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

In simple harmonic motion, the particle oscillates between \( -A \) and \( +A \), where \( A \) is the amplitude. The total distance and displacement are related as follows: 
Step 1: Number of cycles in 12.5 seconds. The time period of the motion is \( T = 2 \, {s} \). The number of full cycles in 12.5 s is: \[ \frac{12.5}{2} = 6.25 \, {cycles}. \] 
Step 2: Total distance traveled \( D \). In one complete cycle, the particle covers a total distance of \( 4A \). Therefore, in 6.25 cycles, the total distance traveled is: \[ D = 4A \times 6.25 = 4 \times 1 \, {cm} \times 6.25 = 25 \, {cm}. \] 
Step 3: Total displacement \( d \). The displacement after 6 full cycles is zero, but after the 0.25 remaining cycle, the particle is at the extreme position (amplitude \( A \)). Therefore, the total displacement is: \[ d = 1 \, {cm}. \] 
Step 4: Calculate the ratio \( \frac{D}{d} \). The ratio of the total distance to the displacement is: \[ \frac{D}{d} = \frac{25}{1} = 25. \] Thus, the correct answer is \( \boxed{25} \).