1. Write the SHM equation

For simple harmonic motion, we can take $x = A\sin(\omega t)$ so that at $x=0$, we may choose $t=0$.

2. Use the given information for motion from $x=0$ to $x=\frac{A}{2}$

When $x=\frac{A}{2},$ we have $A\sin(\omega t)=\frac{A}{2}$ $\sin(\omega t)=\frac{1}{2}$ Thus, $\omega t = \frac{\pi}{6}$ for the first time.

Given this time is $2\,\text{s}$, $\frac{\pi}{6\omega}=2$ So, $\omega=\frac{\pi}{12}\,\text{rad/s}$

3. Find the time to go from $x=\frac{A}{2}$ to $x=A$

At $x=A$, $\sin(\omega t)=1$ which gives $\omega t=\frac{\pi}{2}$ for the first arrival at amplitude.

So,

• time from $x=0$ to $x=A$ is $t_A=\frac{\pi/2}{\omega}$
• time from $x=0$ to $x=\frac{A}{2}$ is $t_{A/2}=\frac{\pi/6}{\omega}=2\,\text{s}$

Hence required time is $t=t_A-t_{A/2}$ $=\frac{\pi/2-\pi/6}{\omega}$ $=\frac{\pi/3}{\omega}$

Now substitute $\omega=\frac{\pi}{12}$: $t=\frac{\pi/3}{\pi/12}=4\,\text{s}$

4. Check with options

The correct option is: $\boxed{4\,\text{s}}$ which is Option A.