Step 1: Understand the motion

Time period: $T = 8 \, \text{s}$.
Therefore, angular frequency:

$ \omega = \frac{2\pi}{T} = \frac{2\pi}{8} = \frac{\pi}{4} \, \text{rad/s}. $

At $t=0$, the particle is at the mean position.

So displacement function:

$ x(t) = A \sin(\omega t). $

Velocity: $v(t) = \frac{dx}{dt} = A\omega \cos(\omega t).$

Step 2: Distance travelled in 1st second (0 to 1s)

Distance travelled = integral of $|v|$.

For $t = 0$ to small values, $\cos(\omega t) > 0$ (since $\omega = \pi/4$, in 1 second $\omega t = \pi/4 \approx 45^\circ$, still positive). So velocity doesn’t change sign in this interval → monotonic increase of $x$.

Hence distance travelled = displacement magnitude:

$ s_1 = x(1) - x(0) = A\sin\left(\frac{\pi}{4}\cdot 1\right) - 0 = A\sin\frac{\pi}{4} = \frac{A}{\sqrt{2}}. $

Step 3: Distance travelled in 2nd second (1 to 2s)

Check velocity sign between $t=1$s and $t=2$s.

At $t=1$, $\omega t = \pi/4$, cos positive. At $t=2$, $\omega t = \pi/2$, cos zero. So velocity gradually reduces to zero without changing sign. So again monotonic increase in displacement.

Thus distance = displacement change:

$ s_2 = x(2) - x(1). $

$ x(2) = A \sin\left( \frac{\pi}{4} \cdot 2 \right) = A\sin\frac{\pi}{2} = A. $

$ s_2 = A - \frac{A}{\sqrt{2}} = A\left(1 - \frac{1}{\sqrt{2}}\right). $

Step 4: Ratio $s_1 : s_2$

$ \frac{s_1}{s_2} = \frac{\tfrac{A}{\sqrt{2}}}{A\left(1 - \tfrac{1}{\sqrt{2}}\right)} = \frac{\frac{1}{\sqrt{2}}}{1 - \frac{1}{\sqrt{2}}}. $

Simplify denominator: $1 - \tfrac{1}{\sqrt{2}} = \frac{\sqrt{2}-1}{\sqrt{2}}$.

So:

$ \frac{s_1}{s_2} = \frac{\tfrac{1}{\sqrt{2}}}{\tfrac{\sqrt{2}-1}{\sqrt{2}}} = \frac{1}{\sqrt{2} - 1}. $

Multiply numerator and denominator:

$\frac{1}{\sqrt{2} - 1} = \sqrt{2}+1.$

Thus ratio:

$ s_1 : s_2 = (\sqrt{2}+1) : 1. $

Step 5: Match with options

They ask ratio of distances in 1st and 2nd second = $1 : (\sqrt{2}-1)$ or $2:1$ etc.

We got $(\sqrt{2}+1):1$.

So equivalently $1 : (\sqrt{2}-1)$? Let’s check:

$(\sqrt{2}+1):1$ = multiply numerator and denominator by $(\sqrt{2}-1)$:

$ (\sqrt{2}+1):1 = 1:(\sqrt{2}-1). $

Yes, same.

✅ Final Answer:

$ \boxed{1 : (\sqrt{2} - 1)} $

Option A

Simple Harmonic Motion

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