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Lately, I've been trying to find the period of an angle included in the following differential equations, but only could with the basic model:

Basic or original: $$\mathrm{For}\ (\Phi (0), \Omega (0))=(\Phi_{o},0),\ \frac{d^2\Phi}{dt^2}= \frac{g}{\ell_{o}}\sin{\Phi}-\frac{g}{\ell_{o}}\zeta\ \mathrm{sgn\ \Phi}\ ;$$
Modified: $$\mathrm{For\ the\ same\ initial\ conditions},\ \frac{d^2\Phi}{dt^2}= \frac{g}{\ell_{o}}\frac{\sin{\Phi}}{f(\Phi)}-\frac{g}{\ell_{o}}\zeta \frac{\mathrm{sgn\ \Phi}}{f(\Phi)}\ -2\dot{\Phi}^2 \frac{f'(\Phi)}{f(\Phi)}.$$

Where $g$ is gravity, $\ell_{o}$ is the length of the inverted pendulum, $\zeta$ a group of other constants, $\operatorname{sgn}\left(\cdot\right)$ is the signum function, $\dot{\Phi}=\Omega=\frac{d\Phi}{dt}$, $f(\Phi)=\sqrt[3]{1-\eta\cos{\Phi}}$ ($\eta$ is another constant) and $f'(\Phi)=\frac{df(\Phi)}{d\Phi}$.

And so, the method I used to get the period was basically this:

Let $F(\Phi)= \frac{g}{\ell_{o}}\sin{\Phi}-\frac{g}{\ell_{o}}\zeta\ \mathrm{sgn\ \Phi}$ , then the diff. eq. reduces to $\frac{d^2\Phi}{dt^2}=F(\Phi).$ And now I just proceed. \begin{align} \int \frac{d^2\Phi}{dt^2}d\Phi &= \int F(\Phi)\ d\Phi\\ \frac{1}{2}\dot{\Phi}^2 &= \int F(\Phi)\ d\Phi\ +C\\ \dot{\Phi} &= \frac{d\Phi}{dt} = \sqrt{2\int F(\Phi)\ d\Phi +C}\\ \frac{T}{4}=\int_{t_{o}}^{t_{1}}dt &= \int_{0}^{\Phi_{o}}\frac{d\Phi}{\sqrt{2\int F(\Phi)\ d\Phi +C}}\\ T &=2\sqrt{2} \int_{0}^{\Phi_{o}}\frac{d\Phi}{\sqrt{\int F(\Phi)\ d\Phi +C}}. \end{align}

This worked for the basic model; but didn't for the modified one. The issue was the integral of $F(\Phi)$ since in the modified version it included all terms divided by $f(\Phi)$ and also the $\dot{\Phi}^2 \frac{f'(\Phi)}{f(\Phi)}$ one too. Can someone tell me any easier way to attain the period of this modified system? Or what approximation could I use to make it easier to deal with?

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    $\begingroup$ Welcome. I suggest a clearer title so that other users can easily decide at a glance on the home page whether or not they can answer / are interested in the question $\endgroup$
    – FShrike
    Jun 27, 2022 at 22:11
  • $\begingroup$ Is it intentional that the second equation is not symmetric wrt. the rest position? Compare with an oscillator with quadratic (air-like) friction where the equation is $m\ddot x+c|\dot x|\dot x+kx=0$. $\endgroup$ Mar 17 at 13:09
  • $\begingroup$ It is symmetric because, although $\dot\Phi^2>0$, the $f'/f$ term is positive for $\Phi>0$ and negative for $\Phi<0$ due to the sine in $f'$. You can check it by differentiation of $f(\Phi)=\sqrt[3]{1-\eta\cos\Phi}$. Also, I didn't mention that $\eta\in(0,1)$. $\endgroup$
    – Conreu
    Mar 17 at 14:04
  • $\begingroup$ For further information, this term you mention is accounting for the centrifugal force which changes direction constantly depending on $\Phi$ as for $\Phi>0$ the skier would turn to the left and viceversa for $\Phi<0$. $\endgroup$
    – Conreu
    Mar 17 at 14:08

2 Answers 2

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I just found a really really good approximation, but couldn't find its analitical expression in the form of an integral...

For anyone wondering, here it is:

Since from the last integration step we see that the period is expressed in terms of the initial angle, I thought about taking the average of $\ell$, $\frac{1}{\ell}$ and $\ell'(\Phi)$ and setting integration limits at $-\Phi_{o}$ and $\Phi_{o}$ (the two maximum angles at which the inverted pendulum swings). Only averages I could find were the ones for $\ell'(\Phi)$ and $\frac{1}{\ell}$. So, briefly:

$$\langle \ell'(\Phi)\rangle = \frac{\ell_{o}}{2\Phi_{o}} \int_{-\Phi_{o}}^{\Phi_{o}} f'(\Phi)\ d\Phi = \frac{\ell_{o}}{2\Phi_{o}}(f(\Phi_{o})-f(-\Phi_{o}))=0$$ $$\bigg\langle \frac{1}{\ell} \bigg\rangle = \frac{1}{2\ell_{o}\Phi_{o}} \int_{-\Phi_{o}}^{\Phi_{o}} \frac{1}{f(\Phi)}\ d\Phi = \frac{1}{\ell_{o}\Phi_{o}} \int_{0}^{\Phi_{o}} \frac{1}{f(\Phi)}\ d\Phi$$

