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

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{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}

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

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

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.

• You have two prior versions of this same question. What are you not getting out of them? Even in this version, it does not look like you have correctly managed to get the reference point done correctly. It is not clear what it is you want us to help with, and I think this question is also going to not be fruitful. Commented Jun 21, 2023 at 4:31
• @naturallyInconsistent Could you please help me a little? If you take the pivot as the reference point, you get those equations for sure. Commented Jun 21, 2023 at 4:43
• @naturallyInconsistent I am not getting why, by doing the same approximation, I am not able to get the approximated equations when I take the pivot as the reference point. Commented Jun 21, 2023 at 4:44
• That is part of the thing. I do not think you should be getting a wrong result, and so I think you have done something wrong somewhere else, so you won't be getting any helpful help if you don't show us how you obtained these. Commented Jun 21, 2023 at 4:48
• @naturallyInconsistent Just added how I ended up getting the problematic equation. Thanks for trying to make this post fruitful and for your patience, by the way. I truly appreciate it. Commented Jun 21, 2023 at 4:55

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 $$-Mg\left(\ell+a\right)\left(1-\cos\vartheta\right)$$
• Thank you very much! One small thing, as $\theta$ increases, the term you presented in the last part increases tho. Does it have to be the opposite sign? Commented Jun 21, 2023 at 6:49