Question

A particle oscillates simple harmonically along a straight line with period \(8\) s and amplitude \(8\sqrt{2}\) m. It starts from mean position then the ratio of the distances travelled by it in the \(2^{nd}\) second and the first second of its motion is \(\left(\sin 45^{\circ }=\frac{1}{\sqrt{2}}\right)\) \((\sin 90^{\circ }=1)\)

• A. \(\sqrt{2}\)
• B. \(2\)
• C. \((\sqrt{2}-1)\)
• D. \(\sqrt{3}\)

Hint:

Write x = A sin(wt) and find the position at t = 1 s and t = 2 s.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Starting from the mean position, \(x=A\sin\omega t\) with \(\omega=\frac{2\pi}{T}=\frac{2\pi}{8}=\frac{\pi}{4}\) rad/s.

Step 2: Position at 1 s
\(x_1=8\sqrt{2}\sin 45^{\circ}=8\) m.

Step 3: Position at 2 s
\(x_2=8\sqrt{2}\sin 90^{\circ}=8\sqrt{2}\) m.

Step 4: Distances and Ratio
Distance in first second \(=8\) m. Distance in second second \(=8\sqrt{2}-8\) m. Ratio \(=\frac{8\sqrt{2}-8}{8}=\sqrt{2}-1\), option (C).

Final Answer:
The ratio of the distances is root 2 minus 1. \[ \boxed{\text{(C) }(\sqrt{2}-1)} \]