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

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 2 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

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 ′

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=281.6 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 440 𝑠 =0.78 𝑚

Longitudinal Waves in Air Traveling sound wave https://sites.google.com/site/physicsflash/home/sound Pressure and Displacement Nodes/Antinodes

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)

Standing waves in Air – open/open end (2) Animation – Pressure wave node at both ends http://faraday.physics.utoronto.ca/IYearLab/Intros/StandingWaves/Flash/sta2fix.html Result: Pressure wave node at both ends Pipe length is some multiple of ½ wavelength!

Standing waves in Air – open/open end (3) Allowed widths 𝐿= 𝜆 2 𝐿=𝜆 𝐿= 3𝜆 2 In general 𝐿= 𝑛𝜆 2 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

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

Standing waves in Air – open/closed end (2) Animation – Pressure wave node at end, antinode at other http://faraday.physics.utoronto.ca/IYearLab/Intros/StandingWaves/Flash/sta1fix.html Result Pressure wave node at end, antinode at other Pipe length is some odd multiple of ¼ wavelength

Standing waves in Air – open/closed end (3) Allowed widths 𝐿= 𝜆 4 𝐿= 3𝜆 4 𝐿= 5𝜆 4 In general 𝐿= 𝑛𝜆 4 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

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….

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

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

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

Problem 26 – Coke bottle Open/closed fundamental Closed 1/3 way up 𝑓 1 = 𝑣 𝜆 1 = 𝑛𝑣 4𝐿 = 1 343 𝑚 𝑠 4 ∙ 0.18 𝑚 =476 𝐻𝑧 Closed 1/3 way up 𝑓 1 = 𝑣 𝜆 1 = 𝑛𝑣 4𝐿 = 1 343 𝑚 𝑠 4 ∙ 0.12 𝑚 =715 𝐻𝑧

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

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

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 =0.555 𝑓 1 ′ =0.555 ∙540=300 𝐻𝑧

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 ′ = 330 440 𝐿 ′ =0.75 ∙0.73𝑚=0.5475 𝑚

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 440 𝑠 =0.78 𝑚

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≈ 331+0.60𝑇 =343.6 m/s Allowed frequencies 𝑓 𝑛 = 𝑣 𝜆 𝑛 = 𝑛𝑣 2𝐿 𝑛=1,2,3…. Length is 𝐿= 1𝑣 2 𝑓 1 = 343.6 𝑚/𝑠 2 ∙ 262 /𝑠 =0.656 𝑚 Wavelength 𝜆=2𝐿=1.31 𝑚 same inside and outside tube

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

Problem 34 – Assume open-open <?> Write n and (n+1) harmonics in terms of fundamental 𝑓 𝑛 = 𝑛𝑣 2𝐿 =𝑛 𝑓 1 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, 704 𝑛=1,2,3,4…. ????!!

Problem 34 – Assume open-closed <?> Write n and (n+2) odd harmonics in terms of fundamental 𝑓 𝑛 = 𝑛𝑣 4𝐿 =𝑛 𝑓 1 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, 616 𝑛=1,3,5,7…. success!!