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<salmonegg@sbcglobal.n
science forum beginner
Joined: 13 Sep 2005
Posts: 28
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 Posted: Tue Sep 20, 2005 5:40 pm Post subject:
Oboes vs clarinets
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This is a restatement of a previous post but I now have a better
understanding of how to ask the question. It is based upon a link given in a
reply by ddavis@pb.com.au; http://www.phys.unsw.edu.au/~jw/pipes.html.
If I understand correctly, a clarinet using a uniform diameter bore will
have the same fundamental tone as an oboe using a conical bore that is twice
as long. That is, an oboe has the same fundamental tone as that of a flute
of the same length or a clarinet of half the length.
Why is the fundamental resonance of an oboe that of a half-wave resonator
instead of a quarter-wave as one would expect from a closed-open pipe?
Before looking at the the link, I thought that the fundamental tone of the
oboe would be the same as for a clarinet of the same length, but that it
would support both even and odd quarter-wave harmonics.
The only thing I can think of is that the open end of a conical pipe does
not reflect odd quarter-wave harmonics well enough so as to cancel amplitude
well at the reed end of an oboe. Why it can cancel for a even quarter-wave
(integral half-wave) is also a mystery.
Bill |
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Angelo Campanella
science forum Guru Wannabe
Joined: 08 May 2005
Posts: 226
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| Posted: Wed Sep 21, 2005 1:01 pm Post subject:
Re: Oboes vs clarinets
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salmonegg@sbcglobal.net wrote:
| Quote: | If I understand correctly, a clarinet using a uniform diameter bore will
have the same fundamental tone as an oboe using a conical bore that is twice
as long. That is, an oboe has the same fundamental tone as that of a flute
of the same length or a clarinet of half the length.
Why is the fundamental resonance of an oboe that of a half-wave resonator
instead of a quarter-wave as one would expect from a closed-open pipe?
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First I must confess that I am not a musician in the sense that I play
an instrument. I do not have intimate knowledge of "technique".
Having said that, I suspect that all these instruments can be made to
resonate in a 1/4 wave or 1/2 wave mode shape depending on the end
conditions. The taper has to do more with the open end condition. All
depend on the dynamic impedance at the mouth piece. Part of the
musician's technique has to be total control of the acoustic impedance
at the mouth piece end. Nothing I know of in teaching and learning
instruments is based on sound engineering principles. Rather, all
"method" has been determined empirically throughout the life of the
instrument type (centuries by now). It is up to the acoustical community
(all of us) to reverse engineer the instrument systems to determine
what's really going on.
| Quote: | Before looking at the the link, I thought that the fundamental tone of the
oboe would be the same as for a clarinet of the same length, but that it
would support both even and odd quarter-wave harmonics.
The only thing I can think of is that the open end of a conical pipe does
not reflect odd quarter-wave harmonics well enough so as to cancel amplitude
well at the reed end of an oboe. Why it can cancel for a even quarter-wave
(integral half-wave) is also a mystery.
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The closed end condition (1/4, 3/4, 5/4) does not favor even harmonics
as does open-both-ends (1/2, 1, 1-1/2, 2...).
Also, we are concerned with *radiated* sound; a mode can be there, but
if it does not radiate sound, it does not count in the popular sense. It
seems reasonable that the taper will favor radiation of sound traveling
to the open end. Experience with loudspeaker and microwave horns (having
considerable taper) is that they radiate well above an initial cutoff
frequency which is about 1/4 wave.
Coming back to an earlier point, whether the instrument tube is
operated as a 1/4 (closed end) or 1/2 wave (open end) resonator is under
the direct control of the musician and the mouthpiece he chooses to
install. Clearly, in the case of the flute, the musician has chosen to
use no mouthpiece at all, thus allowing an open end condition.
In the case of the clarinet, the mouthpiece has a rather tiny aperture
and stiff reed, presenting a high acoustic impedance, allowing only
enough flow of pulsed air to create excitation, thus presenting in the
main a closed end condition.
In the case of the oboe, The acoustic impedance of the frail double
reed may not be as high as that of the clarinet reed, thus presenting a
soft impedance, allowing a more open than closed end condition, the
musician accordingly allowing his mouth, throat and lungs to participate
in the resonance... "breathing" the tone. If so, it will take more
training, sophisticated techniques and experience to play the oboe.
