# Number of degrees of freedom of the coupled pendulum problem

In Chapter 4 from the book Theoretical Mechanics of Particles and Continua by A. L. Fetter and J. D. Walecka, it is solved the problem of a coupled pendulum system while considering small oscillations.

There, they say the number of degrees of freedom needed to describe the Lagrangian, are the infinitesimal displacements from equilibrium $\eta_1$ and $\eta_2$, corresponding to each pendulum mass.

My question is: why there are needed two degrees of freedom? Isn't the spring that is attached to both masses a constraint of motion that reduces the degrees of freedom to only one?

Actually, they explicitly write the following equation:

$d-d_{0}=\eta_{2}-\eta_{1}$

which is the equation of the change in length of the spring.

Thank you for any answers or suggestions!

If the spring were perfectly rigid it would reduce the number of degrees of freedom as a constraint. Because it is not, you need the positions of both pendulums to calculate the stretch of the spring. \begin{align}d &= \sqrt{l^2(\cos\theta_1 - \cos\theta_2)^2 + (l\sin\theta_1 - l\sin\theta_2 + d_0)^2} \\ &\approx l(\theta_1 - \theta_2) + d_0.\end{align}
• But given the equation $(d-d_0)=\theta_1 -\theta_2$, Can't I just rearrenge it to $\theta_{1}=(d-d_0)+\theta_2$ so I can write the Lagrangian in terms of only $\theta_2$? – Saavestro Sep 5 '16 at 22:45
• Nope, when you do that $d$ and $\theta_2$ become your two variables. – Sean E. Lake Sep 5 '16 at 22:48
• Sure, I thought $d-d_0$ was a constant, but because $d$ is a function of the angles it can be taken as a generalized coordinate too... It is clear now I think, thank you! – Saavestro Sep 5 '16 at 22:51
There are naively three degrees of freedom: the length $d$ of the spring, and the angles $\theta_1$ and $\theta_2$. We also have the constraint $$d - d_0 = l(\theta_1 - \theta_2).$$ You can use this constraint to eliminate any one of these variables, leaving two independent degrees of freedom.