Question

A particle performing linear S.H.M. has period 8 seconds. At time t = 0, it is in the mean position. The ratio of the distances travelled by the particle in the \( 1^{\text{st}} \) and \( 2^{\text{nd}} \) second is (\( \cos 45^\circ = 1/\sqrt{2} \))}

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

Hint:

Distance in the $n^{th}$ interval is $x(n) - x(n-1)$.

Answer 1

The Correct Option is A

Step 1: Displacement Equation
Since it starts from mean position: $x = a \sin(\omega t)$.
$\omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4}\text{ rad/s}$.
Step 2: Distances
Distance in 1st sec ($x_1$): $x(1) = a \sin(\frac{\pi}{4} \cdot 1) = \frac{a}{\sqrt{2}}$.
Displacement at 2nd sec ($x(2)$): $x(2) = a \sin(\frac{\pi}{4} \cdot 2) = a \sin(\frac{\pi}{2}) = a$.
Distance in 2nd sec ($d_2$): $x(2) - x(1) = a - \frac{a}{\sqrt{2}} = a \left( \frac{\sqrt{2}-1}{\sqrt{2}} \right)$.
Step 3: Ratio
Ratio $= \frac{a/\sqrt{2}}{a(\sqrt{2}-1)/\sqrt{2}} = \frac{1}{\sqrt{2}-1}$.
Final Answer: (A)