1. Write the SHM equation

Since the body starts from the origin, we can take $x = A\sin(\omega t).$

Given period, $T=2\text{ s}$ so angular frequency is $\omega = \frac{2\pi}{T} = \frac{2\pi}{2} = \pi\ \text{rad/s}.$

2. Use energy relation in SHM

Total energy in SHM is $E = \frac{1}{2}m\omega^2A^2.$

Potential energy at displacement $x$ is $U = \frac{1}{2}m\omega^2x^2.$

Hence kinetic energy is $K = E-U = \frac{1}{2}m\omega^2(A^2-x^2).$

Therefore, $\frac{K}{E} = 1-\frac{x^2}{A^2}.$

We are given $K = 75\%\text{ of }E = \frac{3}{4}E.$

So, $1-\frac{x^2}{A^2} = \frac{3}{4}.$

Thus, $\frac{x^2}{A^2} = \frac{1}{4} \,\Rightarrow\, \frac{|x|}{A} = \frac{1}{2}.$

3. Find the earliest time

Since $x=A\sin(\omega t),$ we need $\sin(\omega t)=\frac{1}{2}$ for the first time after starting.

Thus, $\omega t = \frac{\pi}{6}.$

Using $\omega=\pi$, $t = \frac{\pi/6}{\pi} = \frac{1}{6}\text{ s}.$

4. Check options

• A: $\frac{1}{6}$ s ✅
• B: $\frac{1}{4}$ s ❌
• C: $\frac{1}{3}$ s ❌
• D: $\frac{1}{12}$ s ❌

So the correct option is A.