A particle is executing simple harmonic motion with time period 2 s...

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}}

Options

A $\frac{16}{5}$
B $10$
C $\frac{15}{4}$
D $25$

Correct answer

D. $25$

Step-by-step solution

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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