Standing Waves in Sound Tubes Physics Mrs Coyle

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 1 st harmonic : L= ½ λ =>λ=2 L => f = v/(2 L) 2 nd harmonic f 2 = 2 f 1 3 rd harmonic f 3 = 3 f 1

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

Example 1 A pipe is open on both ends and is 1. 00 m 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. 00 m, b) 172 Hz

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

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

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. 4 Hz, b) 290 Hz