# I Musical Acoustics: Beats

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1. Jul 29, 2016

### DaydreamNation

I have a question about 'beats' between two sound waves of slightly different frequencies. Basically, I think I understand everything on this page (and have read a few textbooks on this): http://www.animations.physics.unsw.edu.au/jw/beats.htm . And in practice, I work with this concept all the time. However, I still do not entirely understand why the beat frequency is the absolute value of the difference between the two frequencies. I understand the mathematical explanation on the page that I linked. However, I still do not grasp, physically, why the peaks or troughs of two waves moving at different frequencies would coincide that many times per second.

Let me elaborate. Hartmann writes in Principles of Musical Acoustics: "
Suppose that we measure the two waves from the two tuning forks (at 500 Hz and 501 Hz) when each is at a positive maximum. They are “in phase.” Then the sum will be as loud as it
can be. As time progresses the two waves will become out of phase because the
forks have different frequencies. After half a second the 500-Hz wave will have
executed 250 complete cycles and it will be at a positive maximum again. However,
the 251-Hz wave will have executed 250.5 cycles, and the extra half a cycle will
mean that this wave is at a negative maximum. The positive maximum and negative
maximum will cancel and the sum will be quiet. After a full second, the 500-Hz
wave will have gone through 500 cycles and be at a positive peak; the 501-Hz wave
will have gone through 501 complete cycles and also be at a positive peak. As a
result, the waves will once again be perfectly in phase and there will be a maximum
again. This example shows that there is one maximum and one minimum per second
when the frequency difference is 1 Hz."

So far so good.

Then he gives the example of waves moving at 500 Hz and 510 Hz:
Now there are ten beats per second. Ten beats per second is rather fast but one can
still hear them as a rapidly varying loudness. Again, we choose to start the clock
when both waves are together. After 50 ms (0.05 s) the 500-Hz wave will have gone
through 25 cycles but the 510-Hz wave will have gone through 25.5 cycles. The
waves will be out of phase and will cancel. After 100 ms the two waves will have
gone through 50 and 51 complete cycles and will be in phase again.

This makes sense to me. However, I think that this only works because 10 is the greatest common factor of the two numbers he has chosen. If we combine sound waves moving at 179 Hz and 189 Hz, the beat frequency would still be 10 Hz but the physical explanation would be much less clear. After 100 ms, the waves would have gone through 17.9 cycles and 18.9 cycles, respectively, meaning that neither one would be at its peak. So why would there be a 'maximum' here? In fact, I don't think the two waves' peaks would coincide more than once per second. So why do we get 10 beats/s in this case (other than because trigonometry says so)?

2. Jul 29, 2016

### Staff: Mentor

I think the absolute value is used to simply make the frequency always positive as there is no notion of a negative frequency.

To see why you could experiment with a graphing calculator:

https://www.desmos.com/calculator

try sin(10x) vs (sin(500x)+sin(510x)) and see what you get.

Last edited: Jul 29, 2016
3. Jul 29, 2016

### andrewkirk

The maximum will occur at around 100ms +0.1 x 1000ms/179 = about 100.6ms because that's when the 179Hz wave is next at a peak (it is 0.1 cycles later than the synchronisation time), and the 189Hz wave will still be very close to in sync with it, so that will be very close to a peak.

If the ratio of the frequency of the waves to the frequency difference is large (call it $R$), the first peak of one wave after the synchronisation time will occur at a time when the other wave is very close to peak. If the waves peak when phases are at multiples of $2\pi$ and the phase (modulo $2\pi$) of the waves at synchronisation is $-\theta$, the phase of the lower freq wave at the time of the higher freq wave's next peak will be $\frac{-\theta}{R}$, which will be very close to zero, hence very close to a peak.

4. Jul 30, 2016

### DaydreamNation

At this point, the 179 Hz wave would have completed 18 cycles and the 189 Hz wave would have completed approximately 19.006 cycles. When the 179 Hz wave completes 36 cycles (another maximum if I'm understanding you), the 189 Hz wave would have completed approximately 38.011 cycles (meaning it will be at a slightly different phase than it was at the previous maximum and, thus, the combined amplitude of the two waves will be slightly different than at the previous maximum, both of which would be different from the combined amplitude at 0s. And so on: at each of the 10 maxima and minima, the combined amplitude will be slightly different. However, if the frequencies of the two waves were 180 Hz and 190 Hz instead, the combined amplitude at each maximum would be exactly the same because both waves would begin a cycle every 100 ms. I would also think that the 'zero' points (where there is destructive interference) will be closer to zero in the latter case. This should mean that the beating produced by two waves whose frequencies have a common factor will be more even than the beating produced by two waves whose frequencies do not. When I try to listen for this with sine waves, I feel like I can hear it a little bit now. Is this actually true? If so, it seems significant but I don't see it in any textbooks!

I'm not sure I follow all of this. When you say "ratio of the frequency of the waves to the frequency difference", which wave's frequency do you use for the numerator? Or are there two Rs? And, again, does "phase of the waves at synchronisation" refer to the phase difference between the two waves or the phase of the higher-frequency wave or the phase position of the lower-frequency wave? What if both of them synchronise at phase 0?

I understand both of these things. I realise that a calculator or audio demonstration will show me the beating, and I think I understand the trigonometric explanation given on the page that I linked. I just want to understand how the interference patterns actually work, physically.

5. Jul 30, 2016

### andrewkirk

That is correct.
Use the higher frequency, as that is the wave that will have the nearest peak to the sync time.
It's the phase of both of them modulo $2\pi$, which will be the same number since they are synchronised at that point.

I've written a few formulas that may help explain and expand on this. I won't post them yet because I'm considering whether it might make a worthwhile Insights article. If not I'll post them here. Otherwise I'll post a link.

6. Jul 31, 2016

### andrewkirk

Here is the Insights article that goes into the phenomenon in a little more detail.

7. Jul 31, 2016

### DaydreamNation

That looks great! Thanks!