Coming back to the flute, thus, if one fits a hard single reed to it,
it may operate as a 1/4 wave instrument, while if they fit a soft double
reed to it, it would operate as a 1/2 wave instrument.
The taper may have only a secondary role... perhaps in broad banding
the tonal range. I have indeed heard slurring clarinet passages (opening
notes of "Rhapsody in Blue") but none of such for oboe passages (I can't
recall any), which slurring definitely requires a broad band capability,
I think.
Am I way off base here?
Angelo Campanella |
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Gordon
science forum Guru Wannabe
Joined: 12 May 2005
Posts: 115
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| Posted: Wed Sep 21, 2005 3:39 pm Post subject:
Re: Oboes vs clarinets
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salmonegg@sbcglobal.net wrote:
| Quote: | This is a restatement of a previous post but I now have a better
understanding of how to ask the question. It is based upon a link
given in a reply by ddavis@pb.com.au;
http://www.phys.unsw.edu.au/~jw/pipes.html.
If I understand correctly, a clarinet using a uniform diameter bore
will have the same fundamental tone as an oboe using a conical bore
that is twice as long. That is, an oboe has the same fundamental tone
as that of a flute of the same length or a clarinet of half the
length.
Why is the fundamental resonance of an oboe that of a half-wave
resonator instead of a quarter-wave as one would expect from a
closed-open pipe?
Before looking at the the link, I thought that the fundamental tone
of the oboe would be the same as for a clarinet of the same length,
but that it would support both even and odd quarter-wave harmonics.
The only thing I can think of is that the open end of a conical pipe
does not reflect odd quarter-wave harmonics well enough so as to
cancel amplitude well at the reed end of an oboe. Why it can cancel
for a even quarter-wave (integral half-wave) is also a mystery.
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Here are some good books that relate music and physics and are light on
mathematics:
"The Science of Musical Sound" by John R. Pierce (Scientific American,
1983).
"The Science of Sound" by Thomas D. Rossing (2nd ed., Addison-Wesley, 1990).
"Science and Music" by James H. Jeans (Cambridge U. Press, 1937; Dover,
1968). |
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ttonon@ailr.com
science forum beginner
Joined: 23 Sep 2005
Posts: 3
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| Posted: Fri Sep 23, 2005 6:48 pm Post subject:
Re: Oboes vs clarinets
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| Quote: | The taper may have only a secondary role... perhaps in broad banding
the tonal range. I have indeed heard slurring clarinet passages (opening
notes of "Rhapsody in Blue") but none of such for oboe passages (I can't
recall any), which slurring definitely requires a broad band capability,
I think.
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I believe the taper is *THE* reason why the tapered, straight sided
tube can support even overtones, even though it's end conditions are
opened/closed. If I recall correctly, there are only two shapes that
allow such overtones: the straight tube and the tapered tube with
straight sides; tubes with tapered, curved sides cannot. This result
is mathematical/physical, with idealized formulations and linear
response. When it comes to the real world, the clarinet, for instance,
does contain even overtones, and I believe this is due to
nonlinearities in the reed response. But in short, the answer to the
original question is mathematical, and I regret that I do not have a
physical, intuitive explanation why this is so.
Best regards,
Tom |
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<salmonegg@sbcglobal.n
science forum beginner
Joined: 13 Sep 2005
Posts: 28
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| Posted: Fri Sep 23, 2005 11:00 pm Post subject:
Re: Oboes vs clarinets
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On 9/23/05 1:48 PM, in article
1127508494.203908.72370@g44g2000cwa.googlegroups.com, "ttonon@ailr.com"
<ttonon@ailr.com> wrote:
| Quote: | I believe the taper is *THE* reason why the tapered, straight sided
tube can support even overtones, even though it's end conditions are
opened/closed. If I recall correctly, there are only two shapes that
allow such overtones: the straight tube and the tapered tube with
straight sides; tubes with tapered, curved sides cannot. This result
is mathematical/physical, with idealized formulations and linear
response. When it comes to the real world, the clarinet, for instance,
does contain even overtones, and I believe this is due to
nonlinearities in the reed response. But in short, the answer to the
original question is mathematical, and I regret that I do not have a
physical, intuitive explanation why this is so.
|
I am in the process of solving the small signal linear partial differential
equation (PDE) appropriate for a tapered cone. It should be a text book
exercise although it was not when Lord Rayleigh may have done it. I will
report on that when finished.
