Modeling elastic tree branch (Double torsional pendulum?) I’m trying to model “bending tree branch like motion” and it seems, that it can be described with some kind of «upward facing torsional pendulum” I guess. 
The construction is facing upward and start moving if something hits or bends it, then it tries to return to its original position.
The system looks something like this. 
I need an equation of motion for this, maybe you can point me to a specific topic covering such idea, or give a direct answer? Thanks!
 A: General dynamics problems like this are often most simply solved using Lagrangian mechanics.
Let's suppose that at rest, the branches are both vertical. Let $\theta_1$ represent the deviation from vertical of the bottom branch, and let $\theta_2$ represent the deviation from vertical of the top branch. These are our two dynamical variables here.
To greatly simplify the calculation, let's assume $\theta_1$ and $\theta_2$ are small, so we will neglect any second-order or higher terms in these variables. For a full treatment, you'll have to include some trigonometric functions that my analysis here will have the luxury of neglecting.
First, let's compute the kinetic energy of the system as a function of these variables. Assuming unit length of the branches, the position of the bottom bob is given by $(\sin \theta_1, \cos \theta_1) \approx (\theta_1, 1)$, and the position of the top bob is similarly given by $(\theta_1 + \theta_2, 2)$. Thus, the velocity vectors are given by
$$v_1 = (\dot \theta_1, 0)$$
$$v_2 = (\dot \theta_1 + \dot \theta_2, 0)$$
Suppose both bobs have mass $m$. Then, the kinetic energy of your system will be:
$$K = \frac{1}{2}m \left(|v_1|^2 + |v_2|^2\right) =
\frac{1}{2}m \left(
2\dot \theta_1^2 + \dot \theta_2^2 + 2\dot \theta_1 \dot \theta_2
\right)
$$
Similarly, the potential of your system will be given by
$$T = \frac{1}{2}k(\theta_1^2 + \theta_2^2)$$
I'm neglecting gravity here since the change in vertical coordinate is second-order in the $\theta$ variables anyway - a more nuanced treatment would include it.
This means the Lagrangian here is given by:
$$L = T - V = \frac{1}{2}k(\theta_1^2 + \theta_2^2) - \frac{1}{2}m \left(
2\dot \theta_1^2 + \dot \theta_2^2 + 2\dot \theta_1 \dot \theta_2
\right)$$
Let's stick this into the Euler-Lagrange equation for the final equations of motion. We have
$$\frac{\partial L}{\partial \theta_1} = k \theta_1$$
$$\frac{\partial L}{\partial \dot\theta_1} = -2 m \dot \theta_1 - m \dot \theta_2$$
yielding
$$2\ddot \theta_1 + \ddot \theta_2 = -\frac{k}{m} \theta_1$$
and
$$\frac{\partial L}{\partial \theta_2} = k \theta_2$$
$$\frac{\partial L}{\partial \dot\theta_2} = -m \dot \theta_1 - m \dot \theta_2$$
yielding
$$\ddot \theta_1 + \ddot \theta_2 = -\frac{k}{m} \theta_2$$
Subtracting these results, we find that
$$\ddot \theta_1 = \frac{k}{m} \left( -\theta_1 + \theta_2 \right)$$
and thus
$$\ddot \theta_2 = \frac{k}{m} \left( \theta_1 - 2 \theta_2 \right)$$
These are our final two equations of motion. They seem to make sense - the first one indicates that $\theta_1$ follows a standard sinusoidal motion, with an added term for the effect that a non-zero $\theta_2$ has on encouraging an acceleration in $\theta_1$. The second one indicates something similar for $\theta_2$.
