While doing the video, I wanted to know exact location of the video at any given time to make the animation scientifically correct. But rather than make this about the spherical case, let's make it more general. Then, using the Conservation of energy we first have,
$\displaystyle\frac{ds}{dt}=\sqrt{2g(y_0-y)}$
where $y_0$ denotes the y-coordinate of the sliding particle at time $t=0$.
Using the condition we derived for the Tautochrone curve in the video,
$\displaystyle \frac{ds}{dy}=\frac{T_0}{\pi}\sqrt{\frac{2g}{y}}$
Using the above two equations, we get
$\displaystyle \frac{dy}{dt}=\frac{\pi}{T_0}\sqrt{y(y_0-y)}$
In my python code for the video, I used python's odeint to numerically solve this equation. But thinking about it, there was a much more easier way to do this.
Using the pendulum analogy we saw in the video, we can multiply the 'force equation' of the pendulum by $l$, the length of the pendulum to get
$\displaystyle \frac{d^2s}{dt^2}+\frac{g}{l}s=0$
which is just the equation of the Simple Harmonic motion as noted in the video. Solving this for the initial condition that $s=s_0$ at $t=0$ and $s=0$ at $t=T_0$, we easily see that
$\displaystyle s=s_0\cos\left(\frac{\pi}{2}\frac{t}{T_0}\right)$
where we have used the expression $T_0=(\pi/2)\sqrt{l/g}$ to eliminate the $g/l$ term.
Also from the condition that we derived for tautochrones, we know that $s$ is directly proportional to $\sqrt{y}$. This immediately gives
$\displaystyle y=y_0\cos^2\left(\frac{\pi}{2}\frac{t}{T_0}\right)$
which gives the position of the particle at time any time $t$. For the spherical co-ordinate, this reduces to
$\displaystyle \cos\left(\frac{\theta}{2}\right)=\cos\left(\frac{\theta_0}{2}\right)\cos\left(\frac{\pi}{2}\frac{t}{T_0}\right)$
This means, we could have simply used this explicit expression rather than numerically solving a differential equation in the animation!!
Suddenly then, I realized that this will not work when animating the (polar) Isochrone. But then it dawned on me that the polar velocity is constant by definition of the polar isochrone. Great!!
Until then
Yours always
Me
