Equation (9.38) can be solved for
to obtain

where

(10.44) |

We interpret as a

Since the mouthpiece of a clarinet is nearly closed,
which
implies
and
. In the limit as
goes to
infinity relative to
, (9.42) reduces to the simple form of a
rigidly capped acoustic tube, *i.e.*,
.
If it were possible to open the reed wide enough to achieve
matched impedance,
, then we would have
and
, in
which case
, with no reflection of
, as expected. If
the mouthpiece is removed altogether to give
(regarding it now as a
tube section of infinite radius), then
,
, and
.

[How to cite this work] [Order a printed hardcopy] [Comment on this page via email]

Copyright ©

Center for Computer Research in Music and Acoustics (CCRMA), Stanford University