Cycloid has Tautochrone Property

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Theorem

Let a wire $AB$ be curved into the shape of an arc of a cycloid such that:

$A$ is at the cusp
$B$ is the highest point of the arc

and inverted so that its cusps are uppermost and on the same horizontal line.

Thus $B$ is the lowest point of the arc.

Let $AB$ be embedded in a constant and uniform gravitational field where Acceleration Due to Gravity is $g$.


Let a bead $P$ be released from anywhere on the wire between $A$ and $B$ to slide down without friction to $B$.


Then the time taken for $P$ to slide to $B$ is:

$T = \pi \sqrt {\dfrac a g}$

independently of the point from which $P$ is released.


That is, a cycloid is a tautochrone.


Proof

Brachistochrone.png


By the Principle of Conservation of Energy, the speed of the bead at a particular height is determined by its loss in potential energy in getting there.

Thus, at the point $\tuple {x, y}$, we have:

$(1): \quad v = \dfrac {\d s} {\d t} = \sqrt {2 g y}$


This can be written:

\(\ds \d t\) \(=\) \(\ds \frac {\d s} {\sqrt {2 g y} }\)
\(\ds \) \(=\) \(\ds \frac {\sqrt {\d x^2 + \d y^2} } {\sqrt {2 g y} }\)


Thus the time taken for the bead to slide down the wire is given by:

$\ds T_1 = \int \sqrt {\dfrac {\d x^2 + \d y^2} {2 g y} }$

From Equation of Cycloid, we have:

$x = a \paren {\theta - \sin \theta}$
$y = a \paren {1 - \cos \theta}$

Substituting these in the above integral:

$\ds T_1 = \int_0^{\theta_1} \sqrt {\dfrac {2 a^2 \paren {1 - \cos \theta} } {2 a g \paren {1 - \cos \theta} } } \rd \theta = \theta_1 \sqrt {\dfrac a g}$

This is the time needed for the bead to reach the bottom when released when $\theta_1 = \pi$, and so:

$T_1 = \pi \sqrt {\dfrac a g}$


Now suppose the bead is released at any intermediate point $\tuple {x_0, y_0}$.

Take equation $(1)$ and replace it with:

$v = \dfrac {\d s} {\d t} = \sqrt {2 g \paren {y - y_0} }$

Thus the total time to reach the bottom is:

\(\ds T\) \(=\) \(\ds \int_{\theta_0}^\pi \sqrt {\dfrac {2 a^2 \paren {1 - \cos \theta} } {2 a g \paren {\cos \theta_0 - \cos \theta} } } \rd \theta\)
\(\ds \) \(=\) \(\ds \sqrt {\dfrac a g} \int_{\theta_0}^\pi \sqrt {\dfrac {1 - \cos \theta} {\cos \theta_0 - \cos \theta} } \rd \theta\)
\(\text {(2)}: \quad\) \(\ds \) \(=\) \(\ds \sqrt {\dfrac a g} \int_{\theta_0}^\pi \dfrac {\sin \frac 1 2 \theta \rd \theta} {\sqrt {\cos^2 \frac 1 2 \theta_0 - \cos^2 \frac 1 2 \theta} } \rd \theta\) Half Angle Formula for Cosine and Half Angle Formula for Sine

Setting:

$u = \dfrac {\cos \frac 1 2 \theta} {\cos \frac 1 2 \theta_0}$

and so:

$\d u = -\dfrac 1 2 \dfrac {\sin \frac 1 2 \theta \rd \theta} {\cos \frac 1 2 \theta_0}$

Then $(2)$ becomes:

\(\ds T\) \(=\) \(\ds -2 \sqrt {\dfrac a g} \int_1^0 \frac {\d u} {\sqrt {1 - u^2} }\)
\(\ds \) \(=\) \(\ds 2 \sqrt {\dfrac a g} \Big [{\arcsin u}\Big]_0^1\)
\(\ds \) \(=\) \(\ds \pi \sqrt {\dfrac a g}\)

That is, wherever the bead is released from, it takes that same time to reach the bottom.

Hence the result.

$\blacksquare$


Also known as

The result cycloid has tautochrone property can be seen referred to as the pendulum property of the cycloid.


Also see


Historical Note

The fact that Cycloid has Tautochrone Property was discovered by Christiaan Huygens in $1658$, during his work on developing a reliable and accurate pendulum clock.

The Tautochrone Problem was also solved independently by Niels Henrik Abel in $1823$, using the technique now known as Abel's integral equation.


Sources

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