Standing sound wave in a tube and energy 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?
 A: 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
