13. Given the function $f(x)=2 \sin \omega x(\omega>0)$ is monotonically increasing on $\left[0, \frac{\pi}{4}\right]$, and the maximum value on this interval is $\sqrt{3}$. Then $\omega=$ $\qquad$ .
$=13 \cdot \frac{4}{3}$.
From the problem, we know
$$
\begin{array}{l}
\frac{\pi}{4} \omega \leqslant \frac{\pi}{2}, \text { and } 2 \sin \frac{\pi}{4} \omega=\sqrt{3} \\
\Rightarrow \omega=\frac{4}{3} .
\end{array}
$$