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Re: [Phys-L] [SPAM] Re: sound

Thanks! This is very good and detailed. Never see this in typical physics

Not sure how can explain this in "simple" terms to my high school
students. Most of the book problems deal with the frequency equations for
strings and open/closed pipes. Given one variable (L), find the other (f),
etc. Gets more complicated when dealing with an actual instrument and what
changes to produce different notes! writes:
On 04/12/2013 08:14 AM, Anthony Lapinski wrote:
Thanks for these responses. So what, then, changes in the equation fn =

Let's talk about that.

Lips vibrate faster,


so v changes? But I thought v = 343 m/s (speed
of sound in air).

v does not change.

Or does n change, which creates the higher
frequencies/harmonics? What changes in that equation to predict which
frequencies are played on a bugle or other instrument where L is

First, let's talk about the principles, which do not change.

We can illustrate the principles with an analogous system that is
easier to visualize. Get an aluminum or steel rod or bar that is
at least 2 meters long, and thin enough compared to its length that
it is readily flexible. Grab it by the middle and hold it vertical.
Shake it sideways, so that it flexes. There will be resonances.
These are standing waves of the transverse sound waves.
-- The lowest vibrating mode will be such that a single /half/
wavelength will fit into the length.
-- There is of course an even lower mode, i.e. zero frequency,
i.e. uniform translation of the whole thing.
-- If you shake it faster, you can excite higher modes. Note
that the even-numbered modes have a node in the middle, so if
you want to excite these, you need to hold the rod somewhere
off-center. (Or if you are clever and strong, you can hold it
in the middle and drive it with a torsion maneuver.)
-- It is a linear system, so if you drive it at a single frequency,
it *always* responds at that frequency. This is required by Floquet's
theorem. If you shake it at a non-resonant frequency, it will respond
at that frequency ... but the amplitude of the response will be small.
-- If you drive it with something that contains a superposition of
frequencies, you will get a superposition of responses. In frequency
space, the response is the product of the drive times the response
-- To say the same thing in real-time space (the inverse Fourier
transform of frequency space), the response is the convolution of
the drive waveform with the impulse response kernel. Whack the
rod with a mallet to observe the impulse response.

The bugle is only slightly different. The biggest difference is that
there is nonlinear feedback from the filter (the horn itself) and the
drive (the lips) ... and this cannot be ignored. There may have been
some feedback involved in shaking the bar, but we ignored it.

The drive from the lips can be modeled as a succession of impulses.
A Dirac comb. Interestingly enough, the Fourier transform of a
Dirac comb is a Dirac comb in frequency space. So we are driving
the horn with a superposition of frequencies, all of which are
strict multiples of the nominal lip vibration frequency. We are
then filtering this drive with the response function of the horn.
It contains resonance which are /approximately/ multiples of the
fundamental ... not multiples of the drive, but of the fundamental
set by the size and shape of the horn.

By tightening the lips, you can change the frequency of the drive.
Due to nonlinearity, it will "try" to mode-lock to one of the modes
of the horn, but this is not a law of nature, just a rule of thumb.
You can definitely shift the drive frequency a little bit up or down
from the nominal center of the mode. Remember, the horn *will*
respond at the frequency of the drive (although if the drive is
off-resonance, the response will be small).


To summarize: There are two separate bits of physics here.
-- the drive (lips)
-- the filter (horn)
These interact, but it helps to consider them as mostly independent, and
then treat the interaction as a smallish perturbation.

A bugler (unlike a trumpeter) cannot change the horn. All horn players
change the frequency of the drive. The result is the convolution of the
drive with the filter.


Very minor pedagogical point: In the business, people speak of "open"
and "closed" wind instruments.
-- Open means open at both ends.
-- Closed means open at one end and closed at the other.

I mention this because these definitions are wildly inconsistent with
sense definitions. It is OK to use the jargon terms, but you will have
tell students what you are doing, and remind them about 11 times. Another
possibility is to use different terminology: In mathematics there is a
standard term /clopen/ that refers to something that is partly closed and
partly open.


A treeeemendous resource for all things related to the physics of music
is the UNSW site.

I give this site very high marks for clarity, correctness, depth, and

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