Question

A particle oscillates in straight line simple harmonically with period 8 second and amplitude $4\sqrt{2}\text{ m}$. Particle starts from mean position. The ratio of the distance travelled by it in $1^{\text{st}}$ second of its motion to that in $2^{\text{nd}}$ second is $\left(\sin 45^\circ = 1/\sqrt{2}, \sin\frac{\pi}{2} = 1\right)$}

• A. $1 : 8$
• B. $1 : 4$
• C. $1 : 2$
• D. $1 : (\sqrt{2} - 1)$

Hint:

Be careful! The distance in the 2nd second is not just $x(2)$; it's the gap between $x(2)$ and $x(1)$.

Answer 1

The Correct Option is D

Step 1: Concept
Displacement from mean position is $x(t) = A \sin(\omega t)$, where $\omega = 2\pi/T$.

Step 2: Meaning
$T = 8 \text{ s} \implies \omega = \frac{2\pi}{8} = \frac{\pi}{4}$. $A = 4\sqrt{2}$.

Step 3: Analysis
Distance in 1st second ($d_1$) $= x(1) - x(0) = A \sin(\pi/4) - 0 = A(1/\sqrt{2})$. Distance in 2nd second ($d_2$) $= x(2) - x(1) = A \sin(2\pi/4) - A \sin(\pi/4) = A(1) - A(1/\sqrt{2}) = A(1 - 1/\sqrt{2})$. Ratio $\frac{d_1}{d_2} = \frac{A/\sqrt{2}}{A(1 - 1/\sqrt{2})} = \frac{1}{\sqrt{2}(1 - 1/\sqrt{2})} = \frac{1}{\sqrt{2} - 1}$.

Step 4: Conclusion
The ratio is $1 : (\sqrt{2} - 1)$. Final Answer: (D)