Lagrangian Solution
If the particle's position is parametrized by the arclength s(t) from the lowest point, the kinetic energy is proportional to . The potential energy is proportional to the height y(s). In order to be an isochrone, the Lagrangian must be that of a simple harmonic oscillator: the height of the curve must be proportional to the arclength squared.
where the constant of proportionality has been set to 1 by changing units of length. The differential form of this relation is
Which eliminates s, and leaves a differential equation for dx and dy. To find the solution, integrate for x in terms of y:
Where . This integral is the area under a circle, which can be naturally cut into a triangle and a circular wedge:
To see that this is a strangely parametrized cycloid, change variables to disentangle the transcendental and algebraic parts: define the angle .
Which is the standard parametrization, except for the scale of x, y and θ.
Read more about this topic: Tautochrone Curve
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