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:
Show Hint
- \( \frac{15}{4} \)
- 25
- 10
- \( \frac{16}{5} \)
The Correct Option is B
Approach Solution - 1
To solve this problem, we must first understand the concepts of distance and displacement in the context of simple harmonic motion (SHM).
Step 1: Calculate the Number of Oscillations
The particle has a time period \( T = 2 \) seconds. In 12.5 seconds, the number of complete oscillations, denoted as \( n \), is:
\(n = \frac{12.5}{2} = 6.25\)
Step 2: Understanding Distance and Displacement
In SHM, distance (\( D \)) is the total length covered by the particle, while displacement (\( d \)) is the net change in position from the start to the end point.
Step 3: Calculate Total Distance (D)
Since the particle completes 6 full oscillations in 12 seconds and another 0.25 of an oscillation in the remaining 0.5 seconds, the total distance (\( D \)) covered in one complete oscillation is four times the amplitude, since the particle goes from one extreme to the other and back:
\( D_{\text{one oscillation}} = 4 \times \text{Amplitude} = 4 \times 1 \, \text{cm} = 4 \, \text{cm} \)
Then for 6 complete oscillations plus 0.25 of an oscillation, the \text{distance} 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)
After 6 full oscillations, the particle returns to its starting position, so displacement is zero for that part. During the additional 0.25 oscillation, the particle moves from the equilibrium position to the extreme point:
\( d = \text{Amplitude} = 1 \, \text{cm} \)
Step 5: Calculate Ratio \( \frac{D}{d} \)
The ratio of the total distance to the displacement is:
\(\frac{D}{d} = \frac{25 \, \text{cm}}{1 \, \text{cm}} = 25\)
Therefore, the ratio \( \frac{D}{d} \) is 25.
Approach Solution -2
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 \( \boxed{25} \).
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