Tautochrone Curve - "Virtual Gravity" Solution

"Virtual Gravity" Solution

Perhaps the simplest solution to the tautochrone problem is to note a direct relation between the angle of an incline and the gravity felt by a particle on the incline. A particle on a 90° vertical incline feels the full effect of gravity, while a particle on a horizontal plane feels effectively no gravity. At intermediate angles, the "virtual gravity" felt by the particle is g sin θ. The first step is to find a "virtual gravity" that produces the desired behavior.

The "virtual gravity" required for the tautochrone is simply proportional to the distance remaining to be traveled, which admits a simple solution:

\frac{d^2s}{{dt}^2} = - k^2s
s = A \cos kt \,

It can be easily verified both that this solution solves the differential equation and that a particle will reach s = 0 at time π/(2k) from any starting height A. The problem is now to construct a curve that will produce a "virtual gravity" proportional to the distance remaining to travel, i.e., a curve that satisfies:

g \sin \theta = - k^2 s \,

The explicit appearance of the distance remaining is troublesome, but we can differentiate to obtain a more manageable form:

g \cos \theta \,d\theta = - k^2 \,ds \,

or

ds = - \frac{g}{k^2} \cos \theta \,d\theta\,

This equation relates the change in the curve's angle to the change in the distance along the curve. We now use the Pythagorean theorem, the fact that the slope of the curve is equal to the tangent of its angle, and some trigonometric identities to obtain ds in terms of dx:

\begin{align} ds^2 & = dx^2 + dy^2 \\ \ & = \left ( 1 + \left ( \frac{dy}{dx} \right ) ^2 \right ) \,dx^2 \\ \ & = ( 1 + \tan^2 \theta )\, dx^2 \\ \ & = \sec^2 \theta \,dx^2 \\ ds & = \sec \theta \,dx \end{align}

Substituting this into the first differential equation lets us solve for x in terms of θ:

\begin{align} ds & = - \frac{g}{k^2} \cos \theta \,d\theta \\ \sec\theta \, dx & = - \frac{g}{k^2} \cos \theta\, d\theta \\ dx & = - \frac{g}{k^2} \cos^2 \theta \,d\theta \\ & = - \frac{g}{2 k^2} \left ( \cos 2 \theta + 1 \right ) \,d\theta \\ x & = - \frac{g}{4 k^2} \left ( \sin 2 \theta + 2 \theta \right ) + C_x \end{align}

Likewise, we can also express dx in terms of dy and solve for y in terms of θ:

\begin{align} \frac{dy}{dx} & = \tan \theta \\ dx & = \cot \theta \, dy \\ \cot \theta dy & = - \frac{g}{k^2} \cos^2 \theta \,d\theta \\ dy & = - \frac{g}{k^2} \sin \theta \cos \theta \,d\theta \\ & = - \frac{g}{2k^2} \sin 2 \theta \,d\theta \\ y & = \frac{g}{4k^2} \cos 2 \theta + C_y \end{align}

Substituting and, we see that these equations for and are those of a circle rolling along a horizontal line — a cycloid:

\begin{align} x & = r ( \sin \phi + \phi ) + C_x \\ y & = r ( \cos \phi ) + C_y \end{align}

Solving for k and remembering that is the time required for descent, we find the descent time in terms of the radius r:

\begin{align} r & = \frac{g}{4k^2} \\ k & = \frac{1}{2} \sqrt{\frac{g}{r}} \\ T & = \pi \sqrt{\frac{r}{g}} \end{align}

(Based loosely on Proctor, pp. 135–139)

Read more about this topic:  Tautochrone Curve

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