Cycloid has Tautochrone Property

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Theorem

Consider a wire bent into the shape of an arc of a cycloid $C$ and inverted so that its cusps are uppermost and on the same horizontal line.

Let a bead $B$ be released from some point on the wire.


The time taken for $B$ to reach the lowest point of $C$ is:

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

independently of the point at which $B$ is released from.


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:

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

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

This result is 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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