1. Write the SHM equations

Let the two particles have mean positions separated by $X_0$.

We can write their positions as $x_1=A\cos \omega t$ and $x_2=X_0+A\cos(\omega t+\phi),$ where $\phi$ is the phase difference.

2. Find the separation between the particles

Their instantaneous separation is $x_2-x_1=X_0+A\cos(\omega t+\phi)-A\cos\omega t.$

Using the result that the maximum value of $A\cos(\omega t+\phi)-A\cos\omega t$ is $2A\sin\frac{\phi}{2},$ we get

$\text{maximum separation}=X_0+2A\sin\frac{\phi}{2}.$

Given in the question: $\text{maximum separation}=X_0+A.$

So, $X_0+2A\sin\frac{\phi}{2}=X_0+A.$

Therefore, $2A\sin\frac{\phi}{2}=A$ $2\sin\frac{\phi}{2}=1$ $\sin\frac{\phi}{2}=\frac{1}{2}.$

Hence, $\frac{\phi}{2}=\frac{\pi}{6}$ so $\phi=\frac{\pi}{3}.$

3. Match with options

The phase difference is $\boxed{\frac{\pi}{3}}.$

So the correct option is A.

4. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.