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Step-by-Step Solution
Step 1: Express the positions of the two particles
Let the mean (equilibrium) positions of the two particles be at
0 and $X_0$ respectively (since they are separated by $X_0$ along the $x$-axis).
Their displacements from these mean positions can be written as:
$x_1(t) = A \cos(\omega t)$
$x_2(t) = X_0 + A \cos(\omega t + \phi)$
Here, $A$ is the amplitude, $\omega$ is the common angular frequency,
and $\phi$ is the phase difference between their motions.
Step 2: Write the separation between the particles
The instantaneous separation $s(t)$ between the particles is
the difference between their positions:
$s(t) = x_2(t)\;-\;x_1(t)$
Substitute the expressions:
$s(t) = \bigl(X_0 + A \cos(\omega t + \phi)\bigr) - A \cos(\omega t)$
$s(t) = X_0 + A \bigl[\cos(\omega t + \phi) - \cos(\omega t)\bigr]$
Step 3: Simplify the difference of cosines
Use the trigonometric identity for the difference of two cosines:
$\cos \alpha - \cos \beta = - 2 \,\sin\Bigl(\frac{\alpha + \beta}{2}\Bigr)\,\sin\Bigl(\frac{\alpha - \beta}{2}\Bigr)$
In our case, $\alpha = \omega t + \phi$ and $\beta = \omega t$, so:
$\cos(\omega t + \phi) - \cos(\omega t)
= -\,2\,\sin\Bigl(\omega t + \frac{\phi}{2}\Bigr)\,\sin\Bigl(\frac{\phi}{2}\Bigr)$
Hence, the separation becomes:
$s(t) = X_0 - 2\,A\,\sin\Bigl(\omega t + \frac{\phi}{2}\Bigr)\,\sin\Bigl(\frac{\phi}{2}\Bigr)$
Step 4: Identify the maximum separation
The maximum possible value of
$-\,2\,A\,\sin(\omega t + \frac{\phi}{2})\,\sin(\frac{\phi}{2})$
in magnitude is
$2A\,\bigl|\sin(\tfrac{\phi}{2})\bigr|$
(since $|\sin(\omega t + \delta)| \leq 1$ for any angle).
Therefore, the maximum separation $s_{\text{max}}$ is:
$s_{\text{max}} = X_0 + 2A\,\bigl|\sin\bigl(\tfrac{\phi}{2}\bigr)\bigr|$
Given in the problem, this maximum separation is $X_0 + A$. So we set:
$X_0 + 2A\,\bigl|\sin\bigl(\tfrac{\phi}{2}\bigr)\bigr| = X_0 + A
This simplifies to:
$2A\,\bigl|\sin\bigl(\tfrac{\phi}{2}\bigr)\bigr| = A
\quad\Longrightarrow\quad
\bigl|\sin\bigl(\tfrac{\phi}{2}\bigr)\bigr| = \tfrac{1}{2}
Hence,
$\sin\Bigl(\tfrac{\phi}{2}\Bigr) = \pm \tfrac{1}{2}
\quad\Longrightarrow\quad
\tfrac{\phi}{2} = \tfrac{\pi}{6} \;\text{or}\;\tfrac{5\pi}{6}
\quad\Longrightarrow\quad
\phi = \tfrac{\pi}{3} \;\text{or}\;\tfrac{5\pi}{3}
Step 5: Conclude the required phase difference
Among these possible values, the principal value for the phase difference
is often taken to be $\phi = \frac{\pi}{3}$. Thus, the phase difference
between the two SHMs is:
$\boxed{\frac{\pi}{3}}.
Answer: $\frac{\pi}{3}$
