1. SPH3U: Waves & Sound
Musical Instruments 2

2. Wind Instruments
flute
trumpet
trombone
clarinet
saxophone

3. Wind Instruments
Wind Instruments are based on the physics within columns of air. The sound depends on the vibration of the air molecules. Air molecules are forced to vibrate by causing something near them to vibrate -- the initial vibration is created by a reed or by the player’s lips.

4. Wind Instruments
The frequencies produced (pitches) depend on whether the air columns are open or closed and their length.
The lengths of the air columns might be fixed like in a pipe organ, or they might be adjustable like in a trombone or a trumpet.

5. Observation: Boomwhackers
Which Boomwhackers produce high notes, and which ones produce low notes? 
What does adding the cap do?

6. Open-Ended Air Columns
Many musical instruments consist of an air column enclosed inside of a hallow metal tube. If both the ends of the tube are uncovered, it is an example of an ‘open ended’ air column instrument. (Examples include all brass instruments, saxophones, flutes, oboes).

7. Open-Ended Air Columns
As we have already mentioned, a musical instrument has a set of natural frequencies at which it will vibrate. These natural frequencies are known as harmonics, and they are each associated with a standing wave pattern.
In an air column, nodes in the standing wave always form at closed ends, and antinodes always form at open ends.

8. Open-Ended Air Columns
In an open-ended air column then, we will always have an antinode at either end of the column. Thus, the length of the air column is equal to one-half of the wavelength when the air column vibrates at the first harmonic.

9. Open-Ended Air Columns
In an open-ended air column then, we will always have an antinode at either end of the column. Thus, the length of the air column is equal to one-half of the wavelength when the air column vibrates at the first harmonic.

10. Open-Ended Air Columns
If the air column is vibrating at the second harmonic, the column looks like this, and we know that the length of the column is equal to 1 full wavelength.

11. Open-Ended Air Columns
If the air column is vibrating at the third harmonic, the column looks like this, and we know the length of the column is equal to 1.5 full wavelengths.

12. In general we see the following is true for Open Air Columns:
Harmonic
# Waves in Column
# Nodes
# Antinodes
Length / Wavelength relationship
Picture 1
1/2 2
2/2 3
3/2 4
4/2 5

13. Open-Ended Air Columns
Thus, an air column of a given length could cause vibrations that have different wavelengths. Because the speed of sound in air is constant, this means that a change in wavelength will cause a change in frequency. In other words, we could get different musical notes from the same air column.

14. Open-Ended Air Columns
 This is what we see in a trumpet. For the exact same air column, I can get a C, or a G, or a high C. The same air column is producing different standing wave patterns that cause each different note.

15. Open-Ended Air Columns
This is not what we see with the Boomwhackers though. They are all vibrating at their fundamental frequency (1st Harmonic). To calculate the frequency a Boomwhacker will produce let’s consider the following.

16. Open-Ended Air Columns
The Red Boomwhacker has a length of 63cm.
Therefore the wavelength produced is 126cm = 1.26m
If the speed of sound in air is 331m/s
Then the frequency produced is 332/1.26 = Hz. (Note that C is approximately 262 Hz … so this is pretty close!)

17. Example Question 1
The speed of sound waves in air is found to be 340m/s. Determine the fundamental frequency (1st Harmonic) of an open-ended air column which has a length of 67.5cm.

18. Example Question 1 - Solution
The speed of sound waves in air is found to be 340m/s. Determine the fundamental frequency (1st Harmonic) of an open-ended air column which has a length of 67.5cm.
Given/Required Analysis
v = 340m/s * use λ= 2L and v = fλ
L = m thus: v = f (2L)
f = ?
Solution Statement: The frequency is approximately
v = f (2L) Hz
340 = f (2)(0.675)
f = 253 Hz

19. Example Question 2
Determine the length of an open-ended air column required to produce a frequency (1st Harmonic) of 480Hz. The speed of sound waves in air is known to be 340m/s.

20. Example Question 2 - Solution
Determine the length of an open-ended air column required to produce a frequency (1st Harmonic) of 480Hz. The speed of sound waves in air is known to be 340m/s.
Given/Required Analysis
v = 340m/s * use v = fλ and λ = (2)(L)
f1 = 480Hz thus: v = f (2L)
L = ?
Solution Statement: The length of the air column is
v = f (2L) m
340 = 480Hz (2)(L)
L = m

21. In Closed Air Columns
A Closed Air Column is what happens when one of the ends of the tube is covered. For example, this is what we observed when we put a cap on one end of the Boomwhackers.

22. In Closed Air Columns
Sound works the same way in closed air columns as it does in open air columns except for a few important differences. A node is always formed at the closed end of the column. Therefore the standing wave pattern generated will always have a node at one end, and an antinode at the other end.

23. In Closed Air Columns
Thus the wave pattern for the fundamental frequency (1st Harmonic) possible in a closed air column looks like the diagram below. The length of the column will be ¼ the wavelength.

24. The following relationships are also true:
Harmonic
# Waves in Column
# Nodes
# Antinodes
Length / Wavelength relationship
Picture 1
1/4 3
3/4 2 5
5/4 7
7/4 4

25. In Closed Air Columns
** Note: There are no even-numbered harmonics. This is because the second possible frequency produced is three times the fundamental frequency, not twice the fundamental frequency. You can’t get even-numbered harmonics in a closed air column.

26. In Closed Air Columns
 Let’s use what we’ve learned to calculate the frequency the red Boomwhacker if we put the cap on it.

27. In Closed Air Columns
Red Boomwhacker has a length of 63cm.Therefore the wavelength produced is 252cm = 2.52m
If the speed of sound in air is 331m/s then the frequency produced is 331/2.52 = Hz. (This is exactly half of the other value of C we calculated when the end cap was off – which is why it is an octave lower if you put the cap on the Boomwhacker!)