I am doing so in spherical coordinates. I assume there is no circumferential
(phi) variation. I also am assuming that I have the lowest order polar
(theta) mode which corresponds to the first legendre function which is
merely a constant. Moreover, I will assume that time variation is in the
form exp (jwt), Radial variation, I expect, will be of the form
r^n*exp(jkr). Then, the closed-open boundary conditions must be matched.
These assumptions greatly simplify the mathematical task.
After all of this is said, it makes no sense to me that the fundamental mode
of the conical tube is going to be half-wave like a flute instead of
quarter-wave like a clarinet. Physically, I have difficulty understanding
the wave set up by the reed and its reflection from the open end do not
cause a quarter-wave standing wave.
Bill |
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<salmonegg@sbcglobal.n
science forum beginner
Joined: 13 Sep 2005
Posts: 28
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| Posted: Sat Sep 24, 2005 11:10 pm Post subject:
Re: Oboes vs clarinets
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On 9/24/05 5:38 PM, in article dh4riq$b13$1@unlnews.unl.edu, "Edgar A
Pearlstein" <rlst@unlserve.unl.edu epea> wrote:
| Quote: |
salmonegg@sbcglobal.net wrote:
: I am in the process of solving the small signal linear partial differential
: equation (PDE) appropriate for a tapered cone. It should be a text book
: exercise although it was not when Lord Rayleigh may have done it. I will
: report on that when finished.
:
: I am doing so in spherical coordinates. I assume there is no circumferential
: (phi) variation. I also am assuming that I have the lowest order polar
: (theta) mode which corresponds to the first legendre function which is
: merely a constant. Moreover, I will assume that time variation is in the
: form exp (jwt), Radial variation, I expect, will be of the form
: r^n*exp(jkr). Then, the closed-open boundary conditions must be matched.
: These assumptions greatly simplify the mathematical task.
That's the right approach, But start with open-end boundary
conditions, with the cone going from radius r1 to radius r2. The
solution is pretty simple. Then let r2 go to zero. Your "n" will be -1.
That is pretty much what I am trying to do. I think the more interesting |
thing would be to try open-open, closed-open, open-closed, and even
closed-closed. Then, even more interesting, would be to keep r2-r1 constant
and decrease the cone angle until the instrument turns into a clarinet.
If I can hack it, I will try to use Maple to verify my calculation. I am not
adept at Maple.
I may be reading more into the <rlst@unlserve.unl.edu epea> post that I
should. If I let r2 (at the reed) go to zero, with a real reed, then that
end becomes an open end because the bore of the reed is what sets the bore
of the combination. With that, I see no paradox. It is only when the narrow
end is truncated so that the reed aperture is small compared to bore that
that end approximates a closed end as in a clarinet.
Bill |
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Edgar A Pearlstein
science forum beginner
Joined: 11 May 2005
Posts: 3
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| Posted: Sun Sep 25, 2005 12:38 am Post subject:
Re: Oboes vs clarinets
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salmonegg@sbcglobal.net wrote:
: I am in the process of solving the small signal linear partial differential
: equation (PDE) appropriate for a tapered cone. It should be a text book
: exercise although it was not when Lord Rayleigh may have done it. I will
: report on that when finished.
:
: I am doing so in spherical coordinates. I assume there is no circumferential
: (phi) variation. I also am assuming that I have the lowest order polar
: (theta) mode which corresponds to the first legendre function which is
: merely a constant. Moreover, I will assume that time variation is in the
: form exp (jwt), Radial variation, I expect, will be of the form
: r^n*exp(jkr). Then, the closed-open boundary conditions must be matched.
: These assumptions greatly simplify the mathematical task.
That's the right approach, But start with open-end boundary
conditions, with the cone going from radius r1 to radius r2. The
solution is pretty simple. Then let r2 go to zero. Your "n" will be -1. |
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<salmonegg@sbcglobal.n
science forum beginner
Joined: 13 Sep 2005
Posts: 28
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| Posted: Tue Sep 27, 2005 5:50 pm Post subject:
Re: Oboes vs clarinets
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On 9/24/05 5:38 PM, in article dh4riq$b13$1@unlnews.unl.edu, "Edgar A
Pearlstein" <epearlst@unlserve.unl.edu> wrote:
| Quote: |
salmonegg@sbcglobal.net wrote:
: I am in the process of solving the small signal linear partial differential
: equation (PDE) appropriate for a tapered cone. It should be a text book
: exercise although it was not when Lord Rayleigh may have done it. I will
: report on that when finished.
