13 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) $\frac{4}{3}$ Hint: From the problem, we know that $\frac{\pi}{4} \cdot \omega \leqslant \frac{\pi}{2}$ and $2 \sin \frac{\pi}{4} \omega=\sqrt{3}$.