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Resonance in a Closed Tube

Resonance in a Closed Tube. Constant Length, Changing Frequency. Review: Changing Length. First resonance point: ≈ Half of difference ( ½ Δ x) . Decreases as f increases. Antinode to node. x initial ≈ ¼ λ … End correction!. Distance between resonance points:

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Resonance in a Closed Tube

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

  2. Review: Changing Length • First resonance point: • ≈ Half of difference ( ½ Δx). • Decreases as f increases. • Antinode to node. • xinitial ≈ ¼ λ … End correction! • Distance between resonance points: • Constant for same frequency. • Decreases as f increases. • Node to node. • Δx = ½ λ

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

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

  5. What about constant length? • When resonating, net displacement of molecules is zero. • Amplitude at resonance points is a relative maximum, because the sound is loudest. • Constant length: • Constant velocity, b/c constant T. • Changing frequency & wavelength, b/c length changes.

  6. Overtones at Constant Length:

  7. Closed Tube Resonant Frequencies:

  8. Frequency, Wavelength, and Speedof ANY wave, including Sound: v = f λ Know two, find the third! Wavelength calculated as a fraction of L, or L calculated from λ. Speed calculated: v = 331.5 + 0.607 T. Frequency: measured or calculated.

  9. Open Tube Using the analysis of a closed tube as a guide,determine the frequencies at which an open tube of fixed length will resonate.

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