# Anharmonic terms of Lagrangian of spring pendulum with free support

I am trying to find the normal modes of a spring pendulum with moving support. The spring has spring constant $$k$$ and unstretched length $$l_0$$.

Sorry for my bad paint skills. The problem was stated to have three degrees of freedom. Let $$x$$ described the distance of the block of mass $$M$$ from equilibrium. Let the length of the pendulum be described by $$l = l_0 + r$$. Lastly $$\theta$$ is the angle of oscillation. Then the kinetic energy is, $$$$T = \frac{1}{2}M\dot{x}^2 + \frac{1}{2}m(\dot{r}^2+(l_0+r)^2\dot{\theta}^2)$$$$ The potential energy is then, $$$$V = \frac{1}{2}kr^2 - mg(l_0+r)cos(\theta)$$$$ So the Lagrangian is, $$$$L = \frac{1}{2}M\dot{x}^2 + \frac{1}{2}m(\dot{r}^2+(l_0+r)^2\dot{\theta}^2) - \frac{1}{2}kr^2 + mg(l_0+r)cos(\theta)$$$$ In small angle approximation we have, $$$$L = \frac{1}{2}M\dot{x}^2 + \frac{1}{2}m(\dot{r}^2+(l_0+r)^2\dot{\theta}^2) - \frac{1}{2}kr^2 + mg(l_0+r) - \frac{1}{2}mg(1_0+r)\theta^2$$$$ The problem says to ignore anharmonic terms but I am a bit confused as to which terms that would be. I think that the anharmonic terms would be $$mgr\theta^2$$ but I am confused whether or not $$(l_0+r)^2\dot{\theta}^2$$ is anharmonic.

So if I follow my gut then I eliminate $$mgr$$ and $$-\frac{1}{2}mgr\theta^2$$ and the kinetic energy matrix is then $$\begin{pmatrix} M & 0 & 0\\ 0 & m & 0 \\ 0 & 0 & m(l_0+r)^2 \end{pmatrix}$$ and then potential $$\begin{pmatrix} 0 & 0 & 0\\ 0 & k & 0 \\ 0 & 0 & mgl_0 \end{pmatrix}$$ But I'm not sure if the $$m(l_0+r)^2$$ term belongs in the matrix. I feel like it should be $$ml_0^2$$ but I am not sure.

Also I am not 100% sure that I got the Lagrangian correct so please correct me if I am wrong!

The simplest way to keep track of this is to write $$\ell=\ell_0+\epsilon r$$ and use $$\epsilon \theta$$ rather than $$\theta$$. Expanding your Lagrangian in powers of $$\epsilon$$, you can then use $$\epsilon$$ as a counter to keep track of "smallness". The anharmonic terms are those with 3 or more powers of $$\epsilon$$ in the Lagrangian, or two or more powers of $$\epsilon$$ in the equations of motion.
• would that mean $\dot{r} = \dot{\epsilon}r + \dot{r}\epsilon$, or $\dot{r} =\epsilon \dot{r}$? Nov 16, 2021 at 2:15
• @JosephSanders $\epsilon$ is constant. It's just a dummy constant to keep track of the degree of each term when you expand in powers of $\epsilon$. Quadratic terms will be those of degree $\epsilon^2$ etc. Nov 16, 2021 at 14:14