Question 12 of 194

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$25$

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Answer: (D) $25$

A = 1 cm


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