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I know amplitudes cancel (destructive) or combine (constructive) as per image below:

enter image description here

(Source)

But how do frequencies cancel out or combine?

For some context: a question from my textbook

A song is played off a CD. One set of speakers is playing the note at $512$ Hz, but the presence of the second set of speakres causes beats of frequency $4$ Hz to be heard at a point equidistanct from the four speakers. The possible frequencies being playing by the addition speakers are:

Answer: $508$ and $516$ Hz.

I am not sure if I understand the concept correctly, but amplitudes canceling out makes sense as its a matter of e.g. a negative amplitude cancelling out an equal magnitude positive amplitude and combining into one wave (or variations of this depending on the magnitude of each wave).

So to me this seems like a matter of distance/displacements (in the form of Amplitude, or distance above or below the centerline) canceling out.

But I don't see how this works for frequencies.

Frequency is waves/second. So wouldn't playing a $512$ Hz frequency and a $516$ Hz frequency just cause both of them to be heard separately, rather than cancel out to $4$ Hz?

I don't understand how "speeds" can cancel out.

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    $\begingroup$ You might have read this already: en.wikipedia.org/wiki/Beat_(acoustics) this is good as well Beats and Frequencies if you have ever heard a twin engined propellor plane overhead, as well as the normal engine sound , you will have heard beats, as it is virtually impossible to get the engines synchronised. $\endgroup$
    – user108787
    Dec 4, 2016 at 21:09
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    $\begingroup$ No I haven't read it already (I seem to have overlooked the word beats in the textbook question). Thanks for the link, hopefully it will make more sense after reading it. $\endgroup$
    – K-Feldspar
    Dec 4, 2016 at 21:12
  • $\begingroup$ I think (hope :) the post was badly written in that regard to the cancellation part, where would all the energy go if that was the case??? .....The irony of it is, people hear beats all the time, without realising it. Music without beats would be much less interesting. Sorry, partially my fault, I was try to tell you that when I said "as well as" in my coment, I should have been clearer. Anyway, you got good answers :) Regards $\endgroup$
    – user108787
    Dec 4, 2016 at 21:31
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    $\begingroup$ You understand if both frequencies are 512 hz and they are in phase, but what if the second one was 512.001 hz? After 500 seconds they would be out of phase and canceling. After another 500 seconds they would be back in phase and reinforcing. So it would be 1000 seconds between cancelling. That's a "wow-wow" sound at the difference between the two frequencies. Now back to 512 & 516. That's 4 "wow"s per second. $\endgroup$
    – Mike Dunlavey
    Dec 4, 2016 at 21:34
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    $\begingroup$ @CountTo10. Haha sorry. When I said "hopefully it will make sense after reading (the wiki link)" ... I meant hopefully my original textbook question will make sense. Your comment made sense. That is interesting! Weird thing is, I took physics courses in high school and uni but this is the first time I am hearing about beats. Or maybe I was away those days :S ? @ Mike Dunlavey . Thanks for that explanation, that gives me an easy to remember way to make sense of the concept should this come up in an an exam. $\endgroup$
    – K-Feldspar
    Dec 4, 2016 at 21:40

4 Answers 4

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Beats can be thought of as the next level of complication from constructive destructive interference. To demonstrate this best, we should visualize what actually happens when we sum two sine waves of different frequencies:

Beats

There's no magic going on here, this is just straight up addition.

What is happening is that sometimes the two signals are constructively interfering, and sometimes they are destructively interfering. The rate at which they go back and forth between constructive and destructive is defined by the difference in frequencies, and is called the "beat frequency." You can see that there is still a high frequency sine wave there... you still hear the "correct" note (I believe it's the average of the two frequencies), but you also hear what we call an "envelope," making that high frequency go louder and softer. Those are the beats.

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    $\begingroup$ @K-Feldspar Very close. You've got the right idea, though there's a divide by two missing. The beat frequency is always equal to the difference between the two frequencies, so in this case, the beats are charactarized by a sin(1*x) curve. However, if you consider what a sine/cosine curve does, it spends half its cycle positive, and half its cycle negative. So one "cycle" would be two of the oval shaped things: one positive one negative. Other than that detail, you've got it. $\endgroup$
    – Cort Ammon
    Dec 4, 2016 at 21:22
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    $\begingroup$ @K-Feldspar You have it correct. I think the difference between the two images is one of definitions... they may be using the term "beat frequency" to define a different thing than you or I am. Wikipedia has a section that explains why there are two competing definitions: "The frequency of the modulation is (f1 + f2)/2, that is, the average of the two frequencies. It can be noted that every second burst in the modulation pattern is inverted. Each peak is replaced by a trough and vice versa. $\endgroup$
    – Cort Ammon
    Dec 4, 2016 at 22:29
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    $\begingroup$ "...However, because the human ear is not sensitive to the phase of a sound, only its amplitude or intensity, only the magnitude of the envelope is heard. Therefore, subjectively, the frequency of the envelope seems to have twice the frequency of the modulating cosine, which means the audible beat frequency is $f_{beat}=f_2-f_1$" $\endgroup$
    – Cort Ammon
    Dec 4, 2016 at 22:29
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    $\begingroup$ @StockB There are many tools out there, because this is a pretty normal thing to do. This video shows the effect and the name of a website you could go to play around with. $\endgroup$
    – Cort Ammon
    Dec 5, 2016 at 18:48
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    $\begingroup$ @bright-star They are definitely sensitive to relative phases between the two ears, but that's a bit different of an effect. For example, if you hear a 180 degree phase shift, you almost don't notice it. (you usually notice a click if the phase shift occurred at a non-zero displacement). That being said, the mind's ability to identify the source of a sound is a spectacular feat in many ways! One of my favorite ones is one that suggests that we fuse data so well that you can almost hear with your eyes, but that's another story all together! $\endgroup$
    – Cort Ammon
    Dec 5, 2016 at 19:05
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As an alternative explanation, this beat interference is similar to the Moiré pattern which you get for instance if you overlay two sets of line gratings with different spacing.

