Resonance in a Closed Tube
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Resonance in a Closed Tube Constant Frequency, Changing Length
description
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. - PowerPoint PPT Presentation
Transcript of 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.– Constant wavelength, b/c v = f λ
Molecular Movement Represented:
Molecular Movement Represented:
Molecular Movement Represented:Distance between resonance points:
Constant for same frequency.
Decreases as f increases.
Node to node.
Δx = ½ λ
Molecular Movement Represented:First resonance point:
≈ Half of difference ( ½ Δx).
Decreases as f increases. (Duh!)
Antinode to node.
xinitial ≈ ¼ λ … End correction!
Tube Length vs. Wavelength:
L ≈ ¼ λ
L ≈ 5/4 λ
L ≈ ¾ λ
Tube Length vs. Wavelength: L = 1/4 λ, 3/4 λ, 5/4 λ, …odd/4 λ
L ≈ ¼ λ
L ≈ 5/4 λ
L ≈ ¾ λ
Calculating Wavelength:
L ≈ ¼ λλ ≈ 4 L
Δx = ½ λ
λ = 2 Δx
λ ≈ 4/3 L
λ ≈ 4/5 L
Frequency, Wavelength, and Speedof ANY wave, including Sound:
v = f λKnow two, find the third!
Wavelength measured: λ = 2 ΔxSpeed calculated: v = 331.5 + 0.607 T
Frequency: Measured or calculated
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.
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
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
Observations / Conclusions
There appears to be an inverse relationship between frequency and separation distance.
f α 1 / 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?
Open Tube:
Open Tube: Resonant Lengths
Open Tube: Resonant LengthsL = ½ λ, 2/2 λ, 3/2 λ
L = 2/4 λ, 4/4 λ, 6/4 λ, …even/4 λ