:
: I am doing so in spherical coordinates. I assume there is no circumferential
: (phi) variation. I also am assuming that I have the lowest order polar
: (theta) mode which corresponds to the first legendre function which is
: merely a constant. Moreover, I will assume that time variation is in the
: form exp (jwt), Radial variation, I expect, will be of the form
: r^n*exp(jkr). Then, the closed-open boundary conditions must be matched.
: These assumptions greatly simplify the mathematical task.
That's the right approach, But start with open-end boundary
conditions, with the cone going from radius r1 to radius r2. The
solution is pretty simple. Then let r2 go to zero. Your "n" will be -1.
|
Previous correspondence:
*******************************
Subject: Re: Why do conical resonators support even harmonics?
Date: Wednesday, August 17, 2005 6:33 AM
From: Edgar A Pearlstein <epearlst@unlserve.unl.edu>
Newsgroups: alt.sci.physics.acoustics
I don't know of any way to show this other than by solving the
partial diffential equation, which I can do. For many years I taught a
course in the physics of music, and yearned to find a "hand-waving" way to
show the resonances on a conical pipe, but never found one.
Consider the acoustic pressure in an open-open pipe narrower at one
end than the other. The pressure will be higher near the narrow end. (Of
course it is close to zero right at an open end.) And the resonances
will be as for an open-open pipe. Now imagine squeezing the
narrow end down to where it is actually closed; then the pressure will be
quite high at the point. But this is not the same as having a closed-open
pipe, and the resonances will still be as for an open-open pipe.
I don't regard the above as a good argument, though. For if one
started with a closed-open pipe and narrowed one end, one would come to a
different conclusion about the resonances!
salmonegg@sbcglobal.net wrote:
: Aside from the clarinet and oboe having single and double reed
respectively,
: the main difference in construction is that the clarinet has a uniform
bore
: while the oboe has a conical bore. Supposedly, the clarinet will not
: generate strong even harmonics. The clarinet forms an closed boundary
: condition at the mouthpiece and an open boundary condition at the open
end.
: On the other hand, the oboe with similar boundary conditions, along with
the
: saxophone and other conical bore instruments, will readily support even
: harmonics.
: Is there a simple way, short of solving partial differential equations, to
: understand why this is so? Is there some way to determine how easy it is
to
: excite such harmonics? Would that involve acoustic impedance seen at the
: mouthpiece? How does the cone angle affect excitation of the various
: harmonics?
****************************************
I have just completed my zero order approximation describing what makes an
oboe resonate at the same frequencies as that of flute of the same length.
After working through the PDE (partial differential equation), it became
clear why one could explain, in hindsight, what happens WITHOUT actually
solving the PDE.
In essence, I followed the method outlined at the top of this post. The PDE
solution was in the form [exp(j*w*t)]*[A*cos(k*r) + B*sin(k*r)]/r. Where A
and B are arbitrary coefficients selected to meet boundary conditions.
Actually only A or B needs to be specified because they will form the same
ratio independent of amplitude. The wave number k = w/c. The speed of sound
is c.
Consider a solution for local pressure. At the cone apex, A = 0. Otherwise,
the pressure would be infinite. Thus the spatial part of the solution
describes a standing wave in which sin(k*r)/r must be finite or zero at both
ends. This is the same situation as for a flute. Any odd quarter-wave
resonance cannot occur because it leads to infinite pressure. An oboist
would not be able to play pianissimo against such a pressure.
A heuristic approach would note that there are spherical waves in the
conical bore emanating from the apex. To conserve power as the wave expands
from the apex, the amplitude of the wave varies as 1/r. Once that 1/r
behavior kicks in, cos(k*r) modes are suppressed.
In summary, the conical bore does not add even harmonics to a quarter-wave
resonator. It eliminates odd quarter-wave resonances.
Real oboes do not have a conical apex. My next step will be see what happens
as the cone gets turned into a frustum. My guess is that the A/B will grow
from zero to some larger number as more of the apex is trimmed off. I will
see how the math works out, if at all. A difficulty I foresee the pressure
at the mouthpiece end may be high, but how do I go about treating high
differently than infinite?
