Question

A particle executes a simple harmonic motion with a periodic time \(8\) second. At time \(t = 0\), it is at a mean position. The ratio of the distance traveled by a particle in the 2nd and that in the 1st second of its motion is
(\(sin45^{\circ} = cos45^{\circ} = \frac{1}{\sqrt{2}}\), \(sin90^{\circ} = cos0^{\circ} = 1\))

• A. \(\frac{1}{\sqrt{2}}\)
• B. \(\sqrt{2}\)
• C. \(\sqrt{2}-1\)
• D. \(\frac{1}{\sqrt{2}-1}\)

Hint:

Write x = A sin(2 pi t / T) and find displacement at each second.

Answer 1

The Correct Option is C

Step 1: Phase Idea:
In $T=8$ s the phase advances $2\pi$. In 1 s it advances $45^{\circ}$. Starting from the mean position, the particle covers the first $45^{\circ}$ of phase, then the next $45^{\circ}$.

Step 2: Distances:
First $45^{\circ}$: $A\sin45^{\circ}=0.707A$. Next $45^{\circ}$ (to $90^{\circ}$): $A-0.707A=0.293A$.

Step 3: Ratio:
$\dfrac{0.293}{0.707}=0.414=\sqrt2-1$. Option (C).

Final Answer:
Option (C). \[ \boxed{\text{(C) } \sqrt{2}-1} \]