Since for the integration of $\frac{1}{f(\Phi)}=\frac{1}{\sqrt[3]{1-\eta \cos{\Phi}}}$ the solution was undefined at $0$, doing the approximation $\cos{\Phi} ≈ 1-\frac{\Phi^2}{2}$, which is very precise for $|\Phi|<90º$, we could actually get a solution which is not undefined at $0$ and evaluated at which is $0$ too:

$$\Phi\sqrt[3]{\frac{1}{1-\eta}}\ \ {}_{2}F_{1} \bigg(\frac{1}{3},\frac{1}{2};\frac{3}{2};\frac{\eta}{2\eta-2} \Phi^2 \bigg) +C$$

Thus: $$\bigg\langle \frac{1}{\ell} \bigg\rangle=\frac{1}{\ell_{o}}\sqrt[3]{\frac{1}{1-\eta}}\ \ {}_{2}F_{1} \bigg(\frac{1}{3},\frac{1}{2};\frac{3}{2};\frac{\eta}{2\eta-2} \Phi_{o}^2 \bigg)$$

Now, using both averages, the modified differential equation reduces to this:

\begin{align} \frac{d^2\Phi}{dt^2} & =g \bigg\langle \frac{1}{\ell} \bigg\rangle\sin{\Phi}-g \zeta \bigg\langle \frac{1}{\ell} \bigg\rangle\ \mathrm{sgn\ \Phi}\ -2\dot{\Phi}^2 \langle \ell'(\Phi)\rangle \bigg\langle \frac{1}{\ell} \bigg\rangle\\ & =g \bigg\langle \frac{1}{\ell} \bigg\rangle\sin{\Phi}-g \zeta \bigg\langle \frac{1}{\ell} \bigg\rangle\ \mathrm{sgn\ \Phi} \end{align}

Finally, using the same method, mentioned in the original question, with which I found the approximated period of the basic differential equation, I solve the one I previously got. Coincidentally, the equation for the period is just the same for the basic model but the $\frac{1}{\ell}$ converts into $\big\langle \frac{1}{\ell} \big\rangle$:

Basic: $$T = 4\sqrt{\frac{\ell_{o}}{g}} \ln{\Bigg|\frac{\Phi_{o}-\zeta}{\sqrt{2\zeta\Phi_{o}-\Phi_{o}^2}-\zeta}\Bigg|}$$ Modified: $$T = 4\sqrt{\frac{1}{g \bigg\langle \frac{1}{\ell} \bigg\rangle}} \ln{\Bigg|\frac{\Phi_{o}-\zeta}{\sqrt{2\zeta\Phi_{o}-\Phi_{o}^2}-\zeta}\Bigg|}$$

Looks like the plot of the approximated and original period really agrees with these calculations.

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After revisiting this post I forgot to add that I found and exact solution

\begin{align} \frac{\mathrm d^2\Phi}{\mathrm dt^2}&=\frac{g}{\ell_0}\frac{\sin\Phi}{f(\Phi)}-\frac{g}{\ell_0}\zeta\frac{\text{sgn}\,\Phi}{f(\Phi)}-2\dot\Phi^2\frac{f'(\Phi)}{f(\Phi)}\\ \ddot\Phi+2\dot\Phi^2\frac{f'(\Phi)}{f(\Phi)}&=\frac{g}{\ell_0f(\Phi)}(\sin\Phi-\zeta\text{sgn}\,\Phi)\\ f(\Phi)^4\ddot\Phi+2\dot\Phi^2f(\Phi)^3f'(\Phi)&=\frac{g}{\ell_0}f(\Phi)^3(\sin\Phi-\zeta\text{sgn}\,\Phi)\\ \frac{1}{2}\left(2f(\Phi)^4\frac{\dot\Phi\ddot\Phi}{\dot \Phi}+4\dot\Phi^2f(\Phi)^3f'(\Phi)\right) &= \frac{1}{2}\frac{\mathrm d}{\mathrm d\Phi}(f^4\dot\Phi^2)\\ &=\frac{g}{\ell_0}f(\Phi)^3(\sin\Phi-\zeta\text{sgn}\,\Phi)\\ \implies \int_{\Phi_{\text{0}}}^\Phi\frac{1}{2}\frac{\mathrm d}{\mathrm d\Phi}(f^4\dot\Phi^2)\,\mathrm d\Phi &= \frac{1}{2}f(\Phi)^4\dot\Phi^2 \\ &=\int_{\Phi_{\text{0}}}^\Phi\frac{g}{\ell_0}f(\Phi)^3(\sin\Phi-\zeta\text{sgn}\,\Phi)\, \mathrm d\Phi\\ \implies \dot\Phi &= \frac{\mathrm d\Phi}{\mathrm dt} \\ &= \sqrt{\frac{2g}{\ell_0f(\Phi)^4}\int_{\Phi_{\text{0}}}^\Phi(1-\eta\cos\Phi)(\sin\Phi-\zeta\text{sgn}\,\Phi)\, \mathrm d\Phi}\\ \int_0^{\frac{\mathcal T}{4}}\mathrm dt&=\int_0^{\Phi_{\text{0}}}\frac{\mathrm d\Phi}{\sqrt{\frac{2g}{\ell_0f(\Phi)^4}\displaystyle\int_{\Phi_{\text{0}}}^\Phi(1-\eta\cos\Phi)(\sin\Phi-\zeta\text{sgn}\,\Phi)\, \mathrm d\Phi}}\\ &\hspace{-0.6cm}\boxed{\mathcal T=2\sqrt 2\sqrt{\frac{\ell_0}{g}}\int_0^{\Phi_{\text{0}}}\frac{|1-\eta\cos\Phi|^{2/3}\mathrm d\Phi}{\sqrt{\displaystyle\int_{\Phi_{\text{0}}}^\Phi(1-\eta\cos\Phi)(\sin\Phi-\zeta\text{sgn}\,\Phi)\, \mathrm d\Phi}}} \end{align}

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