places the particle at an extreme, giving:

Thus,

Therefore, the ratio is $25$, so the correct option is B.