Question

A particle is executing simple harmonic motion with a time period of 2 s and amplitude 1 cm. If \( D \) and \( d \) are the total distance and displacement covered by the particle in 12.5 s, then the ratio \( \frac{D}{d} \) is:

• A. \( \frac{15}{4} \)
• B. 25
• C. 10
• D. \( \frac{16}{5} \)

Hint:

In SHM, the total distance traveled is the sum of the distances moved in each cycle, while the displacement is the straight-line distance between the starting and final positions.

Answer 1

The Correct Option is B

To address this issue, a foundational understanding of distance and displacement within simple harmonic motion (SHM) is required.

Step 1: Determine the Number of Oscillations

Given a time period \( T = 2 \) seconds, the number of complete oscillations, \( n \), in 12.5 seconds is calculated as:

\(n = \frac{12.5}{2} = 6.25\)

Step 2: Define Distance and Displacement

In SHM, distance (\( D \)) represents the total path length traversed, whereas displacement (\( d \)) signifies the net change in position from the initial to the final point.

Step 3: Calculate Total Distance (D)

The particle completes 6 full oscillations and an additional 0.25 of an oscillation. The distance covered in one complete oscillation is four times the amplitude, as the particle moves from one extreme to the other and back:

\( D_{\text{one oscillation}} = 4 \times \text{Amplitude} = 4 \times 1 \, \text{cm} = 4 \, \text{cm} \)

Therefore, the total distance for 6 complete oscillations plus 0.25 of an oscillation is:

\(D = 6 \times 4 \, \text{cm} + (0.25 \times 4 \, \text{cm}) = 24 \, \text{cm} + 1 \, \text{cm} = 25 \, \text{cm}\)

Step 4: Calculate Displacement (d)

Following 6 full oscillations, the particle returns to its starting point, resulting in zero displacement for this portion. During the subsequent 0.25 oscillation, the particle moves from the equilibrium position to an extreme point:

\( d = \text{Amplitude} = 1 \, \text{cm} \)

Step 5: Compute the Ratio \( \frac{D}{d} \)

The ratio of the total distance to the displacement is:

\(\frac{D}{d} = \frac{25 \, \text{cm}}{1 \, \text{cm}} = 25\)

Consequently, the ratio \( \frac{D}{d} \) equals 25.