$\frac{d}{dx}(f(x)g(x)) \neq f^\prime(x)g^\prime(x)$? Think again!

One of the most unintuitive and frustrating facts about calculus is that \[\frac{d}{dx}(f(x)g(x))\neq f^\prime(x)g^\prime(x).\]
It is said in certain apocryphal tales that the great man himself once made the above mistake. Instead of the pretty but untrue equation above, we have the unsightly but true "product rule":
\[\frac{d}{dx}f(x)g(x)= f(x)g^\prime(x)+f^\prime(x)g(x).\]

Making your own pendulum wave machine

This post is a continuation of a three part post started here.

Pendulum Waves - Mathematical Description cont. 2

If you missed the first two parts to this post, you can start here to get some context.

Pendulum Waves - Mathematical Description cont.

This is the second of a three part post on mathematically explaining the pendulum waves apparatus (below). The first part is located here.

We'll start off from where we ended last time.
We had two possible equations that we could use to continue our derivation with:

\[y(x,t)=A\cos\left(k_{0}x+\omega(x)t\right)\]\[y(x,t)=A\cos\left(k(t)x+\omega_{0}t\right).\]

Let's use the first of the two equations above.

Since we start the pendulum waves by setting all the pendula to maximum amplitude, we have \[y\left(x,0\right)=A=A\cos\left(k_{0}x+\omega\left(0\right)\cdot0\right)\]\[A=A\cos\left(k_{0}x\right)\implies k_{0}=0.\]

Let's figure out what $\omega(x)$ is. We'll forget about $\omega$ as a function of $x$ for a moment and look at $\omega$ as a function of the n-th pendulum instead.

Recall that the total time of the dance is $\Gamma$ during which the longest pendulum performs $N$ oscillations. So the period of the longest pendulum is $T=\frac{\Gamma}{N}$. OK, so we have the period of one pendulum now. How do we figure out the gabajillion other pendulums on the thing? Well, we also know that as the pendulums get shorter, their number of oscilations increases. More specifically, if the longest pendulum has 51 oscillations, then its shorter neighbour will have 52 oscillations, its shorter neighbour will have 53 oscillations, and so forth. So letting $n$ represent the n-th pendulum (counting from 0), the period of any pendulum on the apparatus is $T=\frac{\Gamma}{N+n}$.

To express $\omega$ as a function of $x$, note that the x-position of the pendulum is $x=nd$ (so $n=\frac{x}{d}$ where $d$ is the common spacing between adjacent pendulums. We're almost done now. Now we can see that: \[\omega(x)=\frac{2\pi}{T}=\frac{2\pi}{\frac{\Gamma}{N+\frac{x}{d}}}\] \[\omega(x)=\frac{2\pi \left(N+\frac{x}{d}\right)}{\Gamma}=\frac{2\pi \left(Nd+x\right)}{\Gamma d}\]

Therefore, our complete wave equation is: \[y(x,t)=A\cos\left(\frac{2\pi \left(Nd+x\right)}{\Gamma d}t\right)\]

As a note of interest, we can rewrite the equation above as

\[y(x,t)=A\cos\left(\frac{2\pi t}{\Gamma d}x + \frac{2\pi N}{\Gamma}t\right)\] which corresponds to \[y(x,t)=A\cos\left(k(t)x+\omega_{0}t\right).\]

And we are done! That's the equation describing all of the waves you see in the video above.

Now that we are finished, we can take a nice break and prepare ourselves for the next post, where I'll discuss the properties of the equation we've just found.