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I'm asked to find a system that is modeled (or at least approximated) by x'(t)=\sin (x(t))

Differentiate yields x''(t)=\cos (x(t)) \sin (x(t)) and integrating (care of Wolfram Alpha ) yields $x(t)=2 \cot ^{-1}(e^{-c-t})$ , but I am at a loss to find an actual physical example that follows this relationship

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    @Ross How does that not apply to the question? The question is to find a physical system, not something that happens to be a nice mathematical approximation to one.2011-06-06

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Imagine a bowl whose shape is given in cylindrical coordinates by $z=c-\frac{1}{2g}\sin^2(r)$ for $r\in[0,\frac{\pi}{2}]$ say. Then take a ball and place it at rest anywhere in this bowl. The energy of the ball is $E = \frac{1}{2}mr^{\prime2}+mgz$, so $\frac{1}{2}r^{\prime2} = \frac{E}{m}-g(c-\frac{1}{2g}\sin^2(r)) = \frac{E}{m}-gc+\frac{1}{2}\sin^2(r)$. So if we now choose $c=\frac{E}{mg}$, we end up with $\frac{1}{2}r^{\prime2} = \frac{1}{2}\sin^2(r)$ and $r^\prime = \sin(r)$.

For this to be proper, you'd have to worry about taking the square root in the end and the point $r=0$ and dimensions working out and stuff, but you get the idea.

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    Nice new physical interpretation. That could work.2011-06-06
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An inverted pendulum in a very viscous medium.

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    @Chris Taylor: the acceleration term is on the lhs. More specifically, if you rescale time by the transformation $t = k s/g$, $x(t) = X(s)$, the differential equation becomes $\frac{g}{k^2} X''(s) + X'(s) = \sin(X(s))$. The limit as $k \to \infty$ (or $g \to 0$) is $X'(s) = \sin(X(s))$.2011-06-06