Bill |
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Angelo Campanella
science forum Guru Wannabe
Joined: 08 May 2005
Posts: 226
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| Posted: Wed Sep 28, 2005 1:55 am Post subject:
Re: Oboes vs clarinets
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salmonegg@sbcglobal.net wrote:
| Quote: | A heuristic approach would note that there are spherical waves in the
conical bore emanating from the apex. To conserve power as the wave expands
from the apex, the amplitude of the wave varies as 1/r. Once that 1/r
behavior kicks in, cos(k*r) modes are suppressed.
In summary, the conical bore does not add even harmonics to a quarter-wave
resonator. It eliminates odd quarter-wave resonances.
|
Excellent insight, IMHO!
| Quote: | Real oboes do not have a conical apex.
|
As I recall, they have a small (uniform?) bore tube that communicates
with the oboe body. In that case, an "open-end" environment might exist.
| Quote: | My next step will be see what happens
as the cone gets turned into a frustum. My guess is that the A/B will grow
from zero to some larger number as more of the apex is trimmed off. I will
see how the math works out, if at all. A difficulty I foresee the pressure
at the mouthpiece end may be high, but how do I go about treating high
differently than infinite?
|
Since the bore is small, the energy or mouthpiece force is not all that
large. As I recall, oboe players have a peculiar 'bite' to maintain the
tube excitation. They are not at all relaxed as one sees them perform.
It takes concentration and training to manage that little small area,
high pressure acoustical wave generator.
Angelo Campanella. |
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<salmonegg@sbcglobal.n
science forum beginner
Joined: 13 Sep 2005
Posts: 28
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| Posted: Sat Oct 08, 2005 11:07 pm Post subject:
Re: Oboes vs clarinets
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On 9/27/05 12:50 PM, in article BF5EF0A0.830B%salmonegg@sbcglobal.net,
"salmonegg@sbcglobal.net" <salmonegg@sbcglobal.net> wrote:
| Quote: | I have just completed my zero order approximation describing what makes an
oboe resonate at the same frequencies as that of flute of the same length.
After working through the PDE (partial differential equation), it became
clear why one could explain, in hindsight, what happens WITHOUT actually
solving the PDE.
In essence, I followed the method outlined at the top of this post. The PDE
solution was in the form [exp(j*w*t)]*[A*cos(k*r) + B*sin(k*r)]/r. Where A
and B are arbitrary coefficients selected to meet boundary conditions.
Actually only A or B needs to be specified because they will form the same
ratio independent of amplitude. The wave number k = w/c. The speed of sound
is c.
Consider a solution for local pressure. At the cone apex, A = 0. Otherwise,
the pressure would be infinite. Thus the spatial part of the solution
describes a standing wave in which sin(k*r)/r must be finite or zero at both
ends. This is the same situation as for a flute. Any odd quarter-wave
resonance cannot occur because it leads to infinite pressure. An oboist
would not be able to play pianissimo against such a pressure.
A heuristic approach would note that there are spherical waves in the
conical bore emanating from the apex. To conserve power as the wave expands
from the apex, the amplitude of the wave varies as 1/r. Once that 1/r
behavior kicks in, cos(k*r) modes are suppressed.
In summary, the conical bore does not add even harmonics to a quarter-wave
resonator. It eliminates odd quarter-wave resonances.
Real oboes do not have a conical apex. My next step will be see what happens
as the cone gets turned into a frustum. My guess is that the A/B will grow
from zero to some larger number as more of the apex is trimmed off. I will
see how the math works out, if at all. A difficulty I foresee the pressure
at the mouthpiece end may be high, but how do I go about treating high
differently than infinite?
|
I am just finishing up looking at the subject. It turns out that equations
for resonant frequencies are given in my old edition of the AIP Handbook,
But having gone through the exercise, I now think that I understand the
phenomena much better.
It turns out that the PDE handles very nicely and compares closely the the
heuristic explanation given above. The characteristic equation gives the
correct frequencies for the oboe and clarinet approximations.
The interesting thing, for me, that comes out is that the harmonics for a
truncated cone are not multiples of the fundamental. I note that a saxophone
appears to have more of the apex snipped off than an oboe does so that it
can accommodate the larger clarinet like mouthpiece. This means that the
higher harmonic resonant frequencies will deviate more from integer ratios
than will those of an oboe.
While steady state forced vibration will synch to the reed rather than to
the mode, transient excitation of the mode especially during attack will
provide some dissonance. I have always thought that saxophones sounded a bit
raspier, especially in jazz, than does a clarinet. I would appreciate any
information a clarinet/saxophone/oboe player could pass along on the topic.
Bill |
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