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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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    I don't have a lot of those-are you sure it is not $x''=\sin(x(t))$? In the small angle approximation where $\sin(x)=x$ you have $x''=x$, which is a harmonic oscillator.2011-06-06
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    Actually it would probably be more physical to interpret $x^{\prime\prime} = \sin(x)$ as a pendulum rather than an approximation to a harmonic oscillator.2011-06-06
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    @gfes: that was my thought-when you do the pendulum you make the small angle approximation. But that does not apply to the original question.2011-06-06
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    @Ross, no it is certainly $x'=\sin (x(t))$2011-06-06
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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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    Can this be extended to all $r$ (not just $r \in \left[0,\frac{\pi}{2}\right]$) ?2011-06-06
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    The problem is that if you try to extend it further $z$ starts to decrease as a function of $r$, so the 'bowl' analogy wouldn't be so nice anymore. Mathematically it would still work though.2011-06-06
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    Actually, in this form it's not clear that it's well behaved at the origin either. I'm sure this is fixable but to avoid all this nastyness you can switch to 2D, making the problem less physical, but a lot easier.2011-06-06
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    I think that you may have a problem with square rooting. The system $x'(t)=\sin (x(t))$ is unstable at $x=0$ and stable at $x=\pi$. Perhaps modifying to a "bumpy carpet" with $z\to -z$ ...2011-06-06
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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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    You're right. Very nice.2011-06-06
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    Could you explain further2011-06-06
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    @Zeophlite the equation for a damped, upside-down pendulum is $x''(t) + kx'(t) = g \sin x(t)$ where $k$ is the damping coefficient. If the damping coefficient is high (ie the medium is very viscous) then we can neglect the acceleration term on the rhs, giving $x'(t) = (g/k)\sin x(t)$ and then changing units gets rid of the constant on the rhs.2011-06-06
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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