JEE Mains · Physics · STD 11 - 13. oscillations
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 ________.
- A \(\frac{16}{5}\)
- B \(10\)
- C \(\frac{15}{4}\)
- D \(25\)
Answer & Solution
Correct Answer
(D) \(25\)
Step-by-step Solution
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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