# Musical Acoustics – ½ Wavelength Resonators, Part 2

Last week we discussed the characteristics of the quarter wavelength resonator and its role in musical acoustics (as well as in behind the ear hearing aids and our vocal tracts for those speech scientists out there). Quarter wavelength resonances are found in tubes that are “open” at one end and “closed” at the other end. A feature of a quarter wavelength resonator is that there are odd numbered multiples of the lowest frequency resonance. An example that was given was the articulation of the vowel [a] as in ‘father’ where the first resonance (also called formant) was at 500 Hz, the second was at three times that (1500 Hz), the third at five times that (2500 Hz), and so on. Most brass instruments function as quarter wavelength resonators and this includes the trumpet (but not the coronet), the trombone, the tuba, French horn, and so on. We’ll come back to the coronet shortly. Trumpets are “closed” at the mouth piece end and “open” at the bell end. Our vocal tract, during the articulation of the vowel [a] is “closed” at the vocal cord end “open” at the lips.

Well, what about musical instruments that are “closed” at both ends (or equivalently “open” at both ends? The guitar strings are held tightly at both ends and as such would function as a half wave length resonator. The same is true of the piano – strings are held tightly at both ends. And the same is true of all stringed instruments (e.g., the violin, viola, cello, and bass).

Unlike quarter wavelength resonators that have odd numbered multiples of its first resonance, half wave length resonators have integer multiples (1, 2, 3, 4, …) of its first resonance. The formula looks similar to the quarter wavelength resonator formula from last week but has a few differences. Here is its formula: F = kv/2L. The speed of sound is still given by “v” and for our purposes is 34,000 cm/sec. The length of the tube is still given by L, just as it was for quarter wavelength resonators, but we now only divide by 2 (not 4) so the resonances will be twice as high as if the instrument of the same length were a quarter wavelength resonator – that is, they would all be one octave higher in pitch. And the other difference is that the multiplier is not (2k-1) as it was last week, but only k, so that the resonances are much more tightly packed than a quarter wavelength resonator.

Woodwind musical instruments that are half wavelength resonators have an “octave key” that doubles the frequency and essentially moves the note to the next resonance. Oboes, saxophones, and bassoons all have octave keys, as do recorders but not clarinets. A clarinet has a “register key” that increases the note by one and a half octaves (or a 12th). That is, a register key doesn’t just increase by a factor of 2 (an octave) but increases the note by a factor of 3 which is exactly what would be expected from a quarter wavelength resonator- resonances are 1x, 3x, 5x, and so on, the frequency of the lowest resonance. When I play the note C on a saxophone, by pressing my octave key, I get to the C an octave higher. When I play the note C on a clarinet, when I press the register key, I get a G (one and a half octaves higher). If we used numbers rather than musical notes, a one half wavelength resonator such as an oboe would generate C at 262 Hz and then with the octave key generate C at 512 Hz- one octave higher. On my clarinet, if I played C at 262 Hz, and I hit the “register key”, the note would be G at 3 x 262 Hz or 786 Hz.

So what’s going on?   Both the clarinet and the saxophone seem to be “closed” at the mouth piece end, and “open” at the other end, yet one is a half wavelength resonator and the other is a quarter wavelength resonator. Actually to really confuse you, the clarinet actually functions as a half wavelength resonator in the upper register- it is a quarter wavelength resonator in the lower register. That’s why I love it so much. Actually I love it for its sound; I hate it for its weird acoustics. And why can’t the clarinet by a C instrument like the piano or violin? I hate having to transpose, but that’s another blog!

But back to resonators. Well, I sort of lied to you. It is true that quarter wavelength resonators are open at one end and closed at the other end, and half wavelength resonators are either closed at both ends or open at both ends, but that is assuming that the tube has a relatively uniform inner diameter, like that of a behind the ear hearing aid (and a clarinet).   The width of my clarinet near my mouth piece is roughly the same as the width near the bottom- let’s forget about the small flare of the clarinet near the bottom- that’s just for ornamentation and is not a real flare… more about that later in Part 4 of this blog in two weeks.

The inner diameter of the oboe, saxophone, and bassoon, all change but change in a uniform manner- they are “conical”. When this happens they function as half wavelength resonators despite being open at one end and closed at the other. If my clarinet were to be made of pliable rubber and if I were to gradually flared out of the bore into a conical form, then it too would start to behave as a half wavelength resonator.

There is one other structure that has properties of a quarter wavelength resonator but also may be more complex than we first thought- our outer ear canals.

Our ear canals are about 3 cm long- actually they are a bit shorter but let’s assume that their length is roughly 3 cm (30 mm) long. Ear canals are “open” at the medial side and “closed” by the tympanic membrane at the lateral side, so they should function as quarter wavelength resonators. Using the quarter wavelength resonator formula from last week’s blog F = (2k-1)v/4L, the lowest resonant frequency (set k = 1) is 34,000/4 x 3 = 2833 Hz and indeed this is not too far off the measured real ear unoccluded resonance (REUR) of 2700 Hz. So far so good. However, the ear canal is conical- larger on the lateral side and narrower on the medial side- kind of like a side-ways oboe. If the outer ear canal was functioning as a half wavelength resonator (like an oboe) then using the formula F = kv/2L, we would get a resonance of 34,000/2 x 3 = 5666 Hz. We do have a resonance in the 5600 Hz region which researchers have attributed to the volume of air in the concha. However, it is quite possible that this resonance is not purely a concha-related resonance, but the ear canal functioning somewhat as a half wavelength oboe-like resonance.

If indeed we were to remove the outer ear from KEMAR or just placed a large wad of gum or putty in KEMAR’s ear, indeed the 5600 Hz “concha-related resonance” would indeed disappear. Now, using a probe tube microphone, try this experiment on yourselves. Don’t surgically remove your pinna, but just place a large wad of putty obscuring the concha bowl- it doesn’t always disappear- it certainly doesn’t with me. I have written an article about this and you can find it here. If I was ever criticized by acoustic researchers, this is one article where I received countless emails, but try it for yourselves- in some people it disappears (supporting the concha related volume resonance) and in others it remains (supporting the half wavelength character).

This is an area where musical acoustics informs human acoustics.

Next week in Part 3 of this blog series, we will dive into the wonderful world of impedance and damping!