(Taking Different Reference Points Give Different Conclusion) Approximation for Single Spring Pendulum

Question

Please refer to this post: Approximations for a spring pendulum's equations of motion in 2D

I am doing the same approximation, by letting

$$\theta \rightarrow \epsilon \theta$$ $$a \rightarrow \epsilon a$$

However, if I set my reference point to be the pivot (i.e., at the top where the spring is attached), my equations become

\begin{equation} \begin{aligned} (l+a) \ddot{\theta}+2 \dot{\theta} \dot{a}+g \sin \theta =0 \quad \text{and} \quad \ddot{a}-(l+a) \dot{\theta}^2-g\cos \theta+\frac{k}{M} a =0 \end{aligned} \end{equation}

so, I cannot get the following equation in the end because of the cosine term.

\begin{equation} \begin{aligned} \epsilon \ddot{a}+\epsilon \frac{k}{M} a & =0 \end{aligned} \end{equation}

Taking different reference points should not matter when it comes to getting the same conclusion, but in this case, apparently, it matters. Why is that?

The Lagrangian you get when you take the pivot as the reference point is the following.

$$ \frac{1}{2}M\dot{a}^2 + \frac{1}{2}M(l+a)^2\dot{\theta}^2 -\frac{1}{2}ka^2 + Mg(l+a)\cos{\theta} $$

And then, if you calculate the Euler-Lagrangian equation when it comes to $a$, you will get the following result.

$$ \ddot{a}-(l+a) \dot{\theta}^2-g\cos \theta+\frac{k}{M} a =0 $$

Now, if you do the same approximation method as stated earlier in this post and the original post, you will get the following result. (You neglect the term with $\epsilon^2$.)

$$ \epsilon\ddot{a} + \epsilon \frac{k}{M}a = g $$

It becomes a different equation for some reason.

Answer 1

The offending term comes from your Lagrangian term of $$Mg\left(\ell+a\right)\cos\vartheta$$ In the typical case, there is only the constant $\ell$ and not $\left(\ell+a\right)$, so that we could choose the reference point anywhere, by adding a constant term to the Lagrangian.

In more detail, the problem is that the Lagrangian has the cross-term of $$a\cos\vartheta$$and this is the culprit that makes it such that you do no longer have the freedom to choose the reference point anywhere you like. This cross term is sensitive to where precisely you picked the reference point, and so, if you want to get equations that can even be linearised properly (notice that your equations cannot be satisfied on the zeroth order term), you must pick it to be $$-a\left(1-\cos\vartheta\right)$$ in order to get the correct results. Obviously, this should be done in the general, and thus the correct original Lagrangian term has to be $$-Mg\left(\ell+a\right)\left(1-\cos\vartheta\right)$$