Two spring of force constant $k_{1}$ and $k_{2}$ are configured as the figure given below

The angular frequency of this configuration is

In the given configuration, the block of mass $m$ is connected to two springs with force constants $k_{1}$ and $k_{2}$ on either side. When the block is displaced by a small distance $x$ from its equilibrium position, one spring is compressed by $x$ and the other is stretched by $x$.

The restoring force exerted by the first spring is $F_{1} = -k_{1}x$ and the restoring force exerted by the second spring is $F_{2} = -k_{2}x$. Both forces act in the same direction, opposing the displacement.

The total restoring force $F$ acting on the block is $F = F_{1} + F_{2} = -(k_{1} + k_{2})x$.

This is the standard form of the equation for simple harmonic motion, $F = -kx_{eff}$, where the effective spring constant $k_{eff} = k_{1} + k_{2}$.

The angular frequency $\omega$ of a mass-spring system is given by $\omega = \sqrt{\dfrac{k_{eff}}{m}}$.

Substituting the value of $k_{eff}$, we get $\omega = \sqrt{\dfrac{k_{1} + k_{2}}{m}}$.

Answer: $\sqrt{\dfrac{k_{1}+k_{2}}{m}}$