Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Two particles of different masses $m_1$ and $m_2$ are connected by a massless spring of spring constant $k$ and equilibrium length $d$. The system rests on a frictionless table and may both oscillate and rotate.

I need to find the Lagrangian for this system. I'm not sure if I'm interpreting it correctly, but I think there are 4 degrees of freedom in this problem, $x_1, y_1, x_2, y_2$ or $r_1,\theta_1,r_2,\theta_2$. If I use the former choice I get my Lagrangian to be

$L = \frac{1}{2}m_1(\dot{x_1}^2 + \dot{y_1}^2) + \frac{1}{2}m_2(\dot{x_2}^2 + \dot{y_2}^2) - \frac{1}{2}k(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} -d)^2$.

Does this make any sense? It seems like the EOM would be a mess in this case.

share|improve this question
I dont find any mistake in this. Waiting for more answers. –  user7757 Jul 23 '12 at 5:21

2 Answers 2

This is correct, and you should use the rectangular coordinates until later. The equations of motion aren't a mess, because the system has a center of mass conservation law, so you can linearly mix up the variables:

$$ X = m_1 x_1 + m_2 x_2$$ $$ Y= m_1 y_1 + m_2 y_2 $$

for the center of mass and

$$ x = x_1 - x_2 $$ $$ y = y_1 - y_2 $$

which are the relative coordinates. In terms of this transformation (which is something you should just know), the Lagrangian for the CM becomes that of a free particle, while the Lagrangian for the relative coordinate becomes that of a 2d particle on a spring of finite length

$$ {m (\dot{x}^2 + \dot{y}^2\over 2} + {k (\sqrt{x^2 + y^2} - d)^2\over 2} $$

Where m is the reduced mass. Now you can transform the relative coordinates x,y into polar form $r,\theta$ and the $\theta$ equation is expressing conservation of angular momentum. This reduces to a 1d problem for r with a potential.

$$ V(r) = {k\over 2}(r-d)^2 + {A\over r^2} $$

Where A is a constant for constant angular momentum an effective centrifugal repulsion plus the attractive potential.

share|improve this answer
Thanks. I was learning Hamiltonian Mechanics (I'm asked to find Hamilton's equations after this part) a few months after I took my classical mechanics class which covered lagrangians but not hamiltonians. So I forgot the standard coordinates that are used in a two-body problem. Thanks. –  childofsaturn Jul 24 '12 at 6:58

Your lagrangian is right, but it is needlessly complicated. For two isolated masses, it is always best to move to centre-of-mass and relative coordinates, $$X=\frac{m_1x_1+m_2x_2}{m_1+m_2},$$ $$x=x_2-x_1,$$ and similarly for the $y$s. The kinetic energy is then expressed using the total mass $M=m_1+m_2$ and the reduced mass $m$ such that $\frac{1}{m}=\frac{1}{m_1}+\frac{1}{m_2}$, and your lagrangian becomes quite a bit simpler: $$L=\frac{1}{2}M\left(\dot{X}^2+\dot{Y}^2\right)+\frac{1}{2}m\left(\dot{x}^2+ \dot{y}^2 \right)-\frac{1}{2}k\left(r-d\right)^2$$ for $r=\sqrt{x^2+y^2}$.

share|improve this answer
Sorry for the quasi-duplicate, @Ron - I made this last night but my broadband died. –  Emilio Pisanty Jul 23 '12 at 8:55
Yes I got exactly this. Thank you too. –  childofsaturn Jul 24 '12 at 7:00

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.