Physics 371 April 9, 2002
• Brass Instruments
lip-driven oscillations (feedback)
adjustment of natural modes
mouthpiece and bell
playing a chromatic scale:
slides and valves
pressure distribution
for different modes
adjusting the natural modes of a brass instrument
f
(a)
(b)
(c)
(a) natural mode frequencies
of a cylindrical tube closed
at one end;
(b) upper shift of frequencies
of the lower modes caused
by replacing part of the tube
by a bell;
(c) downward shift of the
higher modes caused by
adding a mouth piece.
it’s a misconception to think of it as an open pipe!
for a periodic excitation, the partials are by
necessity exact multiples of the fundamental
x
x
x
but misconception: brass intruments play in JUST tuning
table on p. 267of Backus shows that e.g. 8th resonance is
as much as 54 cent off
Tuba players have
a sense of humor about
their instrument......
note: Tuba has a
long conical section
a home-built tuba
Tuba
reminder: conical horn
has same modes as OPEN
pipe
f0, 2f0, 3f0,4f0,
for trumpet, pedestal note is
out of tune, but tuba has a larger
conical section than a trumpet
or French horn so that pedestal
note (lowest mode) is more
nearly in tune
note:
if the conical section is
nearly 50% of the
horn length, the
pedestal tone is in tune
(Tuba)
Playing a chromatic scale (sequence of half-steps):
one method: the slide (trombone)
to bridge the gap from mode 2 to mode 3 (a fifth)
requires 7 slide positions Bb-A-Ab-G-Gb-F-E
how long should I make the trombone if I can extend arm by 0.5 m?
Answ: no longer than 3 m when slide is extended -> 2m for pos 1
another method: valves to add length of tubing
entire scale with only three valves - how?
French Horn
Trumpet
example of a piston valve:
valve depressed adds more length (lowers pitch)
need to bridge the gaps in the natural scale:
if first mode is not used - biggest gap is FIFTH
(1)
2
3
4
5
6
7
8
9
fifth fourth major minor
third third
# semitones: 7
5
4
3
lower pitch 1 semitone by adding length l1 to original length lo
2
l2
3
l3
4
l1 +l3
5
l2 +l3
6
l 1 + l 2 +l 3
10
problem: the lengths do not add up properly!
example: take trumpet of length 137 cm
lower pitch by
new length (cm)
1
137x1.059... = 145.1
2
137x(1.059..)2 = 153.8
3
137x(1.059..)3 = 162.9
4
137x(1.059..)4 = 172.6
5
6
7
added # cm
needed:
8.1
16.8
25.9
35.7 but 8.1+25.9 = 34.0
off 1.7cm/172cm = 1%
137x(1.059..)5 = 182.9
137x(1.059..)6 =
solution: thumb slide or add more valves
Schematic airflow through a French horn: F horn on top, Bb horn
on the bottom. Each numbered valve pair is operated by a single lever