1. String and Air Instruments
Review – Standing waves in String Instruments
Examples – String Instruments
Longitudinal Waves in Air
Standing Waves in Air Instruments (open-open)
Standing Waves in Air Instruments (open-closed)
Summary Air Instruments (open-open, open-closed)
Examples – String and Air Instruments

2. Review -Standing waves on a String
String anchored between 2 points and velocity fixed
Allowed opening widths
𝐿= 𝜆 2
𝐿=𝜆
𝐿= 3𝜆 2
In general
𝐿= 𝑛𝜆 𝑛=1,2,3…
Allowed wavelengths
𝜆 𝑛 = 2𝐿 𝑛 𝑛=1,2,3….
Allowed frequencies Velocity is
𝑓 𝑛 = 𝑣 𝜆 𝑛 = 𝑛𝑣 2𝐿 𝑛=1,2,3… 𝑣= 𝑇 𝜇

3. String - Example 12-7 𝑓 𝑛 = 𝑛𝑣 2𝐿 𝑓 𝑛 ′= 𝑛𝑣 2 𝐿 ′
Frequency of highest note
𝑓 𝑛 = 𝑛𝑣 2𝐿
Frequency of lowest note
𝑓 𝑛 ′= 𝑛𝑣 2 𝐿 ′
Ratio
𝑓 𝑛 𝑓 𝑛 ′ = 𝑛𝑣 2𝐿 𝑛𝑣 2 𝐿 ′ = 𝐿′ 𝐿
𝐿′ 𝐿 = 𝑓 𝑛 𝑓 𝑛 ′

4. String - Example 12-8 Same as string v = 440 Hz.
Allowed wavelengths in string
𝜆 𝑛 = 2𝐿 𝑛 𝑛=1,2,3….
𝜆 1 =0.64 𝑚
Velocity in string
𝑣 𝑠𝑡𝑟𝑖𝑛𝑔 = 440 𝑠∗0.64 𝑚= 𝑚 𝑠
Frequency in Air
Same as string v = 440 Hz.
Velocity in Air
vair = 343 m/s
Wavelength in Air
𝜆= 343 𝑚 𝑠 𝑠 =0.78 𝑚

5. Longitudinal Waves in Air
Traveling sound wave
Pressure and Displacement Nodes/Antinodes

6. Standing waves in Air – open/open end (1)
Display as transverse wave (easier to see)
Result:
Pressure wave node at both ends
Pipe length must be some multiple of ½ wavelength! (open/open)

7. Standing waves in Air – open/open end (2)
Animation – Pressure wave node at both ends
Result:
Pressure wave node at both ends
Pipe length is some multiple of ½ wavelength!

8. Standing waves in Air – open/open end (3)
Allowed widths
𝐿= 𝜆 2
𝐿=𝜆
𝐿= 3𝜆 2
In general
𝐿= 𝑛𝜆 𝑛=1,2,3…
Allowed wavelengths
𝜆 𝑛 = 2𝐿 𝑛 𝑛=1,2,3….
Allowed frequencies Velocity is
𝑓 𝑛 = 𝑣 𝜆 𝑛 = 𝑛𝑣 2𝐿 𝑛=1,2,3… 𝑣=343 𝑚/𝑠

9. Standing waves in Air – open/closed end (1)
Display as transverse wave (easier to see)
Result
Pressure wave node at one end, antinode at other
Pipe length is some odd multiple of ¼ wavelength

10. Standing waves in Air – open/closed end (2)
Animation – Pressure wave node at end, antinode at other
Result
Pressure wave node at end, antinode at other
Pipe length is some odd multiple of ¼ wavelength

11. Standing waves in Air – open/closed end (3)
Allowed widths
𝐿= 𝜆 4
𝐿= 3𝜆 4
𝐿= 5𝜆 4
In general
𝐿= 𝑛𝜆 𝑛=1,3,5…
Allowed wavelengths
𝜆 𝑛 = 4𝐿 𝑛 𝑛=1,3,5….
Allowed frequencies Velocity is
𝑓 𝑛 = 𝑣 𝜆 𝑛 = 𝑛𝑣 4𝐿 𝑛=1,3,5… 𝑣=343 𝑚/𝑠

