David Webber, author of MOZART, keen sax player, and occasional clarinettist has always wondered why the clarinet overblows at the twelfth whereas the sax overblows at the octave. He was always told it is because the sax has a conical bore, while the clarinet has a cylindrical bore. Well it has to be that, doesn't it? Because it's the only essential difference between the two. Well now he has actually sat down and done the maths, and the answer drops out beautifully from an analysis of the acoustics of idealised cylindrical and conical tubes. This article will not help you with MOZART: consider it simply a caprice for woodwind fans (who don't mind just a little bit of maths).
The problem is most poignantly illustrated by the clarinet and the soprano sax. The clarinet produces a much lower note than the
similarly sized saxophone and overblows at the twelfth rather than the octave. Why?
The difference is only that of a cylindrical tube (clarinet) vs a conical one (saxophone). How does this make such a difference to the properties of the instruments?
It is also worth noting that the flute (cylindrical) and oboe (conical) for example display no differences of this nature - both have approximately the same lowest note and both overblow at the octave. Any credible theory must also explain this.
Classically, the problem is approached by examining the acoustic properties of idealised tubes. Here we shall present this analysis for open and closed cylindrical and conical tubes.
The analysis for cylindrical tubes is presented in many elementary texts, but these are often shy of presenting the slightly more complicated analysis of conical tubes, and thus may be somewhat unconvincing in their attempts to explain the difference between clarinets and saxophones.
A sound wave in air obeys the general equation
(1)
where c is the (constant) speed of sound. F here is the potential and from it the fluctuating components P of the pressure and U of the velocity can be derived as
(2)
where r is the mean density of the air.
From this point let us consider only longitudinal waves in the instrument, ignoring any which might be transverse to the sides of the tube. In cylindrical and conical tubes these equations reduce slightly differently:
Longitudinal waves in a cylindrical pipe
In the case of longitudinal waves inside a cylindrical instrument, the equations reduce to
(1x)
(2x)
where x is the distance along the instrument measured from the mouthpiece end, and the oscillating air velocity U is directed along the axis of the instrument.
Longitudinal waves in a conical pipe
In the case of longitudinal waves inside an instrument of conical bore the equations reduce to
(1r)
(2r)
where r is the distance from the point of the cone, and the oscillating air velocity U is directed along the instrument, radially from the apex of the cone. (The above form of the wave equation derives from the fact that the conical bore of the instrument may be considered as a sector of a sphere centred on the apex of the cone, so that a spherical co-ordinate frame is appropriate.)
The standing wave solutions of the wave equation are slightly different in each case. Solving the wave equation for standing waves in F and then deriving P and U, we can write them in the forms:
Standing waves in a cylindrical pipe
(3x)
Standing waves in a conical pipe
(3r)
In these solutions
Note that the above are solutions of a given frequency and wavelength (and the general solution is any superposition of these). The wavelength in the above solutions is
(5)
and the frequency is
(6)
Important: none of the potential, the pressure, or the velocity, are actually periodic in distance along a conical pipe. The 1/(kr) factors spoil this periodicity. The combinations (kr).F and (kr).P are periodic, but the velocity U cannot be made periodic even with such a multiplying factor. However, the wave is in every case (co)sinusoidal in time with frequency v and so each of these solutions does correspond to a simple pure note. When we talk about "wavelength" it must be understood, in the case of a conical pipe, that we are referring to the wavelength of the periodic function (kr).P.
The allowed wavelengths are determined by the boundary conditions at the ends of the instrument. Two types of boundary conditions are generally used.
Different instruments are all generally considered to be open at the bell, but either closed or open at the mouthpiece end, according to the form of the mouthpiece and the means of sound production. They are referred to as closed pipes and open pipes respectively.
Reed instruments are modelled as closed pipes; the flute is modelled as an open pipe. In what follows we shall see how the solutions of the wave equations defined above reflect the properties of the instruments experienced by players. We shall examine the four possibilities presented by open and closed cylindrical and conical pipes. We shall define the length of the pipe to be L.
A closed cylindrical pipe (clarinet) | A closed conical pipe (sax, oboe, bassoon) |
Let the mouthpiece end be at x=0 and the the bell end at x=L. Then U(x=0)=0 and P(x=L)=0. From equ(3x) we find sin(b)=0 and cos(kL+b)=0. Thus b=0 and cos(kL)=0. |
Consider the the mouthpiece end to be at the closed apex of the cone at r=0 and the bell end to be at r=L. Then U(r=0)=0 and P(r=L)=0. From equ(3r) we find that the first condition is satisfied by b=0 (though the limit as r goes to zero has to be taken carefully) and the second then gives sin(kL)=0. |
An open cylindrical pipe (flute) | An open conical pipe |
Let the mouthpiece end be at x=0 and the the bell end at x=L. Then P(x=0)=0 and P(x=L)=0. From equ(3x) we find cos(b)=0 and cos(kL+b)=0. If we set b=p/2 to satisfy the first condition, then the second becomes sin(kL)=0. |
To be open at the mouthpiece end, the pipe cannot go all the way to a point. Therefore let us saw it off at some small distance r=s. To keep the tube length as L, let the bell end be at r = s+L. Then P(r=s)=0 and P(r=L+s)=0. Equ(3r) gives sin(ks+b)=0 and sin(k(L+s)+b)=0. Thus b=-ks and sin(kL) = 0. |
The closed cylindrical pipe (clarinet) | All conical pipes and the open cylindrical pipe (saxophone, oboe, bassoon, and flute) |
cos(kL) = 0 implies
kL = p.n/2 for any odd integer n=1,3,5,7,.... and equation (5) gives the wavelength
for any odd integer n=1,3,5,7,....
The wavelength of the fundamental (n=1) is 4 times the length of the instrument, and the first available harmonic (n=3) is one third of that wavelength - a twelfth, not an octave, higher. |
sin(kL)= 0 implies
kL = p.n for any integer n=1,2,3,4,... and equation (5) gives the wavelength
for any integer n=1,2,3,4,...
The wavelength of the fundamental (n=1) is twice the length of the instrument, and the first available harmonic (n=2) is half that wavelength - an octave higher. |
It is instructive to sketch the resultant wave forms, in the four cases.
The closed cylindrical pipe
(clarinet)
The open cylindrical pipe
(flute)
The closed conical pipe
(oboe, bassoon, saxophone)
The open conical pipe
We noted above that the pressure and velocity waves in the conical pipe are not simple periodic functions with the wavelength of the note. Comparing the wave forms of the closed conical pipe and the closed cylindrical pipe, we see that the velocity profiles for the fundamental and the harmonic above actually look very similar, even though the frequencies of the notes produced (for a given length of pipe) are not!
Thus this simple approach yields the principal differences between the clarinet and the other woodwinds. The clarinet's lowest note is much lower than the others (for a given length of the tube) and it overblows at the twelfth, where all the others overblow at the octave. Interestingly overblowing at the twelfth requires both a closed pipe and a cylindrical one. The understanding of the differences between the clarinet and the soprano saxophone, which are almost identical apart from the cylindrical and conical bores, is particularly convincing.