Moiré-pattern of line grating with difference in spacing of 5%https://upload.wikimedia.org/wikipedia/commons/2/26/Moire1_95.png

In this picture the two line gratings would correspond to the two frequencies (512 Hz, 516 Hz) from your speakers and the dark Moiré pattern which has bigger spacing (=lower frequency) would correspond to your beat frequency (4 Hz).

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  • $\begingroup$ You'd get something even closer if you combined differently. Sound has positive and negative components, what you display above does not in the same way. Still, fun to look at. $\endgroup$
    – Yakk
    Dec 6, 2016 at 18:29
  • $\begingroup$ @Yakk: Not sure what you mean. How should I combine differently? I don't think sound having positive and negative components makes a difference. $\endgroup$
    – user1583209
    Dec 6, 2016 at 20:42
  • $\begingroup$ It would be different because sound can interact destructively, whereas your lines cannot. Still a useful visualization, +1 $\endgroup$
    – Kevin
    Dec 6, 2016 at 21:07
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    $\begingroup$ @user1583209 - More appropriate would be a pattern of black, grey, and white, where white acts as the negative. When you combine the black and white you get grey, black and black is black, so on. grey and white is between white and grey, and grey and black is between black and grey. Just black and white fails to have a zero, or rather a negative, just 1 and 0, black and white, on or off. If white and black mix on your image, you get black, not a mid-color. It is still a very nice visualization. $\endgroup$
    – EvSunWoodard
    Dec 6, 2016 at 21:13
  • $\begingroup$ Agreed. I was thinking whether it is possible to get away with black and white only, because if one looks at the interference pattern from a distance (so that individual lines are not visible) you can see black, grey and white patterns. $\endgroup$
    – user1583209
    Dec 6, 2016 at 21:17
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Well, imagine that you superpose two signals (i.e. using two speakers, one emitting a signal at $f_1$ and another at $f_2$). Imagine that these signals are in phase at $t = 0$. Since they have very different frequencies, they will oscillate at very different speeds, and if you add their waveforms just as in the picture, the sum will appear to be random.

If, however, the frequencies are quite similar ($|f_1-f_2|$ is small), then at the beginning the signals will remain approximately in phase for some time, and add constructively. One of the signals will slowly drift though, and at some point they will reach anti-coincidence, and add destructively. It is very easy to understand mathematically, Using $\cos(x) + \cos(y) = 2\cos(\frac{x-y}{2})\cos(\frac{x+y}{2})$. Using these formulas we can find the output amplitude of the two-speaker device : \begin{equation} S(t) = 2\cos\left(2\pi \frac{f_1-f_2}{2} t\right)\cos\left(2\pi\frac{f_1+f_2}{2} t\right) \end{equation}

The shape of this signal is the following (blue) :

beats

The envelope is due to the beats modulation frequency, which equates $|f_1-f_2|$. In your case, it has to be 4 Hz, so : \begin{equation} f_2 = f_1 \pm \mbox{ 4 Hz} \end{equation}

So, if $f_1$ is 512 Hz, then $f_2$ is either 508 Hz or 516 Hz. So, the spectrum is still just two peaks at $f_1$ and $f_2$, and you are right to say that they are the frequencies heard, but the envelope of the signals is periodic with a frequency 4 Hz.

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  • $\begingroup$ Showing the mathematics behind this is the right approach. Gets my vote $\endgroup$
    – Floris
    Dec 5, 2016 at 21:38
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They do not cancel out to 4 Hz! The result of interference of 512 Hz and 516 Hz sound waves is a beat, which is basically a sound wave at average frequency (514 Hz) where the amplitudes change over time with frequency 4 Hz, i.e. you'd hear a sound of 514 Hz changing in volume from loud to silent.

beat

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    $\begingroup$ Technically from loud to silent to loud, to silent and back to loud again at 4Hz. $\endgroup$
    – Mad Physicist
    Dec 5, 2016 at 16:07
  • $\begingroup$ Agreed. I fixed it in the answer. $\endgroup$
    – user1583209
    Dec 6, 2016 at 1:31

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