Frequency estimation on chaotic waveform
Here is a stab.

Start with:

$$ \ddot{u}  - 2\beta \dot{u}  +(\omega^2+\beta^2)(u-s)=0 $$

Take the derivative:

$$ \dddot{u}  - 2\beta \ddot{u}  +(\omega^2+\beta^2)(\dot{u}-\dot{s})=0 $$

Now, $ \dot{s} = 0 $ Wherever it is defined.  So:

$$ \dddot{u}  - 2\beta \ddot{u}  +(\omega^2+\beta^2)(\dot{u})=0 $$

Let $ v = \dot{u} $, and you get

$$ \ddot{v}  - 2\beta \dot{v}  +(\omega^2+\beta^2)v=0 $$

Which is an ordinary second degree linear differential equation.  v will have the same frequency as u.

The most basic (good) differential approximation is:

$$ d[n] = \frac{u[n+1]-u[n-1]}{2} $$

So run that over your data and see what you get and please add a chart to your question with the results.  

I think you may best solve this one by analyzing the zero crossing of those results.  The transitions are going to cause weirdness as your system "falls" towards the new equilibrium.



Followup:

A more traditional approach perhaps.

$$ \ddot{u}  - 2\beta \dot{u}  +(\omega^2+\beta^2)(u-s)=0 $$

Your homogenuous equation is:

$$ \ddot{u}  - 2\beta \dot{u}  +(\omega^2+\beta^2) u=0 $$

with solutions:

$$ u_h = \left[ C_1 \cos( \omega t ) + C_2 \sin( \omega t ) \right] e ^ { \beta t } $$

And your particular equation is:

$$ \ddot{u}  - 2\beta \dot{u}  +(\omega^2+\beta^2) s=0 $$

Assume:

$$ u = a t^2 + b t + c $$

$$ \dot{u} =  2 a t + b $$

$$ \ddot{u} =  2 a  $$

$$ (2a)  - 2\beta ( 2 a t + b )  + (\omega^2+\beta^2) s=0 $$

$$ (2 - 4 \beta  t)  a  - 2\beta  b   + (\omega^2+\beta^2) s=0 $$

$$ a = 0 $$

$$ b = \frac{\omega^2+\beta^2}{2\beta} \cdot s $$

So :

$$ u_p = \left( \frac{\omega^2+\beta^2}{2\beta} \cdot s \right) t $$

For any range of consistent $s$ you now have your general and particular solutions.

Therefore you should only need five points to solve for your five unknowns:  $ \omega, \beta, C_1, C_2,$ and $ c $.

With noise, that goes out the window.  If you take the second derivative, your particular solution will be squashed completely.  Then you should only have to worry about the $s$ transition points.

Line crossings will give you $\omega$, and the envelope will give you $\beta$.