1. 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

2. pressure distribution for different modes

3. (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. adjusting the natural modes of a brass instrument it’s a misconception to think of it as an open pipe! (a) (b)(c) f

4. x x x for a periodic excitation, the partials are by necessity exact multiples of the fundamental 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

5. Tuba players have a sense of humor about their instrument...... note: Tuba has a long conical section

6. reminder: conical horn has same modes as OPEN pipe f 0, 2f 0, 3f 0,4f 0, 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 Tuba a home-built tuba

7. note: if the conical section is nearly 50% of the horn length, the pedestal tone is in tune (Tuba)

8. to bridge the gap from mode 2 to mode 3 (a fifth) requires 7 slide positions B b -A-A b -G-G b -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 one method: the slide (trombone) Playing a chromatic scale (sequence of half-steps):

9. entire scale with only three valves - how? French Horn Trumpet another method: valves to add length of tubing

10. example of a piston valve: valve depressed adds more length (lowers pitch)

11. need to bridge the gaps in the natural scale: (1) 2 3 4 5 6 7 8 9 10 if first mode is not used - biggest gap is FIFTH fifth fourth major minor third third # semitones: 7 5 4 3 lower pitch 1 semitone by adding length l 1 to original length l o 2 l 2 3 l 3 4 l 1 +l 3 5 l 2 +l 3 6 l 1 + l 2 + l 3

12. problem: the lengths do not add up properly! example: take trumpet of length 137 cm lower pitch byadded # cm new length (cm)needed: 1137x1.059... = 145.1 8.1 2137x(1.059..) 2 = 153.8 16.8 3137x(1.059..) 3 = 162.9 25.9 4137x(1.059..) 4 = 172.6 35.7 but 8.1+25.9 = 34.0 off 1.7cm/172cm = 1% 5 137x(1.059..) 5 = 182.9 6 137x(1.059..) 6 = 7 solution: thumb slide or add more valves

13. Schematic airflow through a French horn: F horn on top, B b horn on the bottom. Each numbered valve pair is operated by a single lever