Lagrangian - 2 masses attached by spring

1. Aug 30, 2015

Xyius

1. The problem statement, all variables and given/known data

EDIT: This is a 2D problem, so all of my $x$ variables should be vectors. I just realized this and it may answer my question, I don't know yet though.

Two masses, $m_1$ and $m_2$ are connected by a massless spring of spring constant $k$. The spring is at it's equilibrium length and the masses are both at rest; there is no gravitational field. Suddenly, $m_2$ is given a velocity $v$. Assume that $v$ is so small that the two masses never collide in their subsequent motion. Describe the motion of both masses. What are their maximum and minimum separations?

2. Relevant equations
Lagrange's Equation
COM equations

3. The attempt at a solution
My professor says that we should use center-of-mass coordinates to make the problem easier (He also says assume the length of the spring is zero). The initial kinetic energy expression is given by,

$$T=\frac{1}{2}m_1 \dot{x}_1^2+\frac{1}{2}m_2 \dot{x}_2^2.$$

The center-of-mass transformations are given by,

$$\dot{x}_{1,CM}=\dot{x}_1-\dot{X}$$
$$\dot{x}_{2,CM}=\dot{x}_2-\dot{X}.$$

Plugging these expressions into the initial expression for kinetic energy and doing some simplification gives me the following.

$$T=\frac{1}{2}m_1\dot{x}_{1,CM}^2+\frac{1}{2}m_2\dot{x}_{2,CM}^2+\frac{1}{2}M\dot{X}^2$$

The expression for the potential energy is given by,

$$V = \frac{1}{2}k(x_{2}-x_{1})^2$$

Using the same transformation equations on this expression gives the following.

$$V = \frac{1}{2}k(x_{2,CM}-x_{1,CM})^2$$

Which makes my Lagrangian equal to,

$$\mathcal{L}=T-V=\frac{1}{2}m_1\dot{x}_{1,CM}^2+\frac{1}{2}m_2\dot{x}_{2,CM}^2+\frac{1}{2}M\dot{X}^2-\frac{1}{2}k(x_{2,CM}-x_{1,CM})^2$$

Using Lagrange's equation, I get the three following equations of motion.

$$M\ddot{X} = 0$$
$$m_1\ddot{x}_{1,CM}=-k (x_{2,CM}-x_{1,CM})$$
$$m_2\ddot{x}_{2,CM}=k (x_{2,CM}-x_{1,CM})$$

What is confusing to me is, why is he recommending COM coordinates? If I were to do this problem in laboratory coordinates, would I not get the exact same equations minus the equation for the center-of-mass?? Wouldn't my Lagrangian in that case be,

$$\mathcal{L}=T-V=\frac{1}{2}m_1\dot{x}_{1}^2+\frac{1}{2}m_2\dot{x}_{2}^2-\frac{1}{2}k(x_{2}-x_{1})^2?$$

Which yields the same two differential equations for both masses.

Last edited: Aug 30, 2015
2. Aug 30, 2015

Orodruin

Staff Emeritus
You have introduced CoM coordinates without really taking CoM into account in the sense that you are still keeping both $x_1$ and $x_2$. In the CoM coordinates, you can express one as a function of the other. This is the entire point with introducing CoM coordinates. Note that you are increasing the number of degrees of freedom of the system when writing it as you have up to this point, $x_1$ and $x_2$ are not independent.

3. Aug 30, 2015

Xyius

I guess I can isolate $\ddot{x}_{1,CM}$ and $\ddot{x}_{2,CM}$ and subtract the two differential equations from each other and see what I get.

EDIT:

I get $\ddot{x}=-\frac{k}{\mu}x$
Where $x = x_{2,CM}-x_{1,CM}$
This is real easy to solve as it is just an oscillator. I believe I can answer the rest problem from this (assuming it is correct)

Last edited: Aug 30, 2015