Beats

The phenomenon

When two sinusoidal traveling waves of different frequencies superpose, something quite interesting happens. The figure at the right shows a freeze photo of two sinusoidal travelling waves of slightly different wavelengths going in the same direction (to the right). They have the same amplitude but their wavelengths differ a little (by 10%). We've stopped the time at a point where at the position of the dot on the two waves (about 1/2 wavelength from the origin), the two waves are both at their maxima. As a result, at that point in space, they add up and give twice their individual values. (Shown in the bottom graph.)

But notice what happens as you move further out along the wave. Since they are oscillating with different wavelengths, they get out of phase. After about 5 or 6 oscillations, at the point where the top oscillation has a maximum, the second oscillation has a minimum. As a result, at that point, the sum of the two waves cancel. 

Since they are continually shifting slightly with respect to each other, they alternately go in phase producing the large bumps and out of phase producing cancellations and a zero we see in the bottom graph.

These clusters of waves between the zeros are called a wave packet.

You probably won't be surprised to learn that when signals such as sound or light are used to communicate information, sometimes the information is not in the amplitude or frequency of the individual sinusoidal oscillations, but in the shapes and size of the packets. 

Some questions

This pattern raises some interesting questions.

• What determines the size of the wave packets? The one shown is much larger than the wavelength of the individual wavelengths that make it up — about 10 times as big.
• How do these move as we let time run? We showed what it looks like at a fixed instant distributed over space. What does it look like at a fixed position (like an ear or an eye) as a function of time? 
• What would this sound like if I did it with sound? Before going on, you might want to try listening to an example. Try page (scene) 6 of David Harrison's "Beats" page. (You may have to download the swf file to run it on some computers.)

Doing the math 

There's a lot going on when we add two travelling sine waves together. There are two independent variables and a bunch of parameters. The most straightforward way to figure out what's going on is to simply do the math: write the two oscillations and add them. The trick to managing the sum is the following trigonometric identities:

$$\sin{(A+B)} = \sin{A} \cos{B} + \cos{A} \sin{B}$$

$$\sin{(A-B)} = \sin{A}  \cos{B} - \cos{A} \sin{B}$$

Although this is the straightforward version (and the one you need to remember) we need to transform it into another form to get the most useful result out for analyzing beats. This is slightly messy, but the steps are not hard. We want to add two waves $\sin{a} + \sin{b}$. So let's choose a more useful set of parameters by defining $a$ and $b$ by

$$A+B = a$$

$$A-B = b$$

Solving back for $A$ and $B$ in terms of $a$ and $b$, we get

$$A = (a+b)/2$$

$$B = (a-b)/2$$

Our two trig identifies then become

$$\sin{a} = \sin{\bigg(\frac{a+b}{2}}\bigg) \cos{\bigg(\frac{a-b}{2}\bigg)} + \cos{\bigg(\frac{a+b}{2}}\bigg) \sin{\bigg(\frac{a-b}{2}}\bigg) $$

$$\sin{b} = \sin{\bigg(\frac{a+b}{2}}\bigg) \cos{\bigg(\frac{a-b}{2}\bigg)} - \cos{\bigg(\frac{a+b}{2}}\bigg) \sin{\bigg(\frac{a-b}{2}}\bigg) $$

Adding these gives our desired result:

$$\sin{a} + \sin{b} = 2\sin{\bigg(\frac{a+b}{2}\bigg)} \cos{\bigg(\frac{a-b}{2}\bigg)}$$

The reason that we want to do this is because if our waves are close in frequency, we want to pay attention to the difference (and the average) of the frequencies and this is what our new identity does.

