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JEE Mains · Physics · STD 11 - Oscillations
  • JEE Mains 2025
  • Easy
  • MCQ

A particle is executing simple harmonic motion with time period 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 \(\frac{\mathrm{D}}{\mathrm{d}}\) is ________.

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Answer & Solution

Correct Answer

(D) \(25\)

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Detailed explanation

A = 1 cm

\(\begin{aligned} & \mathrm{n}=\frac{12.5}{2}=6.25 \text { cycles } \\ & \therefore \mathrm{D}=4 \times 6+1=25 \\ & \mathrm{~d}=1 \\ & \frac{D}{d}=25\end{aligned}\)
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