12. Comparison of waves on string and air
Both have
Wavelength – distance between peaks at fixed time
Frequency – rate of repetitions at fixed position (like your ear)
Wave velocity 𝑣=𝑓𝜆
Differences
String wave velocity varies with tension and mass/length 𝑣= 𝑇 𝜇
String has ½- wavelength harmonics 𝑓 𝑛 = 𝑣 𝜆 𝑛 = 𝑛𝑣 2𝐿 𝑛=1,2,3….
Air wave velocity set at 343 m/s (at 20° C)
Air has ½- or ¼- wavelength harmonics 𝑓 𝑛 = 𝑣 𝜆 𝑛 = 𝑛𝑣 2𝐿 𝑛=1,2,3….
𝑓 𝑛 = 𝑣 𝜆 𝑛 = 𝑛𝑣 4𝐿 𝑛=1,3,5….

13. Examples of String and Air Instruments
String Instruments
Guitar
Violin
Piano
Air Instruments
Flute
“Trombone”
Soda bottle

14. Examples Examples Problem 25 – Open & closed, 1st 3 harmonics
Problem 26 – Coke bottle
Problem 27 – Range of human hearing, pipe lengths
Problem 28 – Guitar sounds with fret
Problem 29 – Guitar sounds with fret
Problem 30 – Length of organ pipe
Problem 32 – Flute
Problem 34 – Pipe multiple harmonics

15. Problem 25 – Organ Pipe Open at both ends Closed at one end
𝑓 𝑛 = 𝑣 𝜆 𝑛 = 𝑛𝑣 2𝐿 = 𝑛 343 𝑚 𝑠 2 ∙ 1.12 𝑚 𝑛=1,2,3….
𝑓 𝑛 =153, 306, 459, 612 𝐻𝑧
Closed at one end
𝑓 𝑛 = 𝑣 𝜆 𝑛 = 𝑛𝑣 4𝐿 = 𝑛 343 𝑚 𝑠 4 ∙ 1.12 𝑚 𝑛=1,3,5….
𝑓 𝑛 =77, 230, 383, 536 𝐻𝑧 <<skip even harmonics

16. Problem 26 – Coke bottle Open/closed fundamental Closed 1/3 way up
𝑓 1 = 𝑣 𝜆 1 = 𝑛𝑣 4𝐿 = 𝑚 𝑠 4 ∙ 0.18 𝑚 =476 𝐻𝑧
Closed 1/3 way up
𝑓 1 = 𝑣 𝜆 1 = 𝑛𝑣 4𝐿 = 𝑚 𝑠 4 ∙ 0.12 𝑚 =715 𝐻𝑧

17. Problem 27 – Full-range Pipe Organ
Open/open fundamental
𝑓 1 = 𝑣 𝜆 1 = 1𝑣 2𝐿
Lowest frequency
𝐿= 𝑣 2 𝑓 1 = 343 𝑚 𝑠 2∙ 20 𝑠 =8.6 𝑚 (‼)
Highest frequency
𝐿= 𝑣 2 𝑓 1 = 343 𝑚 𝑠 2∙ 20,000 𝑠 =8.6 𝑚𝑚

18. Pipe Organ – variation in tube length
St Steven’s Cathedral, Vienna

19. Problem 28 – Guitar 𝑓 3 = 𝑣 𝜆 1 = 3𝑣 2𝐿 𝑓 1 ′= 𝑣 𝜆 1 = 1𝑣 2∙0.6∙𝐿
Original frequency of 3rd harmonic (on string)
𝑓 3 = 𝑣 𝜆 1 = 3𝑣 2𝐿
Fingered frequency of fundamental
𝑓 1 ′= 𝑣 𝜆 1 = 1𝑣 2∙0.6∙𝐿
Ratio
𝑓 1 ′ 𝑓 3 = 1𝑣 2 ∙ 0.6∙𝐿 3𝑣 2 𝐿 = 1 3 ∙ 0.6 = 𝑓 1 ′ =0.555 ∙540=300 𝐻𝑧

