A particle is executing simple harmonic motion with time period 2 s and amplitude 2 cm. If D and d are the total distance and displacement covered by the particle in 12.5 s, then is [2025]

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Solution

Q.
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 $\frac{D}{d}$ is [2025]

1 $\frac{15}{4}$

2 25

3 10

4 $\frac{16}{5}$

Ans.
(2)

Time period,

In 12.5 s, 6 total oscillations and 1 quarter oscillation.

So, D = (6 $\times$ 4A) + A = 25A

d = A

$\Rightarrow \frac{D}{d} = 25$

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