Question 60: Oscillations

Question

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

Accepted Answer

$$ x(t) = \cos(\pi t) $$
The particle executes simple harmonic motion (SHM) with amplitude $$ A = 1 	ext{ cm} $$ and period $$ T = 2 	ext{ s} $$. In one complete cycle (2 s), the motion is as follows:
It starts at $$ x = 1 	ext{ cm} $$.
It moves to $$ x = -1 	ext{ cm} $$ (covering a distance of $$ 2 	ext{ cm} $$).
It returns to $$ x = 1 	ext{ cm} $$ (covering another $$ 2 	ext{ cm} $$).
Thus, the total distance traveled in one complete cycle is:
$$ D_{	ext{cycle}} = 2 + 2 = 4 	ext{ cm}. $$
In 12.5 s, the number of cycles is:
$$ \frac{12.5}{2} = 6.25 	ext{ cycles}. $$
For the 6 complete cycles (12 s), the distance traveled is:
$$ D_{	ext{complete}} = 6 	imes 4 = 24 	ext{ cm}. $$
For the remaining 0.5 s, determine the displacement. At $$ t = 12 	ext{ s} $$, the particle is at:
$$ x(12) = \cos(12\pi) = 1 	ext{ cm}. $$
After an extra 0.5 s (i.e., at $$ t = 12.5 	ext{ s} $$), the position becomes:
$$ x(12.5) = \cos(12.5\pi) = \cos\left(12\pi + \frac{\pi}{2}ight) = \cos\frac{\pi}{2} = 0 	ext{ cm}. $$
The distance covered in this interval is:
$$ D_{	ext{extra}} = |1 - 0| = 1 	ext{ cm}. $$
So, the total distance traveled is:
$$ D = D_{	ext{complete}} + D_{	ext{extra}} = 24 + 1 = 25 	ext{ cm}. $$
The net displacement, which is the difference between the final and initial positions, is calculated as follows:
Initial position at $$ t = 0 $$:
$$ x(0) = \cos 0 = 1 	ext{ cm}, $$
Final position at $$ t = 12.5 	ext{ s} $$:
$$ x(12.5) = 0 	ext{ cm}. $$
Thus, the displacement is:
$$ d = |0 - 1| = 1 	ext{ cm}. $$
The ratio of the total distance traveled to the displacement is:
$$ \frac{D}{d} = \frac{25}{1} = 25. $$
Therefore, the correct answer is $$25$$.

Answer

$10$

Answer

$\frac{16}{5}$

Answer

$25$

Answer

$\frac{15}{4}$

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

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