$$\begin{align} S_1(t) &= A\sin(k_1x-\omega_1 t)\nonumber\\ S_2(t) &= A\sin(k_2x-\omega_2 t)\nonumber\\ \end{align}\nonumber $$ If we write the angular frequency difference $\delta\omega = \omega_1-\omega_2$ and the average $\bar\omega = (\omega_1+\omega_2)/2$, then it's easy to show that $$\begin{align} \omega_1 &= \bar\omega + \delta\omega/2\nonumber\\ \omega_2 &= \bar\omega - \delta\omega/2\nonumber\\ \end{align}\nonumber $$ Let's look at the interference of these two waves at some point, say x=0. Then the two waves are described by: $$\begin{align} S_1(t) &= A\sin(\omega_1 t)=A\cos(\bar\omega t + \delta\omega t/2)\nonumber\\ S_2(t) &= A\sin(\omega_2 t)=A\cos(\bar\omega t - \delta\omega t/2)\nonumber\\ \end{align}\nonumber $$ If we add these up using the principle of superposition, we get $$S_{tot} = A\sin(\bar\omega t + \delta\omega t/2) + A\sin(\bar\omega t - \delta\omega t/2)\nonumber$$ Next, we expand each of the 2 sines using $$\sin(a\pm b)=\sin(a)\cos(b)\mp \cos(a)\sin(b)\nonumber$$ to get $$\begin{align} S_{tot} &= A\sin(\bar\omega t + \delta\omega t/2) + A\sin(\bar\omega t - \delta\omega t/2)\nonumber\\ &= A\big[\sin(\bar\omega t)\cos(\delta\omega t/2) + \cos(\bar\omega t)\sin(\delta\omega t/2)\big]\nonumber\\ &+ A\big[\sin(\bar\omega t)\cos(\delta\omega t/2) - \cos(\bar\omega t)\sin(\delta\omega t/2)\big]\nonumber\\ &=2A\cos(\delta\omega t/2)\sin(\bar\omega t)\label{beat} \end{align}\nonumber $$ Now, imagine that the 2 frequencies $f_1$ and $f_2$ are such that the difference $\delta f$ is much smaller than the average. For instance, $f_1=400$Hz and $f_2 = 402$Hz. This gives $\bar f=401$Hz and $\delta f = 2$Hz. This looks like a wave with a slow varying amplitude $2A\cos(2\pi\cdot 1Hz)$ and frequency $\bar f=401$Hz. What you will here is a tone that has the average frequency $401$Hz, with an amplitude that "modulates" such that you will hear a peak at 1 cycle per second.

Below, you will see 2 text windows with frequencies for 2 sounds. Push the start button for each wave and you will hear the wave at that frequency. Change the frequency and you will hear the average modulated by the beat frequency $\delta f/2$.

Lissajous figures

A Lissajous figure is a drawn from 2 oscillating waves. This is named from French physicist Jules Antoine Lissajous who worked on this in the 1850s. So if you have 2 sine waves, say $x_1=A\sin(\omega_1t)$ and $x_2=A\sin(\omega_2t)$, then the Lissajous figure comes from plotting $x_1$ vs $x_2$. As they both change, this plot traces out an interesting curve. If the ratio of $\omega_1$ to $\omega_2$ is 1:1 then the Lissajous figure will trace out a straight line in the $x_1:x_2$ plane (or if you change the relative phase to be $\pi/2$, it will trace out a circle). If the ratio is $2:1$ it will trace out a figure-8, and so on. As the ratio gets more complex, so does the pattern.

When the ratio of frequencies is a rational number (the ratio of 2 integers), then the Lissajous pattern drawn will repeat. This is easy to see: let's say that the ratio of the frequencies is given by $$\frac{f_1}{f_2} = \frac{n}{m}\nonumber$$ For instance, $n=2$ and $m=1$ will draw a figure 8 with the lobes vertical, and $n=1$ $m=2$ will draw a figure 8 with the lbes horizontal. Let's rewrite the above condition as $$\frac{f_1}{n} = \frac{f_2}{m}\nonumber$$ If $n$ and $m$ are the least common factors (for instance, if you start with $f_1=600$ and $f_2=400$, then the ratio is $6:4$ but the least common factor will be $3:2$), then this means that $f_1/n$ must be the fundamental frequency $f_0$, and $f_1$ is the $n$'th harmonic of $f_0$. And similarly $f_2$ is the $m$th harmonic of $f_0$. This means that the Lissajous pattern will repeat at the frequency of the fundamental, $f_0$, or in a time $t_R=1/f_0$.

This also means that if the ratio of frequencies is not a rational number, there is no fundamental frequency that relates them, and the Lissajous figure will never repeat.

In the controls above, you can set the 2 frequencies and hear the tones. For those frequencies, click the "Start" button below and you can see the Lissajous figure for that particular set of frequencies. For instance, try setting the two frequencies so that the difference is much less than the sum, set the relative phase to 90 degrees, and start the Lissajous plot. You will see that the signal is fairly close to a circle, but slowly turns into an ellipse. Another interesting Lissajous is when you set $f_1=400$ and $f_2=300$. The fundamental frequency is $f_0=100$, which means that the horizontal ($f_1$) will take 4 cycles and the vertical $f_2$ will take 3 cycles for the entire Lissajous figure to repeat.

A repeating Lissajous probably maps into why music is so powerful. One can imagine that the brain "likes" (gives you endorphns?) when their is a reasonable repitition of input, and maybe this is why we like chords of $2:1$ (an "octave") and $3:2$ (a "prefect fifth"). Below is a table of the predominant chords in western music.

If both $n$ and $m$ are odd, then at the half time $t_H$, both waves reach a phase of $\pi$ at the same time, which means that the Lissajous figure reverses! But if either $n$ or $m$ is even, then they won't flip signs together and the Lissajour figure won't reverse. But it will still repeat!