# Overtones – What They Are And How To Compute Them

In theory, hitting the middle C on a piano should produce a sound wave with a frequency of 523.25 Hz and nothing else. However, running the resulting audio through a spectrum analyzer, it becomes obvious that there’s much more going on. This is true for all other instruments, from tubas to trumpets, basoons to flutes, contrabasses to violins. Play any note and you’ll get a package of sound waves at different frequencies rather than just one.

First of all: why is that? Let’s focus on stringed instruments. When you plug the string, it goes into its most basic vibration mode: it moves up and down as a whole at a certain frequency f. This is the so called first harmonic (or fundamental). But shortly after that, the nature of the vibration changes and the string enters a second mode: while one half of the string moves up, the other half moves down. This happens naturally and is just part of the string’s dynamics. In this mode, called the second harmonic, the vibration accelerates to a frequency of 2 * f. The story continues in this fashion as other modes of vibration appear: the third harmonic at a frequency 3 * f, the fourth harmonic at 4 * f, and so on. A note is determined by the frequency. As already stated, the middle C on the piano should produce a sound wave with a frequency of 523.25 Hz. And indeed it does produce said sound wave, but it is only the first harmonic. As the string continues to vibrate, all the other harmonics follow, producing overtones. In the picture below you can see which notes you’ll get when playing a C (overtone series): (The marked notes are only approximates. Taken from http://legacy.earlham.edu)

Quite the package! And note that the major chord is fully included within the first four overtones. So it’s buy a note, get a chord free. And unless you digitally produce a note, there’s no avoiding it. You might wonder why it is that we don’t seem to perceive the additional notes. Well, we do and we don’t. We don’t perceive the overtones consciously because the amplitude, and thus volume, of each harmonic is smaller then the amplitude of the previous one (however, this is a rule of thumb and exceptions are possible, any instrument will emphasize some overtones in particular). But I can assure you that when listening to a digitally produced note, you’ll feel that something’s missing. It will sound bland and cold. So unconsciously, we do perceive and desire the overtones.

If you’re not interested in mathematics, feel free to stop reading now (I hope you enjoyed the post so far). For all others: let’s get down to some mathematical business. The frequency of a note, or rather of its first harmonic, can be computed via:

(1) f(n) = 440 * 2n/12

With n = 0 being the chamber pitch and each step of n one half-tone. For example, from the chamber pitch (note A) to the middle C there are n = 3 half-tone steps (A#, B, C). So the frequency of the middle C is:

f(3) = 440 * 23/12 = 523.25 Hz

As expected. Given a fundamental frequency f = F, corresponding to a half-step-value of n = N, the freqency of the k-th harmonic is just:

(2) f(k) = k * F = k * 440 * 2N/12

Equating (1) and (2), we get a relationship that enables us to identify the musical pitch of any overtone:

440 * 2n/12 = k * 440 * 2N/12

2n/12 = k * 2N/12

n/12 * ln(2) = ln(k) + N/12 * ln(2)

n/12 = ln(k)/ln(2) + N/12

(3) n – N = 12 * ln(k) / ln(2) ≈ 17.31 * ln(k)

The equation results in this table:

 k n – N (rounded) 1 0 2 12 3 19 4 24 5 28

And so on. How does this tell us where the overtones are? Read it like this:

• The first harmonic (k = 1) is zero half-steps from the fundamental (n-N = 0). So far, so duh.
• The second harmonic (k = 2) is twelve half-steps, or one octave, from the fundamental (n-N = 12).
• The third harmonic (k = 3) is nineteen half-steps, or one octave and a quint, from the fundamental (n-N = 19).
• The fourth harmonic (k = 4) is twenty-four half-steps, or two octaves, from the fundamental (n-N = 24).
• The fifth harmonic (k = 5) is twenty-wight half-steps, or two octaves and a third, from the fundamental (n-N = 28).

So indeed the formula produces the correct overtone series for any note. And for any note the same is true: The second overtone is exactly one octave higher, the third harmonic one octave and a quint higher, and so on. The corresponding major chord is always contained within the first five harmonics.

# Wavelength (And: Why Is The Sky Blue?)

A very important type of length is wavelength, usually symbolized by the Greek letter λ (in m). It is defined as the distance from crest to crest (one complete cycle) and can easily be calculated for any wave by dividing the speed of the wave c (in m/s) by its frequency f (in Hz):

λ = c / f

What are typical wavelengths for sound? At room temperature, sound travels with a speed of c = 343 m/s. The chamber pitch has a frequency of f = 440 Hz. According to the equation, the corresponding wavelength is:

λ = 343 / 440 ≈ 0.8 m ≈ 2.6 ft

Are you surprised? I bet most people would greatly underestimate this value. Bass sounds are even longer than that. The lowest tone on a four-string bass guitar has a frequency of f = 41.2 Hz, which leads to the wavelength:

λ = 343 / 41.2 ≈ 8.3 m ≈ 27 ft

So the wave coming from the open E string of a bass guitar doesn’t even fit in a common room. In the case of light, the situation is very different. As noted in the introduction, the wavelength of light ranges between 4000 A (violet light) and 7000 A (red light), which is just below the size of a bacterium.

Wavelength plays an important role in explaining why the sky is blue. When light collides with a particle, parts of it are deflected while the rest continues along the initial path. This phenomenon is known as scattering. The smaller the wavelength of the light, the stronger the effect. This means that scattering is particularly pronounced for violet and blue light.

