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For a normal pendulum experiencing simple harmonic motion (in a circular path), the acceleration of the pendulum is as follows:

$$a = \omega^2 x$$

where $\omega$ is angular frequency and $x$ is displacement from equilibrium position.

As for a cycloidal path, my gut feeling is that that acceleration of the pendulum should increase with increasing displacement from the equilibrium position. This allows the pendulum to have the same period no matter how far it was displaced from the equilibrium position. However, I am unable to find any understandable explanations or formulas that can support this, hence the question.

Thanks everyone!

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You've got it backwards. If you write out the net forces on a pendulum bob oscillating in a circular path, you'll find that the acceleration of the pendulum is not $a = \omega^2 x$, but rather $$ \alpha = - \omega^2 \sin \theta, $$ where $\alpha$ is the (angular) acceleration and $\theta$ is the (angular) distance from equilibrium. The minus sign indicates that the acceleration is opposite the direction of the displacement, i.e., the acceleration is always towards the equilibrium position.

Now, for small angles, with $\theta \ll 1$ (in radians), we can make the approximation that $\sin \theta \approx \theta$, and so the above equation becomes $$ \alpha \approx - \omega^2 \theta $$ In other words, the acceleration is approximately proportional to the displacement. This condition is what leads to simple harmonic motion; it follows from the fact that the only functions for which the second derivative is proportional to minus the function (i.e. $f'' \propto - f$) are sines and cosines. However, if we actually use larger angles, the approximation $\sin \theta \approx \theta$ is no longer a good one; and in fact, for large angles, the period of a circular pendulum is not simple harmonic motion. For example, the period changes at larger amplitudes: enter image description here

The cycloidal path, on the other hand, relies on the fact that the acceleration of the pendulum is always directly proportional to the displacement, even for large displacements from the bottom of the arc. It's not too hard to show using free-body diagrams that acceleration due to gravity of a cart moving along a "roller coaster" path is equal to $g$ times the rate of change of the height $y$ with arc length $s$ (i.e., $a = g (dy/ds)$.) If we want the cart to exhibit simple harmonic motion, it must be the case that $a$ is proportional to $s$ as well, i.e., $dy/ds \propto -s$. Actually seeing that a cycloid has this property requires a fair amount of mathematics (though if anyone has a nice derivation of that, please post it as an answer alongside mine—I'd be interested to see it.)

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