When we talk, our vibrating chords oscilate next air molecules which oscilate the next molecules and so on. Hence sound wave travels.

As we know, energy that reaches the destination is not the same one as the starting one because the voice receiver hears is definitely less loud.

Why energy does not reach to the end? The first particle oscilates which in turn makes next air particle to oscilate but why does not the first particle have the power to oscilate the next one with the same amplitude and where is this energy that first particle had transformed into as logically it does not transfer all of its energy to the next particle?

Updated Questions per conversation with @Questor:

Question 1: in your comments, you said: "In a perfectly elastic collision, in a universe composed complete of our atmosphere sound could travel forever.... However, your ability to perceive the sound falls off at 4𝜋𝑟2 due to the inverse square law." then, you also said: "In a 3d space with perfectly elastic collisions, sound does not "travel" infinitely because the intensity of the sound fades as the area of the pressure wave expands, until it reaches a point that it is imperceptible.". Which one is true ? both can't be. Since the intensity decreases with the distance increase.

Question 2: If there was no loss in heat, even with 3d space, we would still have the same effect of decreasing intensity on each point as the distance increases right ? because even if no energy is lost, energy now must spread out over larger area. if correct, then with the heat loss, not only energy is spread out over bigger region each time it spreads, but some of it even gets lost(converted to heat) which causes even less energy spread to the next spherical shell than when we had no heat loss case.

Question 3: In your 1d space example(perfectly elastic example), why would the receiver get the same intensity as the loudspeaker created ? The way I understood elastic collision is, if 2 objects collide, no KE should be lost, it should be the same afterwards the collision. but that doesn't mean the first moving air particle would give the whole of its KE to the next air particle which would cause the first particle to stop immediatelly. Is this what would happen ? but if so, then masses should be equal, right ? but we know that once particle 1 hits the particle 2, particle 1 doesn't stop completely, it bounces off, so it still has some speed. Which means it doesn't give all of its energy to particle 2. So not sure why 1D space, we would get the same intensity to the end even in perfectly elastic collision(with no energy transfered into heat or any other form)

  • 2
    $\begingroup$ When you speak to the person next to you in a room, many other people in the same room can hear you - the sound wave is going in many directions. $\endgroup$
    – Jon Custer
    Jun 1 at 14:52
  • $\begingroup$ Yes but with my logic, how does your logic answer it ? Why does the first particle not give all of its energy to the next particle ? $\endgroup$
    – Giorgi
    Jun 1 at 14:54
  • $\begingroup$ Have you played pool? When you hit a ball with the cue ball, does the cue ball come to a stop while transferring all its energy to the target ball? $\endgroup$
    – Jon Custer
    Jun 1 at 14:55
  • $\begingroup$ Seems like cue ball only transfers some of its energy to the ball it hits(not all of it). This happens in sound waves but why ? Why not giving the whole energy from first particle to the next ? Maybe since sound wave travels in many direction, first particle tries to oscilate not only the next particle but sorrounding particles and if so, it just splits its energy into the sorrounding particles energy and each one gets portion of the original particle. Could this be correct ? @JonCuster $\endgroup$
    – Giorgi
    Jun 1 at 15:00
  • $\begingroup$ Not only does one particle usually only transfer some energy to another in a collision, but they then head off in different directions from the original trajectory. $\endgroup$
    – Jon Custer
    Jun 1 at 15:28

6 Answers 6


There are 2 things going on with sound waves.

  1. No such thing as a lossless collision. Every time an air particle hits another air particle due to a sound wave, part of that sound is converting into heat. This is an energy loss...

This loss is minor, but it does occur.

  1. Spherical expansion... If the world was a 1 dimensional space, then sound would travel much further. But it is not, it is a 3 dimensional space...

What that means is that if you are R meters away from a sound source that sound wave is now effecting a surface of $4 \pi R^2$ (Sound familar?) This is refered to as the Inverse Square Law... The furthere away you are the large of an area that same number of sound units is spread over. Inverse Square Law image credit to http://hyperphysics.phy-astr.gsu.edu/hbase/Acoustic/invsqs.html The overall strength of the signal hasn't decreased (except due to loss from heat).. but the intensity has because that energy is now spread over $4 \pi r^2$ units). This drop off in intensity applies to Gravity, Electricity, Light, and Radiation. Without this drop off in intensity the laws of thermodynamics would be violated as energy would have to be created out of nowhere..... And the universe would be a single giant black hole.

