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I was asked a puzzling question/thought experiment:

Given the source of a sound in a wide open field so acoustics do not play a role, is it possible for a sound to be louder as you move away from it.

My answer was instinctively no. As you move away from sound it dissipates, so it should not be louder.

The response is that, if there is another, weaker source of sound closer to you, then by walking away the source closer to you will lose strength and the source farther will "shine" through better.

This doesn't feel right to me. Surely when you move away from the sound, the farther one dissipates as well as the nearer one? Is this potentially because of the inverse square law?

edit: This question probably applies in a similar way to light. Not quite sure what the right tags should be for this

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  • $\begingroup$ I often hear whistles of trains at night in winter though I can't hear it in Summer. I am used to think that my location is such that I am getting the sound in a similar fashion as in total internal reflection of light in air in the case of mirage in a desert. You may google it whether it may be possible in this way or not. But I am not sure. $\endgroup$ – user22180 Apr 12 '14 at 8:23
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No, but it can sound like it.

If there are two sounds that have similar frequencies, we will only hear the loudest one (this phenomenon is used to help compress music files). This is caused auditory masking. Now, if you have a loud sound far away, and a less loud sound close to you, the one close to you will sound louder, and will mask the farther away one. If you move away from both of them, then because of the inverse square law the near sound will get softer faster than the far-away one, so the far-away sound will no longer be masked. This may sound to you like the farther-away sound is getting louder as you move away from it. (Even though it isn't in reality.)

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The only way that I can envision a sound getting louder as you walk away is if you happen to initially be located at a node where destructive interference causes the waves to be near zero amplitude.

When the waves are exactly out of phase at the point you stand, you will not hear anything. As you move in any direction away from said point, it will get louder. This of course requires either two sound sources or a solid boundary which reflects the original wave with a phase shift (an echo, but you ruled that out as a possibility). And then carefully positioning yourself to begin the experiment so that you are at a destructive node.

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Maybe possible if you can construct "Air Lens"

It is in my mind for such times. Sound wave can refract like light wave. So it possible that if you can shape the pressure of air in an area in a fit form. It could be guide spherical sound wave into point cone. Make it louder at focal point

This is the explanation that my idea occur in reality http://www.youtube.com/watch?v=z7RbHBK58h8

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  • $\begingroup$ Cool link - although I think the explanation given in the video is incomplete. It's the density, not the temperature, that makes the refraction possible. $\endgroup$ – Floris Jun 26 '14 at 17:14
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I think that in general, as you move away from a sound it gets softer due to the dissipation of energy. However, I can see possibilities using exotic configurations of air density where the sound does get louder. For example, imagine that the air density increases while retaining the same bulk modulus. Then, as you move away from the source, the sound velocity will drop, allowing a buildup of sound pressure, just like in a sonic boom.

Another, more subjective possiblity, is that the sound experiences nonlinear effects such as frequency-dependent modulation and disperson. In that case, a powerful sound source emmitting sound above your hearing threshold would be inaudible up close, but due to nonlinear effects, more of that high energy sound is modulated downwards to audbile frequencies.

From OPs Comment

If your sources (A and B) have amplitudes $a,b$ at $r_A=r_b=1$ then assuming we move a distance $\delta$ from both A and B simultaneously we have the formulas:

$I_A(\delta)=\frac{a}{(r_A+\delta)^2},I_B(\delta)=\frac{b}{(r_B+\delta)^2}$ assuming the inverse square energy dissipation. What we want to know is how the intensities change as we move away from both sources, if $r_A>r_B,\frac{a}{r_A^2}>\frac{b}{r_B^2}$. One way to look at this is the ratio of the two intensities as you move away:

$\frac{I_A(\delta)}{I_B(\delta)} = \frac{a(r_B+\delta)^2}{b(r_A+\delta)^2}=\frac{a}{b}(\frac{r_B+\delta}{r_A+\delta})^2, \\ \lim\limits_{\delta\rightarrow \infty}\frac{a}{b}(\frac{r_B+\delta}{r_A+\delta})^2 = \frac{a}{b}$

Therefore, if you precieve A as louder than B, then moving away from both will never let B be louder than A, let alone let the sound overall get louder.

