that a guy that spouts off about Cooper pairs and the like has never heard of a standing wave. But since that's apparently what you're claiming, here goes: Of course I've heard of standing waves. 1) You are right about the sound increasing in level when a reflective surface is close by, if certain conditions are met. Specifically, the surface has to be within a small fraction (typically given as 1/(2*pi)) of a wavelength of the receiver at the frequency of interest. However, if I'm not mistaken, the gain is only 3 dB and not 6 dB. What exactly are you referring to as the "receiver" in this instance? The cable? What we're talking about are the conditions at the boundary of a reflective surface. Which will effectively be the conditions of whatever object(s) happen to be at that surface. And it is ideally 6dB, not 3dB. From Master Handbook of Acoustics by Everest: The sound pressure on a surface normal to the indident waves is equal to the energy-density of the radiation in front of the surface. If the surface is a perfect absorber, the pressure equals the energy-density of the incident radiation. If the surface is a perfect reflector, the pressure equals the energy-density of both the incident and reflected radiation. Thus the pressure at the surface of a perfectly reflecting surface is twice that of a perfectly absorbing surface. Pressure doubles. Power quadruples (as it goes up as the square of pressure). So, for pressure, 20 x log 2 = 6dB. For power, 10 x log 4 = 6dB. So that takes care of the case where one is outside, with only the ground nearby, and no walls. On the other hand, Leisure7 was talking about listening in a room. Doesn't matter whether inside or outside. In both cases we're talking about the conditions at a reflective surface. Doesn't matter if that reflective surface is outside or inside. 2) Standing waves can occur when the excitation frequency corresponds to the distance between opposing boundary surfaces (i.e, walls). When a standing wave exists (which is quite likely when playing music in a listening room), then at the boundaries (i.e., on the floor) there exists a pressure maximum and a velocity minimum. A cable is going to be disturbed far more by velocity than by pressure, so placing it on the floor will reduce its excitation by standing waves compared to placing it at a pressure null. Velocity of what? The wave? We're talking cables here. Until the diameter of the cable begins approaching a wavelength, I don't even see wave velocity coming into play to any degree. Until that point, pretty much all the cable experiences is pressure change. The worst case scenario would be when the cable's diemeter corresponds to 1/4 wavelenth which if you've got say a 1/4" cable, you're looking at what, about 14kHz? Ain't much energy up there. Where I'd be concerned is down at the lower frequencies where there's more energy. But at those frequencies, the wavelengths are so long you can't really produce much of a pressure gradient between one side of the cable and the other with which to move it. se
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