5. As shown in Figure 1, in the Cartesian coordinate system, the vertices of the equilateral triangle $\triangle ABC$ are $B(1,0)$ and $C(3,0)$. A line passing through the origin $O$ intersects sides $AB$ and $AC$ at points $M$ and $N$, respectively. If $OM=MN$, then the coordinates of point $M$ are $\qquad$
5. $M\left(\frac{5}{4}, \frac{\sqrt{3}}{4}\right)$.

Draw a line parallel to the $x$-axis through point $N$ intersecting $A B$ at point $L$. Since $O M=M N$, we have $\triangle O B M \cong \triangle N L M \Rightarrow M L=O B=1$.
Given that $\triangle A B C$ is an equilateral triangle with side length 2, we know that $\triangle A L N$ is an equilateral triangle with side length 1, and $N$ is the midpoint of side $A C$.
Since $A(2, \sqrt{3}), C(3,0)$, then $N\left(\frac{5}{2}, \frac{\sqrt{3}}{2}\right)$.

Thus, $M\left(\frac{5}{4}, \frac{\sqrt{3}}{4}\right)$.