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Resonance in a tube open at both ends

Another source of standing waves is a tube that is open at both ends. In this case, the boundary conditions are symmetrical: an antinode at each end. The resonances of tubes open at both ends can be analyzed in a very similar fashion to those for tubes closed at one end. The air columns in tubes open at both ends have maximum air displacements at both ends ( [link] ). Standing waves form as shown.

Picture is a diagram of the fundamental and three lowest overtones for a tube closed at one end. Fundamental has half of its wavelength in a tube. First overtone has one of its wavelength in a tube, second overtone has one and a half of its wavelength in a tube, third overtone has two of its wavelength in a tube. All have maximum air displacements at both ends of a tube.
The resonant frequencies of a tube open at both ends, including the fundamental and the first three overtones. In all cases, the maximum air displacements occur at both ends of the tube, giving it different natural frequencies than a tube closed at one end.

The relationship for the resonant wavelengths of a tube open at both ends is

λ n = 2 n L , n = 1 , 2 , 3 ,... .

Based on the fact that a tube open at both ends has maximum air displacements at both ends, and using [link] as a guide, we can see that the resonant frequencies of a tube open at both ends are

f n = n v 2 L , n = 1 , 2 , 3... ,

where f 1 is the fundamental, f 2 is the first overtone, f 3 is the second overtone, and so on. Note that a tube open at both ends has a fundamental frequency twice what it would have if closed at one end. It also has a different spectrum of overtones than a tube closed at one end.

Note that a tube open at both ends has symmetrical boundary conditions, similar to the string fixed at both ends discussed in Waves . The relationships for the wavelengths and frequencies of a stringed instrument are the same as given in [link] and [link] . The speed of the wave on the string (from Waves ) is v = F T μ . The air around the string vibrates at the same frequency as the string, producing sound of the same frequency. The sound wave moves at the speed of sound and the wavelength can be found using v = λ f .

Check Your Understanding How is it possible to use a standing wave’s node and antinode to determine the length of a closed-end tube?

When the tube resonates at its natural frequency, the wave’s node is located at the closed end of the tube, and the antinode is located at the open end. The length of the tube is equal to one-fourth of the wavelength of this wave. Thus, if we know the wavelength of the wave, we can determine the length of the tube.

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This video lets you visualize sound waves.

Check Your Understanding You observe two musical instruments that you cannot identify. One plays high-pitched sounds and the other plays low-pitched sounds. How could you determine which is which without hearing either of them play?

Compare their sizes. High-pitch instruments are generally smaller than low-pitch instruments because they generate a smaller wavelength.

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Summary

  • Unwanted sound can be reduced using destructive interference.
  • Sound has the same properties of interference and resonance as defined for all waves.
  • In air columns, the lowest-frequency resonance is called the fundamental, whereas all higher resonant frequencies are called overtones. Collectively, they are called harmonics.

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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