# Two springs connected by a spring

## Homework Statement

Two masses, m1 and m2, are connected to each other by a spring with a spring constant k. The system moves freely on a horizontal frictionless plane. Find the natural frequency of oscillation.

F = -kx
F = ma

## The Attempt at a Solution

Let m1 be the mass on the left, and let m2 be the mass on the right.
Let positive be in the direction of m2.

Let x1 be the displacement of m1 onto the spring, and x2 the displacement of m2 onto the spring.

$$-m_{1}\ddot{x_{1}}=k{x_{1}}$$

$$m_{2}\ddot{x_{2}}=-k{x_{2}}$$

$$m_{2}\ddot{x_{2}}-m_{1}\ddot{x_{1}}=-k(x_{2}-x_{1})$$

$$m_{2}\ddot{x_{2}}-m_{1}\ddot{x_{1}}=-k(\Delta{x})$$

I tried manipulating this into an ODE, but got nowhere.

What if you define a new variable, say, $$q = x_2 - x_1$$?

Well, then I would have

$$m_{2}\ddot{x_{2}}-m_{1}\ddot{x_{1}} = -kq$$

...which, from what I see, doesn't do a whole lot because I still can't factor the left side to do anything.

Well, firstly, the force on the first mass depends on the location of both itself AND the other mass. Think for example if one mass is very far away. Then there would be a big force on the second one.

So your equations of motion are incorrect (total forces should add up to zero).