Applying Hooke's Law to elastic bent string 
There is an elastic spring with elastic constant $k$, natural length $L$ and current elongation $x$ (so the current total length of the string is $L+x$), which is partially inside a tube of length $b < L$. A mass of $m$ kg is rotating at the end, with angle $\beta$ and angular speed $\omega$ (gravitational acceleration $g$).
What is the value of $\omega$ in terms of the other parameters, except $x$?

                                                      
Here is my shot at it
First applying second Newton's Law to m for $\hat{y}$
$$ T_1\cos(\beta) = mg $$
$$ \boxed{T_1 = \frac{mg}{\cos(\beta)}} $$
Then for $\hat{x}$
$$ T_1\sin(\beta) = m\omega^2r = m\omega^2(L-b+x)\sin(\beta) $$
$$ T_1 = mw^2(L-b+x) $$
$$
\begin{equation}
\boxed{\omega^2 = \frac{g}{(L-b+x)\cos(\beta)}}
\end{equation}\tag{1}
$$
Then at N for $\hat{y}$
$$ N = T_1\sin(\beta) $$
$$ \boxed{N = mg\tan(\beta)} $$
Then for $\hat{x}$
$$ T_2 = T_1\cos(\beta) $$
$$ \boxed{T_2 = mg} $$
Now just getting x is missing to substitute it at (1), for which I think using Hooke's Law would be the trick But I am not sure about how to apply it properly
If I assume that it is equal to $T_1$ I would have
$$
\begin{equation}
\boxed{kx = T_1}
\end{equation}\tag{2}
$$
$$ x = \frac{mg}{k\cos{\beta}} $$
And would be ready, but at first I hesitate and think that it is possible that I should consider all the forces acting axially on the string, something like
$$ \begin{equation}
\boxed{kx = T_1+T_2-N\sin(\beta)}
\end{equation}\tag{3} $$
So which would be the right way to apply Hooke's Law here, (2) or (3)? Or other way I am not considering?
 A: As you have shown your self, the tension in the 2 different pieces of the string have different values. What causes these forces called tension to begin with? the stretch/compression is something that the string wants to resist, so it creates a force to go back to it's normal/equilibrium state. So if you have 2 different tensions, that would mean you can look at each piece as a spring.
Now you must ask how does the modelling of series springs work? When everything is stable and the mass is being rotated, you can think of the bit of string in the tube as an extended spring, whose extended length is b. lets call the natural length of that spring len1. so the rest is another spring in series with a natural length of L - len1.
How would this effect the new spring's constants (K)? It is out of scope for me to dig into this, but you must know that:
$$\frac{1}{K_t}=\frac{1}{K_1} + \frac{1}{K_2} (1)$$
Here Kt is the total/equivalent constant for the whole spring, and K1 and K2 are the constants for 2 springs in series.
Let's put what we have into mathematical equations for ease of understanding:
$$K_1 = αK (2)$$
$$K_2 = βK (3)$$
$$K_t = K (4)$$
$$K_t = \frac{K_1 K_2}{K_1+K_2} = \frac{αβ}{α+β}K (5)$$
EDIT (Clarification): equation(4) and (5) come from 2 different facts, (4) is from the fact that we know the total value of K to be, well, "K". (5) is the fact that the equivalent K for the whole system is the original value, Kt. equating the 2 will give you the expression
$$\frac{αβ}{α+β}K = K(6)$$
$$\frac{αβ}{α+β} = 1(7)$$
How can we define α and β? if one spring has half the length of the other, that means that it's constant would be higher, twice actually. meaning α/β = 2, but for your example:
$$\frac{α}{β} = \frac{len2}{len1} (8)$$
Where α is for the string with len1, and β is for the other spring. This is intuitive, if you have difficulty understanding why this is the case, think about how compressing a small spring is much harder when the spring is longer (same material/physical characteristics) So if you have 10cm long spring, and cut 1cm piece, I think you can clearly imagine how harder it gets to compress that tiny bit. That means higher constant (K).
Now let's look at the bit of the string in the enclosed area:
$$T_2 = αKΔx_1 (9)$$
$$T_1 = βKΔx_2 (10)$$
And we ofc know that the total change in length is equal to the required x. writing Δx's in terms of other variables and adding them will yield
$$Δx_1 + Δx_2 = x (11)$$
$$=\frac{T2}{αK} + \frac{T1}{βK} = x (12)$$
How do we calculate α and β?
If you look at the question now, we have equations (5), (6), (7), (8) and (9), 5 equations and we have 5 variables, α, β, Δx1, Δx2 and x. going in depth with that will probably go beyond the focus of your question, but leave a comment if that does not work out.
Hope this helps, and apologies for improper use of mathjax, still a noob.
