# Spring with changing equilibrium

Suppose that we have two cars on a track, each with a different mass. Now suppose that the cars are connected with a spring. We smack one car. I would like to write down the equations of motion for this scenario and solve them but I am just a lowly mathematician who has never really studied physics. My attempt yielded this ODE $$m_1x_1''(t)-m_2x_2''(t)= -k(x_1(t)-x_2(t))-c_1x_1'(t)+c_2x_2'(t)$$ where $c_1$ and $c_2$ are the coefficients of friction, $m_1$ and $m_2$ are the masses, and $k$ the spring constant. Is this correct? How do I solve?

• The best way of dealing with these types of problems is by Lagrangian mechanics. Write down the Lagrangian of the system, and either use the Euler-Lagrange equations or vary the action. This will immediately yield the equations of motion. – Danu Oct 1 '13 at 14:09
• Did you ever make a free body diagram for your situation? I think it will help. – ja72 Oct 1 '13 at 19:30

The two cars each have their own degrees of freedom, such that if the spring force is $F_s$ then \begin{aligned} m_1 \ddot{x}_1 & = +F_s - c_1 \dot{x}_1 \\ m_2 \ddot{x}_2 & = -F_s - c_2 \dot{x}_2 \end{aligned}
You already mentioed that the spring force is $F_s = -k \left( x_1-x_2 \right)$