2
$\begingroup$

What happens to the frequency when two sound waves add up to a complex wave?

Example: I play a G3 on the guitar at 196 hertz and at the same time on the piano A4 440 hertz. What would be the "total frequency"? ( if there is such a thing)

Very thankful for any information.

$\endgroup$
  • 3
    $\begingroup$ This graph may help to visualize what's going on. $\endgroup$ – aRockStr Oct 3 at 19:42
  • $\begingroup$ Very good. Why not formulate an answer? $\endgroup$ – Gert Oct 3 at 19:54
  • $\begingroup$ If the two waveforms are not subjected to non-linear processes such as those having harmonic and/or intermodulation distortion, then no other components will be generated. $\endgroup$ – Richard Fry Oct 3 at 20:39
  • $\begingroup$ @Gert Thanks. That's why it's a comment. $\endgroup$ – aRockStr Oct 4 at 3:49
2
$\begingroup$

The waves do just 'add-up' their amplitudes, so that where two peaks combine you get a larger peak, where a peak combines with a trough the two amplitudes cancel, and where the two waves are both zero you get zero amplitude. Because the two waves have different frequencies, the alignment of their peaks and troughs varies over time in a way that isn't equal to either of their frequencies but is related to their relative values. You end up with a sound that has no well-defined frequency but seems to combine indictions of both the original frequencies plus a third effect that tends to be lower in frequency than either of the other two.

The effect depends on the relative amplitudes and frequencies of the two waves. You might find it easiest to get a mental picture if you imagine a few extreme cases. For example, consider a very low frequency wave with a large amplitude combined with a high frequency wave of small amplitude. The result will look like the original wave with tiny ripples on it. You can observe the effect on the surface of a large body of water in windy conditions when you can see long wavelength undulations with short wavelength ripples super-imposed.

An interesting effect occurs when you combine two sound waves with similar amplitudes. The alignment of their peaks and troughs varies from being almost exactly in phase to almost exactly out of phase over a time period that is equal to the difference between their frequencies, so you hear their two sounds overlaid by a periodic waxing and waning of volume known as 'beats'. If you know what to listen for you can use the sound of the beats to tune two neighbouring guitar strings, fretting the lower string so that it should play the same note as the higher one. If you tune the strings slightly out at first and pluck them together you can hear the beat, and you can adjust the tension until the period of the beat gets longer and eventually disappears.

When the wavelength of one wave is an integer multiple of the other, or the two are in a simple ratio like 2:3 you tend to get an overall effect that is pleasing to the ear. Such ratios are known as harmonics. Generally, when you pluck a string on a guitar it does not vibrate purely at its nominal frequency, but in a complex pattern that is the superposition of a number of harmonics. If you pluck the string carefully at its mid point then the resulting motion is dominated largely by its nominal frequency, whereas if you pluck it near either end you will activate more overtones.

$\endgroup$
-1
$\begingroup$

Good question!

When adding sinusoidal waves, there's a handy trig identity we can use

$$\sin(\alpha)+\sin(\beta)=2\cos(\frac{\alpha-\beta}{2})\sin(\frac{\alpha+\beta}{2})$$

So given that a standard sinusoidal wave can be expressed as

$$y=A\sin(kx-\omega t)$$

To add two waves of differing frequency (but the same amplitude and phase), we can take

$$y_1+y_2=A\sin\left(k_1x-\omega_1 t\right)+A\sin\left(k_2x-\omega_2 t\right)$$ $$\alpha=k_1x-\omega_1t \quad\quad\quad \beta=k_2x-\omega_2t$$ $$\alpha-\beta=\left(k_1-k_2\right)x-\left(\omega_1-\omega_2\right)t \quad\quad\quad \alpha+\beta=\left(k_1+k_2\right)x-\left(\omega_1+\omega_2\right)t$$ $$y_1+y_2=2A\cos\left(\frac{k_1-k_2}{2}x-\frac{\omega_1-\omega_2}{2}t\right) \sin\left(\frac{k_1+k_2}{2}x-\frac{\omega_1+\omega_2}{2}t\right)$$

This is where some lecture notes from RIT's Physics 207 really elegantly explain what's going on

$$\frac{a_1+a_2}{2}=a_{avg}$$ $$\frac{k_1+k_2}{2}=k_{avg} \quad\quad\quad \frac{\omega_1+\omega_2}{2}=\omega_{avg}$$

Which means that what we're really looking at is the following

$$y_1+y_2=2A\cos\left(\frac{\Delta k}{2}x-\frac{\Delta \omega}{2}t\right) \sin\left(k_{avg}x-\omega_{avg}t\right)$$

The sine term is just a wave of our original form $y=A\sin(kx-\omega t)$, but more specifically it's the average of our two waves.

The cosine term acts as a coefficient on the amplitude, causing the average wave from the sine term to increase and decrease with amplitude depending on the difference between our two components of wave speed $k$ and $\omega$.

Thus what we'd end up with is something like this (taken from the aforementioned RIT lecture notes): sum of two waves

Noteworthy is that it doesn't matter whether you interpret this as the high-frequency wave being given an amplitude equal to the $y$ value of the low-frequency wave at each point, or the low-frequency wave being given an amplitude equal to the $y$ value of the high-frequency wave at each point. The effect is the same. This corresponds to the commutative property of addition, which also applies to wave addition, and the commutative property of multiplication, which applies to the multiplied sine and cosine terms in the final equation. Either wave could be "bending" the other, and the effect would be the same.

If you want to consider waves that differ in phase, things get a bit more complicated, but I'm sure someone else would be more than happy to give details on that :)

$\endgroup$
  • 3
    $\begingroup$ The figure illustrates the phenomenon of beats. But OP asked about frequencies that differed a lot. There will not be an average frequency then. $\endgroup$ – Pieter Oct 3 at 20:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.