This problem is from Introduction to Classical Mechanics by David Morin. Specifically it is problem $6.25$ (Spring on a T) in the Lagrangian Mechanics section, which is available on his webpage.
My solution is as follows:
Let $\theta$ be the angle between the short rod an the vertical line and let $x$ be the current length of the spring. Then we have
$r=(l\sin\theta-x\cos\theta, l\cos\theta+x\sin\theta)$ and $\dot{r}=((l\dot{{\theta}}-\dot{x})\cos\theta+x\dot{\theta}\sin\theta,(\dot{x}-l\dot{\theta})\sin\theta+x\dot{\theta}\cos\theta)$
Therefore the Lagrangian is given by $$L=\frac{1}{2}m((l\omega-\dot{x})^2+\omega^2x^2)-\frac{1}{2}kx^2$$
Then the Euler-Lagrange equation for $x$ is $$m\ddot{x}=(m\omega^2-k)x$$
Which gives that $x(t)=Ae^{s_1t}+Be^{s_2t}$ where $s_{1,2}=\pm\sqrt{\frac{m\omega^2-k}{m}}$.
Is this correct?
I've checked the maths and they seem ok. Further, you can test the solution by other way: analyzing the physical meaning of the solution:
This parameter has to be determining the behavior of the system.
$s_{1,2}=\pm\sqrt{\frac{m\omega^2-k}{m}}$
With $\omega>\sqrt{k/m}$ the solution makes $x$ to diverge. It's the case when the frequency is such that the spring is unable to produce a force strong enough to make the mass veer to the center. From the rotating frame we can say that the centrifugal force overcomes the restitution force of the spring.
With $\omega<\sqrt{k/m}$ the parameter is purely imaginary and the solution is oscillatory, with maximum frequency for $\omega=0$, as expected as the force from the spring has not to overcome the centrifugal one.
With $\omega=\sqrt{k/m}$ you've found the special value that the problem says. In this case, the mass is at constant $x$.
So, the physical meaning of the solution says you did it well.