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Resonance in a Closed Tube

Resonance in a Closed Tube. Constant Frequency, Changing Length. Movement of Air. Net displacement of molecules is zero. Amplitude at resonance points is a relative maximum, because the sound is loudest. Constant frequency, b/c the pitch doesn’t change. Constant velocity, b/c constant T.

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Resonance in a Closed Tube

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  1. Resonance in a Closed Tube Constant Frequency,Changing Length

  2. Movement of Air • Net displacement of molecules is zero. • Amplitude at resonance points is a relative maximum, because the sound is loudest. • Constant frequency, b/c the pitch doesn’t change. • Constant velocity, b/c constant T. • Constant wavelength, b/c v = f λ

  3. Molecular Movement Represented:

  4. Molecular Movement Represented:

  5. Molecular Movement Represented: • Distance between resonance points: • Constant for same frequency. • Decreases as f increases. • Node to node. • Δx = ½ λ

  6. Molecular Movement Represented: • First resonance point: • ≈ Half of difference ( ½ Δx). • Decreases as f increases. (Duh!) • Antinode to node. • xinitial ≈ ¼ λ … End correction!

  7. Tube Length vs. Wavelength: L ≈ ¼ λ L ≈ ¾ λ L ≈5/4λ

  8. L ≈ ¼ λ L ≈ ¾ λ L ≈5/4λ Tube Length vs. Wavelength:L = 1/4λ, 3/4λ, 5/4λ, …odd/4λ

  9. Calculating Wavelength: Δx = ½ λλ= 2 Δx L ≈ ¼ λλ≈ 4 L λ≈ 4/3 L λ≈ 4/5 L

  10. Frequency, Wavelength, and Speedof ANY wave, including Sound: v = f λ Know two, find the third! Wavelength measured: λ = 2 Δx Speed calculated: v = 331.5 + 0.607 T Frequency: Measured or calculated

  11. Open Tube Using the analysis of a closed tube as a guide,determine the lengths of an open tube that will resonate. Answer: last page.

  12. Example Data: Closed Tube Resonant Lengths (cm): • 466.1 Hz: 23.0, 60.2, 95.8, 132.9 • 500.0 Hz: 19.0, 53.1, 85.8, 121.0, 154.7 • 1000. Hz: 8.7, 23.9, 42.2, 58.7, 74.6, 92.9, 109.5, 126.3, 143.1, 158

  13. Example Data: Calculations Average separation of resonant points: • 466.1 Hz: 36.6 cm • 500.0 Hz: 33.9 cm • 1000. Hz: 16.6 cm

  14. Observations / Conclusions There appears to be an inverse relationship between frequency and separation distance. f α1 / Δx , “f is inversely proportional to Δx.” f = k / Δx , where k is the proportionality constant[slope is the proportionality constant in a a direct relationship]. To prove inverse relationship: • Calculate k for each frequency. • Are the values of k equal?

  15. Open Tube:

  16. Open Tube: Resonant Lengths

  17. Open Tube: Resonant LengthsL = ½ λ, 2/2λ, 3/2λL = 2/4λ, 4/4λ, 6/4λ, …even/4λ

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