Unless you are looking directly at the Sun, all the light you see when looking at the sky is scattered light coming from the particles in the atmosphere. Since blue light tends to scatter so easily, the sky ends up in just this color. But why not violet? This is a legitimate question. After all, due to its smaller wavelength, violet light is even more willing to scatter. While this is true, it is also important to note that the sun’s rays don’t contain all the colors in the same ratio. In particular, they carry much less violet than blue light. On top of that, our eyes are less sensitive to violet light.

(This is an excerpt from my Kindle book: Physics! In Quantities and Examples)

# Estimating Temperature Using Cricket Chirps

I stumbled upon a truly great formula on the GLOBE scientists’ blog. It allows you to compute the ambient air temperature from the number of cricket chirps in a fixed time interval and this with a surprising accuracy. The idea actually quite old, it dates back 1898 when the physicist Dalbear first analyzed this relationship, and has been revived from time to time ever since.

Here’s how it works: count the number of chirps N over 13 seconds. Add 40 to that and you got the outside temperature T in Fahrenheit.

T = N + 40

From the picture below you can see that the fit is really good. The error seems to be plus / minus 6 % at most in the range from 50 to 80 °F. # Another Home Experiment – Wind Speed and Sound Level

Recently I told you about my home experiment regarding impact speed and sound level. I did another experiment with my sound level meter, this time I was interested in finding out how the sound level varies with the wind speed. So I took my anemometer (yep, that’s a thing) to measure the wind speed and at the same time noted the sound level. I collected some data points and plotted them. Here’s the result: As you can see the fit is not that bad (the adjusted r-square is 0.91).

So the sound level grows with the wind velocity to the power of 0.22, meaning that if the wind speed increases by a factor of twenty-five, the sound level doubles. According to the empirical formula, the noise from the wind inside a category 1 and 2 hurricane is comparable to the sound level at a rock concert. This is of course assuming that the formula holds true past the 12 m/s range over which it was determined (which is not necessarily the case, but for now the best guess).

# Space Shuttle Launch and Sound Suppression

The Space Shuttle’s first flight (STS-1) in 1981 was considered a great success as almost all the technical and scientific goals were achieved. However, post flight analysis showed one potentially fatal problem: 16 heat shield tiles had been destroyed and another 148 damaged. How did that happen? The culprit was quickly determined to be sound. During launch the shuttle’s main engine and the SRBs (Solid Rocket Boosters) produce intense sound waves which cause strong vibrations. A sound suppression system was needed to protect the shuttle from acoustically induced damage such as cracks and mechanical fatigue. But how do you suppress the sound coming from a jet engine?

Let’s take a step back. What is the source of this sound? When the hot exhaust gas meets the ambient air, mixing occurs. This leads to the formation of a large number of eddies. The small-scale eddies close to the engine are responsible for high frequency noise, while the large-scale eddies that appear downstream cause intense low-frequency noise. Lighthill showed that the power P (in W) of the sound increases with the jet velocity v (in m/s) and the size s (in m) of the eddies:

P = K * D * c-5 * s2 * v8

with K being a constant, D the exhaust gas density and c the speed of sound. Note the extremely strong dependence of acoustic power on jet velocity: if you double the velocity, the power increases by a factor of 256. Such a strong relationship is very unusual in physics. The dependence on eddy size is also significant, doubling the size leads to a quadrupling in power. The formula tells us what we must do to effectively suppress sound: reduce jet velocity and the size of the eddies. Water injection into the exhaust gas achieves both. The water droplets absorb kinetic energy from the gas molecules, thus slowing them down. At the same time, the water breaks down the eddies.

During the second Space Shuttle launch (STS-2) a water injection system was used to suppress potentially catastrophic acoustic vibrations. This proved to be successful, it reduced the sound level by 10 – 20 dB (depending on location), and accordingly was used during every launch since then. But large amounts of water are needed to accomplish this reduction. The tank at the launch pad holds about 300,000 gallons. The flow starts at T minus 6.6 seconds and last for about 20 seconds. The peak flow rate is roughly 15,000 gallons per seconds. That’s a lot of water!

The video below shows a test run of the sound suppression system:

http://www-pao.ksc.nasa.gov/nasafact/count4ssws.htm

# Home Experiment – Impact Speed and Sound Level

A while ago I got my hands on a sound level meter and pondered what to do with it. Sound level versus distance from source? Too boring, there’s already a formula for that (see here: Intensity: How Much Power Will Burst Your Eardrums?). What I noticed though is that I’ve never seen a formula relating impact height or speed to sound level, that seemed interesting. So I bought a small wooden sphere at a local store and dropped it from various heights, at each impact recording the maximum sound level. I dropped the sphere from 8 different heights and to reduce the effect of random fluctuations 20 times from each height. So in total I collected 160 data points. I’m not so sure if my neighbors were happy about that.

I calculated the impact speed v from the drop height h using the common v = sqrt (2 * g * h). As you might know, this formula neglects air resistance. However, I’m not concerned about that. The wooden sphere was small and massive and only dropped from heights below about 1 ft. The computed impact speed shouldn’t be off by more than a few percent.

Here’s the resulting plot of impact speed versus sound level (in decibels): The fit turned out to be fantastic and implies that if you increase the impact speed by a factor of five, the sound level doubles. What’s the point of this? I don’t know, but it’s a neat graph and that’s good enough for me.