4. In the plane, $\overrightarrow{A B_{1}} \perp \overrightarrow{A B_{2}},\left|\overrightarrow{O B_{1}}\right|=\left|\overrightarrow{O B_{2}}\right|=1, \overrightarrow{A P}=\overrightarrow{A B_{1}}+\overrightarrow{A B_{2}}$, if $|\overrightarrow{O P}|<\frac{1}{2}$, then the range of $|\overrightarrow{O A}|$ is $\qquad$
Establish the coordinate system as shown, let $P(a, b), O(x, y)$, then
$$
\left\{\begin{array}{l}
(x-a)^{2}+y^{2}=1, \\
x^{2}+(y-b)^{2}=1, \\
(x-a)^{2}+(y-b)^{2}<\frac{1}{4},
\end{array}\right.
$$
(1) + (2) - (3) yields $\frac{7}{4}<x^{2}+y^{2} \leqslant 2$.

Therefore, the range of $|\overrightarrow{O A}|=\sqrt{x^{2}+y^{2}}$ is $\left(\frac{\sqrt{7}}{2}, \sqrt{2}\right]$.