0
$\begingroup$

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

enter image description here

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, \begin{equation} T = \frac{1}{2}M\dot{x}^2 + \frac{1}{2}m(\dot{r}^2+(l_0+r)^2\dot{\theta}^2) \end{equation} The potential energy is then, \begin{equation} V = \frac{1}{2}kr^2 - mg(l_0+r)cos(\theta) \end{equation} So the Lagrangian is, \begin{equation} 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) \end{equation} In small angle approximation we have, \begin{equation} 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 \end{equation} 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!

$\endgroup$
0

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ would that mean $\dot{r} = \dot{\epsilon}r + \dot{r}\epsilon$, or $\dot{r} =\epsilon \dot{r}$? $\endgroup$ Nov 16, 2021 at 2:15
  • $\begingroup$ @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. $\endgroup$ Nov 16, 2021 at 14:14
  • $\begingroup$ okay that makes sense, thank you! $\endgroup$ Nov 16, 2021 at 19:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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