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Applications of Resonance

Applications of Resonance. Harmonics and Beats. Beats. Not that kind This is due to the interference of two waves of similar frequencies. They interfere in a way that you hear alternating loudspots and softspots. Harmonics.

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Applications of Resonance

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  1. Applications of Resonance Harmonics and Beats

  2. Beats • Not that kind • This is due to the interference of two waves of similar frequencies. • They interfere in a way that you hear alternating loudspots and softspots.

  3. Harmonics • We are going to look at three situations, strings, open ended tubes and closed ended tubes. • Strings can have standing wave created by plucking them or by finding the resonant frequency with a tuning fork. • The ends of the strings do not vibrate, therefore they must be….. • Nodes • The simplest wave vibrations is when you have an antinode at the center of the string. That would show half a wavelength. Therefore the wavelength would be

  4. If we remember our wave equation • V=freq x wavelength • Therefore the f = v/wavelength • For the simplest wave f1 = v/2L • This is called the fundamental frequency – the lowest frequency of vibration of a standing wave. • V is the speed of the waves on the string and not the speed of the sound wave.

  5. Lets look at the harmonic series • N stands for the harmonic number • N=2 the wavelength= L the f2=2f1 It is called the second harmonic or 1st overtone • N=3 wavelength=2/3L f3=3f1 This is the 3rd harmonic and 2nd overtone • N=4 wavelength= 1/2L f4=4f1 • N=5 wavelength= 2/5L and so on

  6. Lets see the math!!!! Yee Haw! • fn= n(v/2L) where n=1,2,3 and so on • Lets practice!

  7. What about open ended pipes you ask? • Standing waves can be setup as a column of air in a pipe like in an organ. • If the pipe is open at an end, an antinode exists. • Therefore 2 open ends have two antinodes. • This allows for all of the same harmonics available to the string • fn= n(v/2L) where n=1,2,3 and so on • In this case v is the speed of sound in the pipe.

  8. In a closed end? • Well… Thanks for asking… Since there is that closed end, there must be a node at that end. • Therefore this limits the number of harmonics that are possible. • The fundamental frequency consists of ¼ of a wave pattern. • Therefore for closed ended only odd harmonic numbers are allowed. • fn= n(v/4L) where n=1,3,5

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