Let's adopt polar coordinates. Fix the rotating body at the origin. Everything is happening in a plane. I assume that both springs rotate with the rotating body, i.e the follow the movement. Thus, they have the same angular velocity $\theta$.
Let's take the moment of inertia of the rotating body at the origin as J. I also assume a symmetry of the system in the way that both springs are equal and the masses start out at the same distance, noted $r$. The lagrangian of the system is then
$$ \mathcal{L} = \frac{1}{2}M\left(\dot{r}^2 + r^2\dot{\theta}^2\right) + \frac{1}{2}M\left(\dot{r}^2 + r^2\dot{\theta}^2\right) + \frac{1}{2}J\dot{\theta}^2 - \frac{1}{2}kr^2 - \frac{1}{2}kr^2 $$
$\theta$ being a cyclic variable, the total angular momentum is conserved
$$ \frac{\partial \mathcal{L}}{\partial\dot{\theta}} = (2Mr^2+J)\dot{\theta} = c $$
where $c$ is a constant. This is a result you can find also with newtonian mechanics. No force is creating torque on the system since the spring force is orthogonal to the motion. Then, the total moment of inertia with the help of Steiner's theorem is $2Mr^2 + J$ and the angular momentum follows as above. This constant will help us answer your question. To know whether $\dot{\theta}$ will slow down and accelerate etc... we need to know the behaviour of $r(t)$. This is given by the euler-Lagrange equations for $r$, which again you could get with a newtonian yet lengthy force diagram study :
$$ \frac{d}{dt}\frac{\partial{\mathcal{L}}}{\partial \dot{r}} - \frac{\partial \mathcal{L}}{\partial r} = 0 $$ which after carrying it out gives
$$ 2M\ddot{r} - 2Mr\dot{\theta}^2 +2kr = 0 \text{ or } \ddot{r} = r\dot{\theta}^2 - \omega_0^2 r = (\dot{\theta}^2 - \omega_0^2)r $$
where $\omega_0^2 \equiv \frac{k}{M}$. This differential equation is hard to solve, even if we substitute $\dot{\theta}$ from the conserved quantity, but it tells us two things from the last differential equation :
- If the centrifugal part is greater than the spring part, the movement obeys approximately diverging exponential $\ddot{r} \sim z r$ with $z > 0$. Intuitively, this means that the mass attached to the spring is under such a strong rotation, that the spring is not enough to hold it on a circular motion, and thus the masses go always further apart.
- If both contributions are equal, it means that we're in a steady regime where the spring force is just enough to hold the mass on a circle, and this there is no oscillation and the angular velocity is constant.
- If the spring force dominates the centrifugal part, then we have an oscillatory behavior and your intuition is correct.
Note however that for the first point, as $r$ goes away divergently, the conserved quantity tells us that the angular velocity should diminish, thus allowing the mass to come back since the spring force would overtake the centrifugal part. So all in all, the mass would come back, but I don't know how oscillatory this would be. The best thing would be to numerically solve the above differential equation to study all the cases.
