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I've learnt about the simple pendulum, and while the regular curriculum only uses the linear approximation of $\sin\theta$ to obtain $\ddot\theta+\omega_0^{2}\theta=0$. I tried to find out about a purely analytical solution, without approximations (though I knew about Taylor approximations for sine), so I obtained: $T=4\sqrt\frac{l}{g}K(\sin\frac{\theta_0}{2})$ so that $T$ increases with amplitude. I also tried to obtain $\theta(t)$ but ended up with an admittedly messy expression, via: $$\ddot\theta+\omega_0^{2}\theta=0\\ \dot\theta\ddot\theta+\omega_0^{2}\theta\dot\theta=0 \\ \dot\theta^{2}=\sqrt2 \omega_0 \sqrt{\cos\theta -\cos\theta_0} $$ (with the help of initial conditions: no $\dot\theta_0$, $\theta_0=\theta(t=0)$) with integration I obtained: $$\omega_0 t=K(\sin\frac{\theta_0}{2})-F(\phi,\sin\frac{\theta_0}{2})$$ with $$\sin\phi=\frac{\sin\frac{\theta(t)}{2}}{\sin\frac{\theta_0}{2}}$$ Is there any better closed solution and physical input+method for $\theta(t)$?

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The simple pendulum is the one that uses the small angle approximation. Using a massless rod isn't necessary as any moment of inertia can be expressed as $mr^2$ for some $r$. So I'll go through a quick derivation of the equations of motion and compare the exact solution of the approximate (locally linearized) equation and I'll go over how to get to the elliptic integral for the real equation.

Also, I believe you might have made an error going from the second order equation to the first order equation, but I see no work, so I can't actually tell. The end looks similar to what wikipedia has, but that isn't based on the linearized equation, so that would be highly unexpected. Further, you say that the equation is for unspecified $\dot{\theta}_0$ but the square root implies that $cos\theta\geq cos\theta_0$ so if I send a pendulum at a high speed from the horizontal it's orientation would become complex. Something seems odd.


Looking at the equation for the undamped pendulum - massless rod fixed about a point, the other end fixed to a point mass - where $\theta$ is the angle from the vertical, we can write the Lagrangian, and then the equations of motion.

$$ L = ml^2{\dot{\theta}}^2 - mgl (1-cos\theta)\\ \frac{d}{dt}\bigg[\frac{\delta L}{\delta \dot{\theta}} \bigg] - \frac{\delta L}{\delta \theta}=0 \\ ml^2\ddot{\theta} + mgl~sin\theta = 0 \\ \ddot{\theta} + \frac{g}{l}~sin\theta = 0 $$

From here, we can make the approximation, $sin\theta\approx\theta$ that for small $\theta$ which produces an equation that you have, where $l{\omega_0}^2=g$. This is a second order linear equation, there is no first order term, the possible solutions are $\theta=0$ or $\theta=exp(rt)$ and if you substitute the second, you'll find that $r^2+{\omega_0}^2=0$ so $r$ is a pure imaginary number and the solutions are sines and cosines of frequency $\omega_0$. This fact is independent of initial condition. Importantly, the period is constant if you use this equation, it's linear, single variable.


More exciting is solving the original, I'll rewrite with a small modification, multiplying by $\dot{\theta}$,

$$ \dot{\theta}\ddot{\theta} + \dot{\theta}\frac{g}{l}sin\theta = 0 $$

From here, I'll just do it, let C be a constant of integration.

$$ \frac{d}{dt} \bigg[\dot{\theta}^2/2 - \frac{g}{l}cos\theta\bigg] = 0 \\ \dot{\theta}^2/2 - \frac{g}{l}cos\theta = C \\ \frac{d\theta}{dt} = \sqrt{ 2C+2\frac{\strut g}{l}cos\theta} \\ \int d\theta\bigg/\sqrt{ 2C+2\frac{\strut g}{l}cos\theta} = \int{dt} $$ The last part's a toughie. But the only thing left I can say is that $C$ is probably proportional to the total energy of the system since it's conserved and it has $s^{-2}$ in it.

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  • $\begingroup$ Using the supstitution $\sin{\frac{\theta}{2}} = u$ you can express $\theta(t)$ in terms of sn function link $\endgroup$
    – DrLRX
    Nov 5, 2017 at 22:39

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