See Inverse Square Law for the source of the image, and some fun diagrams. + Links to more thorough explanations.

Edit: OP's edits should probably lead to a new question... but as the confusion was caused by me.

3) A perfectly elastic collision does not mean that both particles are reflected off of each other and then proceed to travel in opposite directions. All it means is that none of the kinetic energy is wasted as heat, sound--:)--, etc.... In-fact it is possible that all of the energy of the collider is transmitted to the object that it hits.

A good example of this in action is a newton's cradle Newton's cradle with perfectly elastic collisions

Which also serves as a good example of how/why vibrations travel in dense mediums (the balls of the cradle) vs non-dense mediums (the balls at the end). As the balls don't need to move to transfer for energy... Which explains why vibrations travel faster thru dense mediums water, bone, metal then it does thru a less dense medium (air).

  1. You got me... The pressure wave technically will go forever however once it has dispersed enough that it cannot displace an atom a single Planck it could be said that the sound no longer exists because it cannot be perceived (Admittedly this will happen long before then). So while my statement is not technically accurate. it is efectively accurate.

  2. That is correct, that is why sound falls off extremely fast... Though if you put it into a linear trench (the Mariana trench for example, or 2 cans and a piece of string) sound can travel a lot further.

Extra Credit (Also because I don't remember this that well). This has been treating waves as though they are only longitudinal when in fact many of them (Light, em-fields, ocean waves, ripples in a pond) are transverse waves.. To better understand these a good place to start is this article on Longitudinal Waves and this one on Transverse Waves yes they are on wikipedia... No I will not summarize the articles here. Wikipedia has been around since before stack exchange and will be around long after stackexchange is a smouldering wreckage.

  • $\begingroup$ Great answer. So with sound waves, not only elastic but inelastic collisions happen. But if you assume that only elastic collisions happen, it still would decrease over distance because in elastic, when collided, particle 1 did not give full energy to particle 2. Particle 2 would do the same(not giving the full energy to particle 3) and overaly energy decreases. Is this incorrect ? $\endgroup$
    – Giorgi
    Jun 1 at 16:49
  • $\begingroup$ @Chemistry This is incorrect. In a perfectly elastic collision, in a universe composed complete of our atmosphere sound could travel forever.... However, your ability to perceive the sound falls off at $4 \pi r^2$ due to the inverse square law. $\endgroup$
    – Questor
    Jun 1 at 18:48
  • $\begingroup$ Take a piece of string, tie two cans... Sound will "travel" (remain perceptible) much further along that 1 dimensional space (the string) then they can thru a 2 or 3 dimensional space because it no longer experiences a fall off from the inverse square law as the wave expands... In a 1d world, with perfectly elastic collisions, a wave can travel for ever and have the same intensity at the start as at the end. $\endgroup$
    – Questor
    Jun 1 at 18:49
  • $\begingroup$ In a 3d space with perfectly elastic collisions, sound does not "travel" infinitely because the intensity of the sound fades as the area of the pressure wave expands, until it reaches a point that it is imperceptible. $\endgroup$
    – Questor
    Jun 1 at 18:52
  • $\begingroup$ well, then I don't get what elastic collision is. en.wikipedia.org/wiki/Elastic_collision here it shows images that one particle colliding with another with an angle, none of them stay stationary after the collision but the 2nd particle doesn't gain full, same speed from the first one if second was stationary before the collision. if not, my logic must be correct. strange $\endgroup$
    – Giorgi
    Jun 1 at 19:31

For a spherical source, have a look at the Gauss law. With the radius from the source increasing, so does the area through which this energy passes ( 4πr² ). The integral of the energy over all of this surface has to be constant for each radius and so the same area in different radii has a decreasing amount of energy passing through it. This is a direct consequence of conservation of energy and the geometry of 3D space.

  • $\begingroup$ The Gauss law is about electricity and not sound. But your analysis nonetheless looks correct to me since the pressure has to be conserved through geometrically increasing surface along r². $\endgroup$
    – dan
    Jun 1 at 16:29
  • 1
    $\begingroup$ Thanks, that was lost in translation; what I meant is called Divergence Theorem in english and Gaußscher Integralsatz in german. $\endgroup$ Jun 2 at 17:14

The flaw in your logic is that one particle does not just "excite" just one other particles and transfer all the energy to that particle.
So if more than one particle is excited there you have the mechanism for the energy "spreading out" as the sound moves away from a source.

