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My son's teacher asked me to tell the kids something about the pendulum and its relation to falling through the earth and oscillating (they have been reading Alice in Wonderland).

Since both can be approximated by a harmonic oscillator, I thought I'd present them some more equivalent systems, in particular that of a point mass moving in a parabolic valley, the most straightforward quadratic potential. I quickly realized that this is not actually a harmonic oscillator: let $x$ be the horizontal position of a point particle of mass $m$ in a uniform downward pointing gravitational field with constant of gravity $g$, and consider a valley centered at the origin with height profile $h(x)$.

If I didn't make any mistakes, the Lagrangian of this system is

$$L(x,\dot x) = \frac12 m\dot x^2\left(1 + h'(x)^2\right) - mgh(x).$$

The Euler-Lagrange equation is (factoring out $m$)

$$\ddot x\left(1 + h'(x)^2\right) - \dot x^2h'(x)h''(x) + gh'(x) = 0.$$

For a quadratic height function, like $h(x) = \frac12x^2$, this becomes

$$\ddot x\left(1 + x^2\right) - \dot x^2x + gx = 0.$$

I don't really know how to solve this, but the solution, while obviously (from the physical problem) oscillatory, is not harmonic.

On the other hand, there must be a solution for $h(x)$ for which we do get a harmonic solution.

My questions (assuming the preceding is correct):

  • Could it have been clear from the outset that this is not a harmonic oscillator? Obviously we have a potential that is quadratic in the configuration space coordinate, but it seems that the dependence of the kinetic term on $x,\dot x$ is not the correct one.
  • For which $h(x)$ would we get a harmonic oscillator?
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    $\begingroup$ Very cool! Another unexpected harmonic oscillator is a train running through a mineshaft drilled at an angle through the earth (missing the center - think Paris -> London instead of London -> Sydney). The period of oscillation is independent of the end points! $\endgroup$ – Floris May 31 '17 at 0:14
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    $\begingroup$ See also this earlier answer which addresses exactly that scenario. $\endgroup$ – Floris May 31 '17 at 17:47
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Because the harmonic oscillator is characterized by the period being insensitive to the amplitude we can understand this problem as being equivalent to the well known (and solved) Tautochrone (or isochrone) problem.

The result is a cycloid written parametrically as \begin{align*} x &= \frac{\sin (2\theta) + 2\theta}{8} \\ \\ y &= \frac{1 - \cos (2\theta)}{8} \;, \end{align*} and the Wikipedia article shows a construction for building is tautochrone pendulum.


I'm afraid you'll have to be creative to make a fruitful line of discussion for a presentation to school kids out of this.

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  • $\begingroup$ Thanks, this is a very clever approach. Don't worry about the presentation, I will at most attempt to draw the shape of the well. $\endgroup$ – doetoe May 30 '17 at 23:25
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    $\begingroup$ Nice answer to the second question. It didn't address the first part: "could it have been clear from the outset...?" The answer is "yes" because the mass is moving along the parabolic trajectory. The rate at which it "picks up" energy while going a given distance along the arc is not a parabolic function of displacement along the arc. $\endgroup$ – Floris May 31 '17 at 0:12

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