Anyway, just a couple out there thoughts.

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  • $\begingroup$ Do you think you could address the part about the weaker but closer source of sound? $\endgroup$ – Cruncher Nov 22 '13 at 19:06
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    $\begingroup$ Is this assuming that both sources of sound are equidistant from you? I was concerned about the case when the weaker sound was closer than the farther sound, then moving a fixed distance away from both $\endgroup$ – Cruncher Nov 22 '13 at 20:52
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    $\begingroup$ Actually, what this shows is that, for any difference between rA and rB, it will be irrelevant if you move sufficiently far away. It will eventually reach a/b. Which is greater than 1 if a > b. However, for all d such that (rB+d/rA+d)^2 < b/a, then the value is actually less than 1, and b IS louder than a. $\endgroup$ – Cruncher Nov 22 '13 at 21:13
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    $\begingroup$ The starting point is only > 1 if the actually louder sound starts closer than the actually quieter sound, which is the opposite of what I asked. That is, a whisper beside your ear, is louder than an amplifier a mile away. But then walk 10 feet back from the whisper, and the amplifier is much louder. The 10 feet is insignificant to the mile, but very significant to the inch $\endgroup$ – Cruncher Nov 22 '13 at 21:24
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    $\begingroup$ @cruncher I see, I was assuming that the closer sound still sounded fainter. If it starts off louder then indeed it will eventually become relatively weaker than the intrinsically louder sound. My misunderstanding. $\endgroup$ – user31580 Nov 22 '13 at 21:49
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Some good points made in different answers. I just want to add my two cents. The short answer is "yes - it can appear louder, and it can be louder".

First - appearance. If you have a loud point source far away, and another source closer by, it is possible that the close source clouds the faraway sound. Imagine a faraway train and a nearby radio playing music. While you are close to the radio, you cannot hear the train because of the music. As you move further from the radio (and the train), the train noise will start to drown out the radio. As you surmised, this is a simple inverse square law problem.

Second - interference. If the source of sound is extended (not a point source) then there will be constructive interference at some points, and destructive interference at others. An incredible example of this is the Game of Life sound synthesis system that uses 192 speakers to focus sound "anywhere". In that system, walking away can increase the intensity of the sound you hear.

Third - refraction. During a windless evening the density of air above a body of water can be higher than the density elsewhere - because the air cools down more quickly. The difference in density causes changes in the speed of sound, and makes the body of air act as a giant lens. This is most often observed as "sound carries over water" - that is, the sound of a party on the other side of the lake can be heard at night when you might not hear it during the day. But depending on the shape of the "moisture lens", it is possible that the sound actually gets louder as you move further away, since you might be moving into the focal point of the lens. The link that @Thaina pointed to explains this.

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Yes, a sound could become louder as you move away. One way for this to occur is if the sound waves are created in such as way as to focus to a point. As the sound is focusing the loudness will increase and after it passes through the focus the loudness will decrease with distance. @Thaina had the right idea by suggesting that an air lens could focus the sound but it would probably be easier to used a phased-array of speakers. This other post discusses how to focus sounds: How can you focus sound.

I think your intuition is leading you astray because you are assuming the emitter of the sound must be a point source which is radiating in all directions. Instead try to consider many beams of sound (like from a parabolic speaker) setup as a phased-array could give you focused sound.

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A scenario that conceivably could work in some situations:

Think of how passive sonar behaves--specifically, convergence zones. Given the right gradient in the sound transmitting media you get a situation where sounds radiated over a range of angles are all focused back at the same point. As you get close to that distance from the source the sound energy goes way up. Assuming the sound is loud enough you can also hear it by a direct path you have the situation you specified--it gets louder as you move away.

Since I have heard of such bending in atmosphere in theory it could happen. The only situations I have heard of such bending involved sufficiently great distances that there was no direct path, though.

What exactly conditions would be required, and whether they could exist on Earth I'll have to leave to those more knowledgeable than I.

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