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?