The mathematical way of treating this spreading out is to assume a point source and say that all the sound energy generated per second, $P$, by the source passes through a sphere of radius $R$ centered on the point source.
Thus the intensity (power per unit area) of the sound at a distance $R$ from the source is $\frac{P}{4\pi R^2}$ and as $R$ increases the intensity of the sound decreases.

You might then consider the production of a parallel beam of sound but that will still spread out due to diffraction.

Air is not a perfect fluid and as it has viscosity this will lead to a loss of energy as heat.

  • $\begingroup$ The problem is sound wave is elastic where masses of the objects are equal(air particle masses are equal). If so, look here - upload.wikimedia.org/wikipedia/commons/c/c6/… it shows that first particle gave all its speed to next particle hence gave it all. I don't get why this is not the case in sound wave $\endgroup$
    – Giorgi
    Jun 1 at 15:32
  • $\begingroup$ @Chemistry - the link you have given does not work. However I assume that the diagram/animation does not fully represent the true picture because the "extra" motion of the air molecules as a result of the sound wave is superimposed on the random thermal motion of the air molecules. $\endgroup$
    – Farcher
    Jun 1 at 15:42
  • $\begingroup$ Since sound waves are elastic collisions, and their particles' masses are equal, I thought this was happening - upload.wikimedia.org/wikipedia/commons/c/c6/… , but it turns out they don't collide like this and collide it with an angle in which case first particle doesn't give its whole kinetic energy to the next particle, but some portion of it. Is this logic correct ? $\endgroup$
    – Giorgi
    Jun 1 at 15:54
  • $\begingroup$ @Chemistry - Thanks for the link which does not represent the actual motion of air molecules which do not all have the same mass and do not only undergo head on collisions. As you have correctly written the energy transfer is partial and you need to consider the motion of a molecules and a group of molecules which surround it. This gif is misleading as it show dots oscillating without collision and all in all it is not possible to get a true visualisation of the actual motion of the air molecules. $\endgroup$
    – Farcher
    Jun 1 at 16:07
  • $\begingroup$ why don't they have the same mass ? $\endgroup$
    – Giorgi
    Jun 1 at 16:20

It's a combination of factors.

  1. The inverse square law. As the sound travels further from its source, it spreads over an increasingly large area. As the source only provided finite energy, this naturally means that the energy must become dissipated as the sound travels further.

  2. Sound is an interaction between particles. As such, with each interaction, some of the sound energy converted into heat energy instead. This is inevitable because sound propagation is highly chaotic as each particle interaction will result in particles moving in new directions dependent upon initial positions and momenta of both the striking and receiving particles.

Additionally, there is inherent background noise (if only due to thermal effects), which creates a small amount of interference and will ultimately make the sound indiscernible to even the most sensitive of apparatus after it fades to some point.


So to draw a bigger circle,

  1. the air molecules vibrate to produce sound.
  2. sound is a sort of mechanical wave (obviously) and travels in a medium (now air).
  3. molecules of air vibrate due to a source providing them energy to vibrate but this collision between molecules is not in reality a perfectly plastic collision when the particles collide some amount of energy is given to the next particle while the rest of the energy is released in the form of heat this continues and the endpoint has a less energetic sound received.
  4. when the molecules collide the enthalpy increases the free energy to be -ve has to release energy an exothermic collision so again energy released here is heat.
  5. The inverse square rule also holds true so the combination of these two things results into a sound waves energy to decrease

To prove my point we take the fact that sound wave travel fastest in solids and slowest in gas since solids have fewer ∆s relative to gases so the free energy for solids is more -ve comparatively. Take two ice cubes at the same temp and let them melt ice cube A melts normally but B melts with a source of sound (as heavy as you can) note the time taken

  • $\begingroup$ Your answer could be improved with additional supporting information. Please edit to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Jun 2 at 5:21
  • $\begingroup$ Welcome to Physics! Instead of posting several answers sequentially that build on each other, you should edit your original answer to include the additional information that you want to include. $\endgroup$ Jun 2 at 17:06

This is because of dampening of the sound wave which happens because

First our voice is not the only one, there are many Waves which keep on interfering and also the medium also contains many dampening agents


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