Standing Waves in Sound Tubes Physics Mrs. Coyle.

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Presentation transcript:

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: Musical instruments (flutes, clarinets etc) Bottles

Resonance in Sound Tubes Open-Open End Fundamental Frequency 1st harmonic : L= ½ λ =>λ=2L => f = v/(2L) 2nd harmonic f 2 = 2f 1 3rd harmonic f 3 = 3f 1

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=> f n = nv/(2L) f n = nf 1 What does this equation remind you of ? Resonance in Sound Tubes Open-Open End

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

Resonance in Sound Tubes Closed-Open Fundamental Frequency 1 st harmonic L= ¼ λ => λ=4L => f = v/(4L) 2 nd harmonic f 2 = 3f 1 3 rd harmonic f 3 = 5f 1

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

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 4 th harmonic and find the 4 th resonant frequency? Answers: a) 41.4Hz, b) 290Hz