14. Given the line $l: y=\sqrt{3} x+4$, the moving circle $O: x^{2}+y^{2}=r^{2}(1<r<2)$, and the rhombus $A B C D$ with one interior angle of $60^{\circ}$, where vertices $A, B$ are on the line $l$, and vertices $C, D$ are on the circle $O$, find the range of the area $S$ of the rhombus $A B C D$ as $r$ varies.
Given the line $C D: y=\sqrt{3} x+m$,
we solve the system $\left\{\begin{array}{l}y=\sqrt{3} x+m, \\ x^{2}+y^{2}=r^{2}\end{array}\right.$
$$
\begin{array}{l}
\Rightarrow 4 x^{2}+2 \sqrt{3} m x+m^{2}-r^{2}=0 \\
\Rightarrow|C D|=\sqrt{4 r^{2}-m^{2}}
\end{array}
$$

and $M\left(-\frac{\sqrt{3} m}{4}, \frac{m}{4}\right)$, thus $|M N|=\left|-\frac{\sqrt{3} m}{4}-\left(\frac{\sqrt{3} m}{12}-\frac{4 \sqrt{3}}{3}\right)\right|=\left|-\frac{\sqrt{3} m}{3}+\frac{4 \sqrt{3}}{3}\right|$,
then $|C D|=|M N| \Rightarrow 4 r^{2}-m^{2}=\frac{1}{3}\left(m^{2}-8 m+16\right)$
$$
\Rightarrow r^{2}=\frac{1}{3}\left(m^{2}-2 m+4\right) \in(1,4) \Rightarrow m \in(-2,1) \bigcup(1,4) \text {. }
$$

Therefore, $S_{A B C D}=2 \cdot \frac{\sqrt{3}}{4}|M N|^{2}=\frac{\sqrt{3}}{2} \cdot \frac{1}{3}(m-4)^{2} \in\left(0, \frac{3 \sqrt{3}}{2}\right) \cup\left(\frac{3 \sqrt{3}}{2}, 6 \sqrt{3}\right)$.