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: Equation of Motion:
Starting at the mean position, \(x=A\sin\omega t\) with \(\omega=\dfrac{2\pi}T=\dfrac{2\pi}8=\dfrac\pi4\ \text{rad/s}\).

Step 2: Position at Each Second:
\(t=0\): \(x=0\). \(t=1\): \(x=A\sin45^{\circ}=\dfrac A{\sqrt2}\). \(t=2\): \(x=A\sin90^{\circ}=A\).

Step 3: Distances:
Distance in the 1st second \(=\dfrac A{\sqrt2}\). Distance in the 2nd second \(=A-\dfrac A{\sqrt2}\). The particle moves in one direction in both seconds (before the extreme position at \(t=2\)), so distance equals displacement.

Step 4: Ratio:
\[ \frac{A\left(1-\frac1{\sqrt2}\right)}{A/\sqrt2}=\sqrt2\left(1-\frac1{\sqrt2}\right)=\sqrt2-1 \]
Option (D) \(1/(\sqrt2-1)\) is the inverse, and (A), (B) are different numbers.

Final Answer:
The ratio is \(\sqrt2-1\), option (C). \[ \boxed{\text{(C) } \sqrt{2}-1} \]