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Standing Waves in Sound Tubes

Standing Waves in Sound Tubes. Physics Mrs. Coyle. Resonance. When an outside force is applied with a frequency equal to or a multiple of the natural frequency of vibration of an object, the object begins to vibrate and its amplitude increases. Resonance in Sound Tubes. Examples:

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Standing Waves in Sound Tubes

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  1. Standing Waves in Sound Tubes Physics Mrs. Coyle

  2. Resonance • When an outside force is applied with a frequency equal to or a multiple of the natural frequency of vibration of an object, the object begins to vibrate and its amplitude increases.

  3. Resonance in Sound Tubes Examples: • Musical instruments (flutes, clarinets etc) • Bottles

  4. Resonance in Sound TubesOpen-Open End Fundamental Frequency 1st harmonic : L= ½ λ =>λ=2L => f= v/(2L) 2nd harmonic f2 = 2f1 3rd harmonic f3 = 3f1

  5. Resonance in Sound TubesOpen-Open End • Number of harmonic matches the number of nodes. (n=1,2,3,…) • Every one node corresponds to ½ λ. • L= n (½λ) => λ=2L/n • f=v λ=> fn = nv/(2L) • fn = nf1 • What does this equation remind you of ?

  6. Example 1 A pipe is open on both ends and is 1.00m long and is at T=20oC. a) What is the wavelength of the lowest resonant frequency? b) What is the fundamental frequency? Answer: a) 2.00m, b) 172 Hz

  7. Resonance in Sound TubesClosed-Open Fundamental Frequency 1stharmonic L= ¼ λ => λ=4L => f= v/(4L) 2ndharmonic f2 = 3f1 3rdharmonic f3 = 5f1

  8. Resonance in Sound TubesOne End Closed-One End Open • Number of harmonic matches the number of nodes. (n=1,2,3,…) • L= (2n-1)λ /4 => λ=4L/(2n-1) • f=v λ=> fn = (2n-1)v/(4L) • fn = (2n-1)f1 • These equations are the same as a string with one free end and one fixed end.

  9. Example 2 A pipe is open on one end and closed on the other is 2.0 meter long. a) What is the lowest resonant frequency? b) Draw the 4th harmonic and find the 4th resonant frequency? Answers: a) 41.4Hz, b) 290Hz

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