20. Problem 29 – Guitar (1) 𝑓 1 = 𝑣 𝜆 1 = 1𝑣 2𝐿 𝑓 1 ′= 𝑣 𝜆 1 = 1𝑣 2 𝐿 ′
Unfingered frequency of fundamental (on string)
𝑓 1 = 𝑣 𝜆 1 = 1𝑣 2𝐿
Fingered frequency of fundamental
𝑓 1 ′= 𝑣 𝜆 1 = 1𝑣 2 𝐿 ′
Ratio
𝑓 1 𝑓 1 ′ = 1𝑣 2𝐿 1𝑣 2𝐿′ = 𝐿′ 𝐿 𝐿′ 𝐿 = 𝑓 1 𝑓 1 ′ =
𝐿 ′ =0.75 ∙0.73𝑚= 𝑚

21. Problem 29 – Guitar (2) 𝜆 𝑠𝑡𝑟𝑖𝑛𝑔 =2∙0.5475=1.095 𝑚 440 Hz
wavelength of 440 Hz fundamental in string
𝜆 𝑠𝑡𝑟𝑖𝑛𝑔 =2∙0.5475=1.095 𝑚
frequency in air same as string
440 Hz
Wavelength in air
𝜆 𝑎𝑖𝑟 = 343 𝑚 𝑠 𝑠 =0.78 𝑚

22. Problem 30 – Organ Pipe 𝐿= 1𝑣 2 𝑓 1 = 343.6 𝑚/𝑠 2 ∙ 262 /𝑠 =0.656 𝑚
Corrected velocity to 21°C
𝑣≈ 𝑇 =343.6 m/s
Allowed frequencies
𝑓 𝑛 = 𝑣 𝜆 𝑛 = 𝑛𝑣 2𝐿 𝑛=1,2,3….
Length is
𝐿= 1𝑣 2 𝑓 1 = 𝑚/𝑠 2 ∙ 262 /𝑠 =0.656 𝑚
Wavelength
𝜆=2𝐿=1.31 𝑚 same inside and outside tube

23. Problem 32 - Flute 𝐿= 1𝑣 2 𝑓 1 = 343 𝑚/𝑠 2 ∙ 294 /𝑠 =0.583 𝑚
Flute open at both ends (open-open)
Allowed frequencies
𝑓 𝑛 = 𝑣 𝜆 𝑛 = 𝑛𝑣 2𝐿 𝑛=1,2,3….
Length is
𝐿= 1𝑣 2 𝑓 1 = 343 𝑚/𝑠 2 ∙ 294 /𝑠 =0.583 𝑚

24. Problem 34 – Assume open-open <?>
Write n and (n+1) harmonics in terms of fundamental
𝑓 𝑛 = 𝑛𝑣 2𝐿 =𝑛 𝑓 𝑓 𝑛+1 = (𝑛+1)𝑣 2𝐿 =(𝑛+1) 𝑓 1
Subtract
𝑓 𝑛+1 − 𝑓 𝑛 = 𝑓 1
So the difference of any 2 harmonics should be the fundamental.
𝑓 1 =440−264=616−440=176 𝐻𝑧
𝑓 𝑛 =176, 352, 528, 𝑛=1,2,3,4…. ????!!

25. Problem 34 – Assume open-closed <?>
Write n and (n+2) odd harmonics in terms of fundamental
𝑓 𝑛 = 𝑛𝑣 4𝐿 =𝑛 𝑓 𝑓 𝑛+2 = (𝑛+2)𝑣 4𝐿 =(𝑛+2) 𝑓 1
Subtract
𝑓 𝑛+2 − 𝑓 𝑛 = 2𝑓 1
So the difference of any 2 harmonics should be twice fundamental.
𝑓 1 =440−264=616−440=2 ∙88 𝐻𝑧
𝑓 𝑛 =88, 264, 440, 𝑛=1,3,5,7…. success!!