Standing Waves in Sound Tubes

PhysicsMrs. 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 EndFundamental Frequency 1st harmonic : L= ½ λ =>λ=2L => f = v/(2L)

2nd harmonic f2 = 2f1

3rd harmonic f3 = 3f1

• 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=> fn = nv/(2L)

• fn = nf1

• What does this equation remind you of ?

Resonance in Sound TubesOpen-Open End

Example 1

A pipe is open on both ends and is 1.00m long and is at T=20oC.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 TubesClosed-Open

Fundamental Frequency1st harmonic L= ¼ λ => λ=4L => f = v/(4L)

2nd harmonic f2 = 3f1

3rd harmonic f3 = 5f1

• Number of harmonic matches the number of nodes. (n=1,2,3,…)

• L= (2n-1)λ /4 => λ=4L/(2n-1)

• f=v λ=> fn = (2n-1)v/(4L)

• fn = (2n-1)f1

• These equations are the same as a string with one free end and one fixed end.

Resonance in Sound TubesOne End Closed-One End Open

Example 2

A pipe is open on one end and closed on theother is 2.0 meter long. a) What is the lowest resonant frequency?b) Draw the 4th harmonic and find the 4th

resonant frequency?

Answers: a) 41.4Hz, b) 290Hz