Example 2 Let the side length of the equilateral $\triangle ABC$ be $2, M$ is the midpoint of $AB$, $P$ is any point on $BC$, and $PA+PM$ are denoted as $s$ and $t$ for their maximum and minimum values, respectively. Then $s^{2}-t^{2}=$ $\qquad$
(2000, National Junior High School Mathematics League)
Solution: Since $P A \leqslant A C$, and when $P$ is at the vertex $C$ of $\triangle A B C$, we get the maximum value of $P M$, which is $P M=C M=\sqrt{3}$.
Thus, $s=P A+P M=2+\sqrt{3}$.
As shown in Figure 1, construct an equilateral $\triangle A^{\prime} B C$, and let $M^{\prime}$ be the midpoint of $A^{\prime} B$. Then,
$\triangle P B M \cong \triangle P B M^{\prime}$.
Therefore, $P M=P M^{\prime}$.
In $\triangle P A M^{\prime}$,
$P A+P M^{\prime}>A M^{\prime}$.
Connecting $C M^{\prime}$, we have
\[
\begin{array}{l}
\angle A C M^{\prime}=60^{\circ}+30^{\circ}=90^{\circ}, \\
A M^{\prime}=\sqrt{A C^{2}+\left(C M^{\prime}\right)^{2}}=\sqrt{7} .
\end{array}
\]

Thus, the minimum value of $P A+P M^{\prime}$ is $\sqrt{7}$, i.e., $t=\sqrt{7}$.
Therefore, $s^{2}-t^{2}=(2+\sqrt{3})^{2}-(\sqrt{7})^{2}=4 \sqrt{3}$.