Gravitational wave of two interacting masses Consider two masses $m_1$ and $m_2$ that are connected by a spring. Mass 1 follows the worldline $x_1(\tau)$ while mass 2 follows $x_2(\tau)$. Note that the argument $\tau$ is the proper time in the rest frame of mass 1 and in the rest frame of mass 2 respectively but we do not use a different symbol for both.
I want to calculate how the system reacts to an incident gravitational wave (GW).
Both worldlines fulfill the geodesic equation:
$$m_1\frac{d^2x_1^\mu}{d\tau^2}=f^\mu_1-m_1\Gamma^\mu_{\nu\lambda}(x_1(\tau))\frac{dx_1^\nu}{d\tau}\frac{dx_1^\lambda}{d\tau}$$
$$m_2\frac{d^2x_2^\mu}{d\tau^2}=f^\mu_2-m_2\Gamma^\mu_{\nu\lambda}(x_2(\tau))\frac{dx_2^\nu}{d\tau}\frac{dx_2^\lambda}{d\tau}$$
where $f^\mu_1$ and $f^\mu_2$ are forces that act on the different masses.
How do we calculate the four-forces? Do we need a specific frame for that?
 A: 
What is the freely falling system? Weber writes in his paper that it is the center of mass between the two masses. Why should the center of mass be freely falling in the presence of a GW?

In an old-fashioned Newtonian picture, you could say a "freely-falling" system has no forces acting on it besides gravity. As responsibly grown-up relativists, we should say a "freely-falling" system means one following a geodesic in space-time. Since gravitational waves are part of the metric, particles moving with a gravitational wave, without being acted on by an external force, are also following geodesics and so are also freely falling.

For me this looks wrong since in a freely falling system Γ=0 and then the geodesic equations would look different?

As implied by my comment above, $\Gamma$ need not be zero for a freely falling observer.
I didn't follow all of your algebra, but I am sure your last term $\partial_\alpha \Gamma^\mu_{\nu \rho} L^\alpha$ cannot be correct. The reason is that you should have an equation between tensors. The left hand side $\frac{D^2 \xi}{D \tau^2}$ is a tensor, and the first term $\sim R x^2$ is a tensor, but $\partial \Gamma L$ is not a tensor. Therefore, this equation can at best hold in one coordinate system. However there's nothing in your calculation that picks out a special coordinate system that explains why you would get a term that's not a tensor.
