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We are given the second-order (non-linear) differential equation describing the pendulum $$\ddot x + \frac{g}{l}\sin(x) = 0.$$ One usually approximates this physical system with the one in which the angle $x\rightarrow 0 $ leading to a second order linear differential equation $\ddot x + \frac{g}{l}x= 0$ which leads to the simple solution of the harmonic oscillator. In case we do not do this approximation, i.e. we allow the pendulum to swing for large angles, what would be the solutions for the equations of motion? Will the pendulum still oscillate on a vertical plane? Will it whirl around the axis set by the thread? Will the system consisting of thread and mass whirl together around the vertical axis? In a book it says it will whirl over the top which I do not understand exactly which motion is meant.

Can you elaborate on one or the other of my questions?

Qmechanic
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user996159
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  • Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. – Community Jul 15 '22 at 15:22
  • You might like my new answer to the linked question https://physics.stackexchange.com/a/718837/123208 It gives more details than Dale's answer here. – PM 2Ring Jul 17 '22 at 18:36

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In case we do not do this approximation, i.e. we allow the pendulum to swing for large angles, what would be the solutions for the equations of motion ?

If the amplitudes are large then we get: $$\theta(t)=\mathrm{am}\left( \frac{v_0 t}{2l} , \frac{4gl}{v_0^2} \right)$$ where $\mathrm{am}$ is the Jacobi amplitude function and $v_0$ is the velocity at $\theta=0$.

Dale
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  • This is cheating a bit. Maybe show some of the steps for getting there and maybe a taylor expansion to relate it to the small angle solution.. maybe. – John Alexiou Jul 15 '22 at 21:41
  • I just plugged it into Mathematica. So there aren’t any steps. If you consider using Mathematica as cheating then there isn’t much I can do about it other than disagree – Dale Jul 15 '22 at 22:35
  • Imagine you walk into class one day and the professor says, today we're learning about simple harmonic motion. And the answer is $x (t) = X \sin \omega t$, class dismissed. You wouldn't feel cheated at all? – John Alexiou Jul 16 '22 at 00:55
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    @JohnAlexiou first, the word “cheated” is a highly provocative word on its own. Second, the “student” came and asked what the solution to the equations of motion were. If you don’t like my answer then write your own. In the meantime, don’t be rude. Your recommendations for improving the question didn’t need the snide “cheating” comment – Dale Jul 16 '22 at 04:15
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Multiplying the equation with $\dot{x}$, we get: $$\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{1}{2}\dot{x}^2-\frac{g}{l}\cos(x)\right) =\left(\ddot{x}+\frac{g}{l}\sin(x)\right)\dot{x}=0.$$ Therefore there exists a constant $E$ (the energy of the pendulum) with: $$\frac{1}{2}\dot{x}^2-\frac{g}{l}\cos(x)=E.$$ This results in an elliptic integral that is pretty hard to compute. But using the theory of Riemann Surfaces, an exact solution to this differential equation can be obtained as: $$x(t)=\arccos(\wp_\Lambda(t-t_0))$$ with the Weierstraß $\wp$-function and a suitable lattice $\Lambda\subset\mathbb{C}$. The derivation in detail is given in Riemann Surfaces by Simon Donaldson (See here) on page 77 to 83.

Samuel Adrian Antz
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    Thanks. While from the mathematical point of view the oscillation is not harmonic, does the motion consist of oscillations done on the xy plane only ? I was intrigued by the claim in the book according to which for small angles the pendulum swings back and forth and for large ones it whirls over the top. Can you explain the difference in the form of motion between small and large angles ? – user996159 Jul 15 '22 at 16:24
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    Elliptic integrals are actually fairly easy to compute to high precision. I give some details (and links) here: https://physics.stackexchange.com/a/718837/123208 – PM 2Ring Jul 17 '22 at 18:39
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If the starting angle is smaller than pi or 180° it will still oscillate but not a harmonic oscillation. If the starting velocity is large enough it will rotate around the axes on top, but slow on top and fast at the bottom. You have no solution with the known functions, so the integration is done numerically. All the motions are in the same plane.

trula
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  • Thanks. In the book one reads "The motion of the pendulum is simple: at low energy, it swings back and forth, and at high energy it whirls over the top". Which motion is meant by ''whirling over the top" ? – user996159 Jul 15 '22 at 15:36
  • I also do not see why there should be a starting velocity. – user996159 Jul 15 '22 at 15:44
  • Wirsing over the top is the circular motion I described. if you take a string with some object on one hand you can easily ty it out, jaus give it enough speed, probably you did this already without thinking of a pendulum. – trula Jul 15 '22 at 15:44
  • The book aims at making the difference between small and large angles. If the circular motion you are talking about is the motion around the vertical axis in the xy plane, I do not see the difference between that motion for large and small angles. Can you explain the difference between the two motions ? – user996159 Jul 15 '22 at 15:48
  • The book talks about high energy and low energy, the circular motion has more energy than any, with a large angle. For large angles it just says, that the harmonic motion does not or only approximately applies. In a schoolbook the could show the graph of s(t) or $\Phi(t)$ for large initial angles but not tell you how to calculate it.So they usually do not. Maybe you get interested in studying physics to learn more than school can provide. – trula Jul 15 '22 at 16:13
  • @user996159 See the animations here: https://en.wikipedia.org/wiki/Pendulum_(mechanics)#Examples – PM 2Ring Jul 17 '22 at 04:27
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The motion is still periodic: it's just not harmonic motion. Possibly the easiest way to see this is by plotting contour curves of constant energy $$ E=\frac{p^2}{2m}-\cos(\theta) $$ Because the motion is restricted to a curve in the $(\theta,p)$ plane, closed curve indicate period motion since, after some time $t$, a curve passing through $(\theta_0,p_0)$ will eventually pass again through $(\theta_0,p_0)$, defining a period of the motion.

enter image description here

What the plot doesn't tell you is how the amplitude is related to the period. For this you use a well known expression which yields an integral that cannot be expressed in terms of elementary functions and must be evaluated numerically, although it is possible to plot $T$ as a function of the maximum amplitude $\theta_{max}$. Plotting $\frac{T}{4}\sqrt{\frac{g}{\ell}}$ as a function of the maximal angle $\theta_{\max}$ yields the curve

enter image description here

so that, for small angles, the period is very nearly that $1$, meaning that $T\approx 2\pi\sqrt{\ell/g}$. The curve rises steadily but slowly until about $\theta_{max}=\pi/2$, at which point the increase with of $T$ with $\theta_{max}$ is much more pronounced. Note that, at $\theta_{max}$ approaches $\pi$, $T\to \infty$.

ZeroTheHero
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