Question 9.5: Which trajectory of the mass of a mathematical pendulum yiel...

We consider Fig. 9.4. From energy conservation, we have

\frac{m}{2} \dot{s}^{2}(y)+gmy = mgh                             (9.20)

or

\dot{s}(y) = \sqrt{2g(h−y)}.                                (9.21)

From this, one can calculate the period by separation of the variables:

\frac{1}{4} T = \int\limits_{0}^{T/4}{dt} =\int\limits_{0}^{s(h)}{\frac{ ds}{\sqrt{2g(h−y)}}}=\int\limits_{0}^{h}{\frac{(ds/dy)dy}{\sqrt{2g(h−y)}}}.                                                 (9.22)

Using the variable u = y/h, (9.22) changes to

\frac{T}{4} = \int\limits_{0}^{1}{\frac{(ds/dy) \sqrt{h}du}{\sqrt{2g(1− u)}}}.                                    (9.23)

We now require that T be independent of the maximum height h:

\frac{dT}{dh}= 0                    for all h.                                     (9.24)

Thus, we get from (9.23) (s^{′} ≡ ds/dy)

\frac{d}{dh} \int\limits_{0}^{1}{\frac{s^{′}\sqrt{h}du}{\sqrt{2g(1 −u)}}}=  \int\limits_{0}^{1}{\frac{du}{\sqrt{2g(1− u)}}} \left(\frac{1}{2}h^{−1/2}s^{′}+\sqrt{h} \frac{ds^{′}}{dh}\right)= 0                     for all h.                                                 (9.25)

With the condition that we keep the dimensionless variable u = y/h constant, we can rewrite the derivative with respect to h as a derivative with respect to y,

\frac{ds^{′}}{dh}=\frac{uds^{′}}{d(uh)}= u \frac{ds^{′}}{dy}= us^{″},                                           (9.26)

and thus, we can transform (9.25) into

\int\limits_{0}^{1}{\frac{du}{\sqrt{8g(1−u)}}(s^{′} + 2ys^{″}) \frac{1} {\sqrt{h}}}= 0                for all h.                                        (9.27)

Any periodic function f (u) satisfying \int_{0}^{1}{ f (u)du} = 0 can generally be expanded into
a Fourier series:

f (u) =\sum\limits_{m=1}^{∞}{[a_{m} sin(2πmu) + b_{m} cos(2πmu)]} .                                        (9.28)

Therefore, from (9.27) it follows that

=\frac{ \sqrt{8gh(1 −u)}}{2y} \sum\limits_{m=1}^{∞}{\left(a_{m}  sin\left(2πm \frac{y}{h}\right) +b_{m}  cos \left(2πm \frac{y}{h}\right)\right)}.                              (9.29)

This holds for all values of h. The left-hand side of (9.29) does not contain h; therefore, the right-hand side must be independent of h too. This holds only for a_{m} = b_{m} = 0 (for all m), as we shall prove now.

To have the right-hand side of (9.29) independent of h, we must have

\sum\limits_{m=1}^{∞}{ \left[a_{m}  sin \left(2πm \frac{y}{h}\right) + b_{m}  cos  \left(2πm \frac{y}{h}\right) \right]}= \frac{constant  · (y/h)h^{1/2}}{\sqrt{8g(1 −y/h)}}                             (9.30)

or

\sum\limits_{m=1}^{∞}{ \left[a_{m}  sin (2πmu) +b_{m} cos  (2πmu)\right]} = \frac{u}{\sqrt{1−u}} \frac{h^{1/2}}{\sqrt{8g}}C.                                        (9.31)

By integrating (9.31) from 0 to 1, we obtain

0 = \frac{h^{1/2}}{\sqrt{8g}}C \int\limits_{0}^{1}{\frac{u}{\sqrt{1− u}}du}= \frac{4}{3} \frac{h^{1/2}}{\sqrt{8g}}C,                                 (9.32)

thus, C = 0. (This reflects the fact that u/\sqrt{1− u} cannot be expanded into a Fourier series à la (9.31).)

Inserting this result C = 0 again into (9.30), we have a_{m} = b_{m} = 0  ∀m, and thus, from (9.29)

s^{″} + \frac{s^{′}}{2y}= 0.                                    (9.33)

From this, one finds by integrating once

\frac{s^{″} }{s^{′}}=− \frac{1}{2y}        ⇒           s^{′} ≡ \frac{ds}{dy} = \widetilde{C} e^{−(1/2) ln  y} =\frac{\widetilde{C}}{\sqrt{y}}.                                 (9.34)

The constant is usually denoted by

\widetilde{C}=\sqrt{\frac{l}{2}},                                      (9.35)

so that we have to solve

\frac{ds}{dy}=\sqrt{\frac{l}{2}} \frac{1}{\sqrt{y}}.                                   (9.36)

This is the differential equation of a cycloid.²

² See W. Greiner: Classical Mechanics: Point Particles and Relativity, 1st ed., Springer, Berlin (2004), Problem 24.4.