5 ms·

Sound intensity actually falls off as 1/r², and here's a source: http://hyperphysics.phy-astr.gsu.edu/hbase/acoustic/invsqs.html#c1 http://hyperphysics.phy-ast

by dkhar 12y ago

Sound intensity actually falls off as 1/r², and here's a source: http://hyperphysics.phy-astr.gsu.edu/hbase/acoustic/invsqs.html#c1 http://hyperphysics.phy-astr.gsu.edu/hbase/acoustic/invsqs.h...

sillysaurus3 12y ago
If sound traveled as a sphere, then it wouldn't be able to circle the Earth. If you imagine a sphere growing from a point on the Earth's surface, most of it will wind up in space, and none of it will reach the opposite side of the Earth. The 1/r^2 falloff is probably just an approximation that holds true for short distances or smaller intensities. In this case, the soundwave was able to follow the curvature of the Earth, which implies that it wouldn't decay as a sphere but rather a plane, which would be a 1/r falloff. I wonder why sound of massive intensity will follow Earth's curvature?
- EGreg 12y ago
  Why don't you think that the sound waves would go in all directions, and be absorbed / redirected at the edges? What you see following the curvature of the Earth is the result of that, and once you factor in the second-order wave reflection, it's still on the order of 1/r^2 isn't it?
  - sillysaurus3 12y ago
    Nah, imagine a point on the earth. Now trace rays from that point in every possible direction. All of those rays will eventually lead to space or to the ground, and none will reach the other side of the Earth. Since molecules are physically further apart the higher you go in the atmosphere, it seems unlikely that the energy would be redirected at the edges, only dispersed. That must mean the wave is literally following wherever the atmosphere is thickest rather than simply being reflected.
    - jonsen 12y ago
      Sound waves can diffract around objects, which is why one can still hear someone calling even when hiding behind a tree. : http://en.m.wikipedia.org/wiki/Difraction http://en.m.wikipedia.org/wiki/Difraction
      - sillysaurus3 12y ago
        Is diffraction the reason why soundwaves could travel around the whole Earth? My previous understanding of diffraction was that obstacles cause waves to propagate in different ways. But the thinning of the atmosphere isn't really an "obstacle." The molecules that soundwaves use to propagate are simply further apart from each other, meaning waves are more likely to disperse and lose energy than to keep traveling or bounce. That would imply the boom from the volcano should disperse into space and go silent rather than travel around the Earth. But since that doesn't happen, it seems like the waves follow wherever the atmosphere is thick. I'm having trouble understanding how diffraction would cause that end result of "waves go where the atmosphere is." If waves could bounce off of the thin atmosphere near space, that would make total sense. But they can't bounce due to thin atmosphere, only disperse, so it seems like there's some other phenomenon in play.
        jonsen 12y ago
        Sound waves are points of high pressure and points of low pressure. At each point of high pressure the pressure tends to spread uniformly in all directions. Likewise at points of low pressure there is sucking from all directions. Opposite directions cancel, i.e. orthogonal to the wave direction. All in all the sum of all points of pressure creates a moving wavefront. Which naturally bends around obstacles. The earth is just a very big obstacle.
        sillysaurus3 12y ago
        Hey, thank you for taking the time to explain this. I really appreciate it. I'm having trouble seeing how that explanation would explain the case at hand. Your explanation is likely correct, and I'm probably just thinking about it incorrectly. Would you mind pointing out the flaw in my logic? In this scenario, a volcano's boom was so loud that it traveled through the atmosphere, all the way around the Earth. Your explanation is perfectly reasonable for thick atmosphere. In thick atmosphere, a soundwave is a pressure differential, and since molecules are densely packed together (since the atmosphere is thick), there's no choice but for the molecules to "slosh around." The high pressure areas will spread to the low pressure areas within the thick atmosphere and create a moving wavefront, exactly like you said. But as the soundwave travels closer to space, the atmosphere becomes thinner. There are fewer molecules for the soundwave to travel through. That means a pressure differential will have less medium through which to traverse. Since there are fewer molecules, there's more room between them to absorb a pressure differential, right? For example, the reason sound travels so well underwater is because water is extremely dense in comparison to the atmosphere, so less energy is needed to travel an equal distance underwater. Correspondingly, near space where the atmosphere is thin, more energy would be needed to traverse an equal distance. That must mean that as the wavefront approaches space, the wavefront should dissipate. Since more energy is required to travel through less atmosphere, then as the atmosphere approaches zero, the energy required for a wavefront to travel one meter should approach infinity, and that's why it seems like the wavefront should dissipate near the edge of the exosphere. But in this case, the wavefront didn't dissipate. The volcano's boom kept on going all the way around the Earth, and it was somehow able to maintain its energy. If the soundwave travels as a sphere from its point of origin, then that sphere should have a hard time traveling all the way around Earth, shouldn't it? So it must not be travelling as a sphere, but something else. You're saying that "something else" is diffraction. I'd like to understand that. How is it that a wavefront of such intensity can approach the exosphere where it should dissipate, yet not dissipate and instead keep traveling all the way around the Earth due to diffraction?

