5. Given that $A, B, C$ are three different points on the circle $O$ (with $O$ as the origin), and $\angle A O B=\frac{2 \pi}{3}$, if $\overrightarrow{O C}=$ $\lambda \overrightarrow{O A}+\mu \overrightarrow{O B}(\lambda, \mu \in \mathbf{R})$,
then when $\omega=\sqrt{3} \lambda+\lambda+\mu$ takes the maximum value, $\frac{\lambda}{\mu}=$ $\qquad$ .
Let's set $A(1,0), B\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right), C(\cos \alpha, \sin \alpha)$, then
$$
\begin{array}{l}
(\cos \alpha, \sin \alpha)=\lambda(1,0)+\mu\left(-\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \Rightarrow\left\{\begin{array}{l}
\cos \alpha=\lambda-\frac{1}{2} \mu, \\
\sin \alpha=\frac{\sqrt{3}}{2} \mu
\end{array}\right. \\
\Rightarrow\left\{\begin{array}{l}
\lambda=\cos \alpha+\frac{\sqrt{3}}{3} \sin \alpha, \\
\mu=\frac{2 \sqrt{3}}{3} \sin \alpha
\end{array} \Rightarrow \omega=\sqrt{3} \lambda+\lambda+\mu=(\sqrt{3}+1)(\cos \alpha+\sin \alpha) .\right.
\end{array}
$$

When $\cos \alpha=\sin \alpha=\frac{\sqrt{2}}{2}$, $\omega$ reaches its maximum value $\sqrt{6}+\sqrt{2}$, at this time $\lambda=\frac{3 \sqrt{2}+\sqrt{6}}{6}$,
$$
\mu=\frac{\sqrt{6}}{3} \Rightarrow \frac{\lambda}{\mu}=\frac{\sqrt{3}+1}{2} .
$$