Interpreting the math

Now let's add together two traveling waves and see how they behave, using our relation. We want to add two traveling waves of the same amplitude. (We've used "$A$" in the above, so let's call the amplitude of the wave "$S$" for size.) We'll assume our waves travel at the same speed (as they would for sound or waves on an elastic string) but have slightly different (angular) frequencies, $\omega_1$ and $\omega_2$. Since their frequencies are different, their wavelengths (and wave numbers) will be different too: $k_1$ and $k_2$. Our sum is therefore

$$y(x,t) = S \sin{(k_1x -ω_1t)} +  S \sin{(k_2x -ω_2t)}. $$

Consider first, the spatial dependence at a fixed time. If we choose $t=0$ we get

$$y(x,0) = S \sin{(k_1x)} +  S \sin{(k_2x)}.$$ 

Using the trig identity we went to such trouble to create, we get

$$y(x,0) = 2S \sin{\bigg(\frac{k_1+k_2}{2}x\bigg)} \cos{\bigg(\frac{k_1-k_2}{2}x\bigg)}.$$

These are the product of two sinusoidal oscillations in space: with wave numbers we'll choose to call an average $k$, and a difference $k$:

$$\langle k\rangle  = \frac{k_1+k_2}{2}$$

$$Δk = k_1-k_2.$$

With these definitions, our equation for the sum of two sine waves becomes

$$y(x,0) = 2S \sin{\bigg(\langle k\rangle x\bigg)} \cos{\bigg(\frac{\Delta k}{2}x\bigg)}.$$

From our analysis, Reading the content in a sinusoidal waves, we know that when we have the spatial dependence of a wave like sin(kx) it corresponds to a wavelength

$$λ =2π/k$$

When we are looking at beats, we tend to have two wavelengths that are very close. We see that $k_1 =\langle k \rangle + Δk/2$ and $k_2 = \langle k\rangle - Δk/2$. If the two wavelengths (and wavenumbers) are close, then we expect $Δk << \langle k \rangle$. The wavelength for$\langle k \rangle$ will be close to the wavelength for $k_1$ and $k_2$, the rapid oscillation seen in all three graphs in the figure at the top of this page. But since $Δk$ is much smaller than the original $k$'s, the wavelength for its oscillation will be much longer than the oscillation of either $k$. This is the oscillation that builds the wave packet (actually two of them).

If we go through the same analysis at a fixed position and let the time run, we'll find that the oscillation in time looks just like the oscillation in space.

Using beats

We did all the above in space, but we could equally well have done it in time. If we have two sound waves coming in, we hear them at the same place as a function of time. So now you have to switch to thinking about beats in time.

If you didn't listen to David Harrison's "Beats" page at the beginning, do so now. Don't forget to turn your sound on! Another good online page to show this is the Online Tone Generator. Set the left ear to 440 Hz and the right to 444 Hz and listen for the oscillations. Then change the right ear to 443, 442, and 441. What happens? (You have to press"Play" to make it take the new values.)

You should hear a tone, modulated by a "wa-wa" — a getting louder and softer. And the closer the two tones are in frequency, the longer the "wa-wa". Can you see how this happens from our math? This is actually a useful way to tune a guitar string if you haven't got an electronic pitch generator (or even if you have). If you fret one string so it should play the same as another string, if they're close, you'll hear the beats. You can then tune one of the strings until the beats get longer and longer until they're not detectable. Then the two strings are properly tuned (relative pitch).

The speed of the packets

If we keep the full "$kx -ωt$" terms in our identity we can see that the result is the product of two $\sin{(kx -ωt)}$ terms, one with the averages of the $k$'s and $\omega$'s and one with the differences. So the are two things moving at once: the little wiggle inside the packets and the packets.

The speed that the little wiggles travel is called the phase velocity and the speed that the packets travel is called the group velocity. Since we know from Reading the content in a sinusoidal wave that the speed of a sinusoidal wave is $ω/k$, we can figure out the speeds of the two parts of the traveling beats. When the differences in the frequencies and wave numbers are small, we can write the velocity of the packets as

$$v_{phase} = \frac{ω}{k},$$ 

$$v_{group} = \frac{dω}{dk}.$$

If the velocity is independent of the frequency, these are the same since $ω$ is just proportional to $k$. But for waves in some media, the speed can be different for different frequencies, for example, light in glass. That's what's responsible for prisms making rainbows! It's also true in quantum mechanics where the speed of an electron wave is related to their energy which determines their frequency (as in photons). This leads to some complex and interesting phenomena!

Workout: Beats

Joe Redish 4/11/16