by dkhar 12y ago

Sound intensity actually falls off as 1/r², and here's a source: http://hyperphysics.phy-astr.gsu.edu/hbase/acoustic/invsqs.html#c1 http://hyperphysics.phy-astr.gsu.edu/hbase/acoustic/invsqs.h...

sillysaurus3 12y ago

If sound traveled as a sphere, then it wouldn't be able to circle the Earth. If you imagine a sphere growing from a point on the Earth's surface, most of it will wind up in space, and none of it will reach the opposite side of the Earth. The 1/r^2 falloff is probably just an approximation that holds true for short distances or smaller intensities. In this case, the soundwave was able to follow the curvature of the Earth, which implies that it wouldn't decay as a sphere but rather a plane, which would be a 1/r falloff. I wonder why sound of massive intensity will follow Earth's curvature?

EGreg 12y ago
Why don't you think that the sound waves would go in all directions, and be absorbed / redirected at the edges? What you see following the curvature of the Earth is the result of that, and once you factor in the second-order wave reflection, it's still on the order of 1/r^2 isn't it?
- sillysaurus3 12y ago
  Nah, imagine a point on the earth. Now trace rays from that point in every possible direction. All of those rays will eventually lead to space or to the ground, and none will reach the other side of the Earth. Since molecules are physically further apart the higher you go in the atmosphere, it seems unlikely that the energy would be redirected at the edges, only dispersed. That must mean the wave is literally following wherever the atmosphere is thickest rather than simply being reflected.
  - jonsen 12y ago
    Sound waves can diffract around objects, which is why one can still hear someone calling even when hiding behind a tree. : http://en.m.wikipedia.org/wiki/Difraction http://en.m.wikipedia.org/wiki/Difraction
    - sillysaurus3 12y ago
      Is diffraction the reason why soundwaves could travel around the whole Earth? My previous understanding of diffraction was that obstacles cause waves to propagate in different ways. But the thinning of the atmosphere isn't really an "obstacle." The molecules that soundwaves use to propagate are simply further apart from each other, meaning waves are more likely to disperse and lose energy than to keep traveling or bounce. That would imply the boom from the volcano should disperse into space and go silent rather than travel around the Earth. But since that doesn't happen, it seems like the waves follow wherever the atmosphere is thick. I'm having trouble understanding how diffraction would cause that end result of "waves go where the atmosphere is." If waves could bounce off of the thin atmosphere near space, that would make total sense. But they can't bounce due to thin atmosphere, only disperse, so it seems like there's some other phenomenon in play.
      - jonsen 12y ago
        Sound waves are points of high pressure and points of low pressure. At each point of high pressure the pressure tends to spread uniformly in all directions. Likewise at points of low pressure there is sucking from all directions. Opposite directions cancel, i.e. orthogonal to the wave direction. All in all the sum of all points of pressure creates a moving wavefront. Which naturally bends around obstacles. The earth is just a very big obstacle.
        sillysaurus3 12y ago
        Hey, thank you for taking the time to explain this. I really appreciate it. I'm having trouble seeing how that explanation would explain the case at hand. Your explanation is likely correct, and I'm probably just thinking about it incorrectly. Would you mind pointing out the flaw in my logic? In this scenario, a volcano's boom was so loud that it traveled through the atmosphere, all the way around the Earth. Your explanation is perfectly reasonable for thick atmosphere. In thick atmosphere, a soundwave is a pressure differential, and since molecules are densely packed together (since the atmosphere is thick), there's no choice but for the molecules to "slosh around." The high pressure areas will spread to the low pressure areas within the thick atmosphere and create a moving wavefront, exactly like you said. But as the soundwave travels closer to space, the atmosphere becomes thinner. There are fewer molecules for the soundwave to travel through. That means a pressure differential will have less medium through which to traverse. Since there are fewer molecules, there's more room between them to absorb a pressure differential, right? For example, the reason sound travels so well underwater is because water is extremely dense in comparison to the atmosphere, so less energy is needed to travel an equal distance underwater. Correspondingly, near space where the atmosphere is thin, more energy would be needed to traverse an equal distance. That must mean that as the wavefront approaches space, the wavefront should dissipate. Since more energy is required to travel through less atmosphere, then as the atmosphere approaches zero, the energy required for a wavefront to travel one meter should approach infinity, and that's why it seems like the wavefront should dissipate near the edge of the exosphere. But in this case, the wavefront didn't dissipate. The volcano's boom kept on going all the way around the Earth, and it was somehow able to maintain its energy. If the soundwave travels as a sphere from its point of origin, then that sphere should have a hard time traveling all the way around Earth, shouldn't it? So it must not be travelling as a sphere, but something else. You're saying that "something else" is diffraction. I'd like to understand that. How is it that a wavefront of such intensity can approach the exosphere where it should dissipate, yet not dissipate and instead keep traveling all the way around the Earth due to diffraction?