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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.

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I dont find any mistake in this. Waiting for more answers. –  ramanujan_dirac 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.

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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. –  dayareishq 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 $r=\sqrt{x^2+y^2}$.

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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. –  dayareishq Jul 24 '12 at 7:00

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