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We know that there is standing wave in a tube if we excite it with a source (ex:speaker). The energy "leaks" from the open end and we hear amplified sound. It seems strange to me:

$Intensity = {power \over area} = {power \over 4 \pi r^2}$ (point source)

If we have a speaker with constant watts and we stand beside the speaker (we have distance r fixed), we would hear a sound that is lower than the speaker with a tube. So what causes this intensity difference?

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  • $\begingroup$ Think about the area that the power is spread over in each case. What does the tube do to this area? $\endgroup$ – probably_someone Dec 24 '16 at 5:47
  • $\begingroup$ Do you mean the tube concentrate it? But in what way? Into a specific direction? $\endgroup$ – Math The Novice Dec 24 '16 at 5:53
  • $\begingroup$ Precisely. Specifically, it concentrates it into the direction the tube points. $\endgroup$ – probably_someone Dec 24 '16 at 5:54
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    $\begingroup$ But when we do that experiment we can clearly hear the sound no matter where we stand, why? $\endgroup$ – Math The Novice Dec 24 '16 at 5:56
  • $\begingroup$ Because the tube is not perfectly soundproof. Sound leaks through the walls. Also, because sound tends to diffract from the opening of the tube. If you had an infinitely long, perfectly soundproof tube that captured the sound from the speaker, you would indeed hear nothing. $\endgroup$ – probably_someone Dec 24 '16 at 5:58
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The resonance in the tube doesn't "amplify" the power being radiated. That would be nonsense, since it would violates the principle of conservation of energy!

However, if you have an open-back loudspeaker vibrating in free air, the sound waves radiated from the front and back of the speaker cone are in antiphase and radiated from the same position in space, so they cancel each other out almost exactly every listening position.

If you put the speaker at one end of a tube, the sounds are now radiated from the two ends of the tube, which re separated in space, by the length of the tube. The relative phase of the sounds is also changed as the sound frequency passes through the resonant frequency of the tube. The combination of those two factors means that at the resonance condition, two sounds cancel out "almost nowhere", instead of "almost everywhere".

The same principle is used in some loudspeaker cabinet designs, where the sound radiated from the back of the speaker is in effect transmitted along a U-shaped "tube" and emitted from a port (i.e. a hole in the case) on the front of the speaker enclosure, with its phase changed so that it doesn't cancel out the sound from the front of the speaker cone.

The image in this Wikipedia link shows that cancellation does not occur, except at a few locations in space (shown in grey on the image). The points marked S1 and S2 correspond to the ends of the tube, from which the sound is being radiated.

https://en.wikipedia.org/wiki/File:Interf.png

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  • $\begingroup$ I disagree with the characterization of the open-back loudspeaker in free air. The reason the cancellation does not, in fact, occur this way is because, in free air, there is no boundary that could cause reflection that would otherwise facilitate the cancellation you allude to. While two observers equidistant from opposite sides of the speaker would indeed record waveforms with opposite phase, each of them would only hear the particular component from the side of the loudspeaker facing them. $\endgroup$ – probably_someone Dec 24 '16 at 6:19
  • $\begingroup$ @probably_someone what you say is only true if the wavelength of the sound is small compared with the dimension of the speaker. Otherwise, the radiation pattern from each face includes diffracting around the edge of the moving speaker cone, which causes the cancellation. $\endgroup$ – alephzero Dec 24 '16 at 6:32
  • $\begingroup$ But the intensity of the diffracted radiation is far less than the intensity of the radiation directly received, so complete cancellation cannot occur. $\endgroup$ – probably_someone Dec 24 '16 at 6:34
  • $\begingroup$ after some pondering on it, I still cannot convinced myself the sound source would interference with "itself". In fact, this phenomenon would have same result with closed-back speaker. $\endgroup$ – Math The Novice Dec 27 '16 at 12:50

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