2. For any two points on the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{4}=1$, if the perpendicular bisector of the line segment joining these two points intersects the $x$-axis at point $P\left(x_{0}, 0\right)$, then the range of $x_{0}$ is $\qquad$
2. $(-3,3)$.

Let the coordinates of $A, B$ be $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$, then $|P A|=$ $|P B|$, that is,
$$
\left(x_{1}-x_{0}\right)^{2}+y_{1}^{2}=\left(x_{2}-x_{0}\right)^{2}+y_{2}^{2} .
$$

Since $A, B$ are on the ellipse, we have
$$
y_{1}^{2}=4-\frac{1}{4} x_{1}^{2}, y_{2}^{2}=4-\frac{1}{4} x_{2}^{2} \text {. }
$$

Substituting (2) into (1) gives
$$
2\left(x_{2}-x_{1}\right) x_{0}=\left(x_{2}^{2}-x_{1}^{2}\right) \text {. }
$$
Since the perpendicular bisector of $A B$ intersects the $x$-axis,
$\therefore x_{1} \neq x_{2}$. Thus,
$$
\begin{array}{l}
x_{0}=\frac{x_{1}+x_{2}}{2}-\frac{3}{4}=\frac{3}{8}\left(x_{1}+x_{2}\right) \text {. } \\
\text { Given }-4 \leqslant x_{1} \leqslant 4,-4 \leqslant x_{2} \leqslant 4, x_{1} \neq x_{2} \text {, } \\
\therefore-8<x_{1}+x_{2}<8 \text {, } \\
-3<x_{0}<3 . \\
\end{array}
$$