I understand echoes are sound reflections. But if foam absorbs sound, and if it reflects off the wall, then that means the waves travel through the foam, and if sound waves get transmitted from a source through the foam and the wall into the next room, doesn't that mean in both cases the sound will have gotten absorbed since it traveled through the foam?

However, this does not appear to be the case, as many sources say the foam only quiets reflected sound, and not sound that will get transmitted through the foam and the wall into the next room. I wonder, if the foam absorbs sound waves that travel through it, why does it only absorb reflected waves, but has zero effect on quieting waves that will transmit through the foam and wall into the next room? If I put thick foam on my walls, wouldn't that make the sound that travels into the next room quieter than if I hadn't put the foam on the walls? Please help, thank you!

  • 1
    $\begingroup$ Small thing to also note is absorbing material is twice as effective for waves reflected from the wall, since the wave travels through the absorbing layer twice (once on the way in, second time after being reflected back from the wall) $\endgroup$
    – antimony
    Commented Apr 5, 2023 at 1:51

3 Answers 3


Found this wonderful explanation online here:


Here is a simple analogy that helps people understand the physics of sound and how it works. Imagine you are building an aquarium to hold water. Would you use glass panels or sponges for the walls of the tank? This is clearly a ridiculous question, but it paints a picture of simple physics that applies here. Sound acts very similar to water when you are trying to control it. If you used sponges as the walls, they would absorb the water but quickly let all of it seep through to the other side. Glass and good seals block the water and keep it in place. Acoustical materials made from soft, squishy things like sponges are going to absorb. Dense, heavy, air-tight materials will block.


This is quite often a misconception among people who are not very familiar with the physical mechanisms of sound absorption and sound insulation.

1. Absorption, Reflections & Insulation

Before going on to try to provide some insight related to your question (which hopefully constitutes an answer to it) I will do my best to provide a high-level explanation of both aforementioned terms.

1.1. Reflections

In a simple case of incidence of a plain wave on an infinite interface (please note that this is just a mathematical formulation which seems to work quite well for the majority of the usual cases) a portion of the incident energy will be transmitted to the "other side" while the rest will be reflected travelling in the opposite way but on the same axis. A simple depiction of that is shown in the following image (source: Acoustics - An Introduction by Heinrich Kuttruff).

Reflection and transmission

In the image, it is assumed that on the right of the interface, the medium is different than that on the left of the interface.

The amount of energy transmitted and consequently that being reflected, which is the former's complement to the sum of unity depends on the impedances of the two media. The greater the mismatch of the impedances, the greater the amount of energy reflected. If the incidence energy is denoted $E_{i}$ and is normalised to unity, $E_{i} = 1$, then the following equations hold

$$ E_{i} = E_{t} + E_{r} = 1 \tag{1} \label{1} $$

where $E_{t}$ is the transmitted energy and $E_{r}$ the reflected energy. This is an expression of the conservation of energy. The reflection coefficient $R$ denotes the fraction of the pressure amplitude of the incident wave after being reflected. It can be used as shown in equation \ref{2}. Its (magnitude) value is constrained in the interval $\left[ 0, 1 \right]$ by definition where the lower bound is achieved for a perfectly "absorbing" (bare with me about this) interface where the upper for a perfectly reflecting one.

$$ p_{r} \left( x, y, t \right) = \hat{p} \left| R \right| e^{ -j k \left[ -x \cos \left( \theta \right) + y \cos \left( \theta \right) \right] + j \phi } ~ e^{j \omega t} \tag{2} \label{2} $$

where $p_{r}$ is the reflected pressure wave and its dependence on the $x$ and $y$ spatial coordinates as well as the temporal $t$ dimension are explicitly declared and $j$ denotes the imaginary unit for which $j^{2} = -1$ is true. $\phi$ denotes the phase shift imposed on the reflected wave which depends also on the impedances (these are complex quantities). The complex reflection coefficient $R$ can be expressed by combining the magnitude and phase shift like

$$ R = \left| R \right| e^{j \phi} \tag{3} \label{3} $$

Furthermore, for the reflection coefficient, the following equation is true

$$ R = \frac{Z \cos \left( \theta \right) - Z_{0}}{Z \cos \left( \theta \right) + Z_{0}} \tag{4} \label{4} $$

where $Z$ denotes the impedance and with $Z_{0}$ is denoted the impedance of the first (left-hand side) medium.

Thus we can see that transmission is mainly dictated by the impedance of the two media. To conclude our discussion on reflection and transmission I show an image depicting a closer-to-reality case (source: Acoustic Absorbers and Diffusers: Theory, Design and Application by Trevor Cox and Peter D' Antonio). If you exchange Layer 3 with the same medium as in Layer 1 you have a usual case where an absorptive panel (or other material) is placed close to the wall but not touching it (as we'll discuss below, this provides better "absorption" than when attached to the wall).

Reflection and transmission in multiple media layers

1.2. Absorption

Now, this is a term that has caused quite some confusion among audio enthusiasts. From the incident wave point-of-view as we showed above everything not reflected is considered as been absorbed! In this treatment absorption, one could easily derive the absorption coefficient from the reflection coefficient. One subtlety is that the absorption coefficient denotes the portion of the energy that is not reflected. Since the reflection coefficient denotes the fraction of the pressure amplitude that is reflected we have to square it to relate it to energy resulting in the equation \ref{5} which provides the relation between the two coefficients

$$ \alpha = 1 - \left| R \right|^{2} \tag{5} \label{5} $$

For us, as a person who wants to "get rid" of the acoustic energy completely (this is to transform it to some other kind of energy since we cannot violate energy conservation) the aforementioned approach is not adequate. For example, in the case of the second image, some of the energy passing through the supposed absorbing panel will hit the wall and return to the room (possibly by passing through the absorbing panel once more).

As you've already stated, you do know that these constitute the reflections heard in a room and those that ultimately result in the room's "signature" reverb. So, what do absorption panels (or other setups available) do? They do what we intend to do; convert acoustic energy to thermal energy! This is the main, but not the only (I'm skipping the other mechanisms here for the sake of simplicity) mechanism that results in what is "officially" termed absorption.

Omitting many details about the actual phenomena taking place to achieve this conversion I'll just state that it depends on the particle velocity (not the speed of sound). That is, the higher the velocity, the higher the conversion from acoustical to thermal energy. Noting that the particle velocity is rather low close to the boundaries and maximum about $\frac{\lambda}{4}$ away from it, with $\lambda$ denoting the wavelength we conclude that those absorptive materials would be more effective when placed at some distance from the wall, or if they are of considerable depth (comparable to the wavelength). Of course, it's not very practical to place panels in the middle of a room so they are usually attached to the walls. An exception to this is the "hanging ceilings" (or clouds I believe they are sometimes called) where you can see some absorptive materials, usually polished to look good, being hung from the ceiling.

Below is a depiction of an absorptive material layer attached to a rigid (very reflective) wall (source: Acoustics - An Introduction by Heinrich Kuttruff).

enter image description here

1.3. Insulation

Not much to say here. This is a very complex topic combining impedance mismatch to reduce transmission of airborne sound as well as vibrations to tackle the issue of structure-borne sound and vibrating structures which re-radiate sound.

The main idea here is to provide as much mass as possible to make heavy constructions that are very hard to move/vibrate. Since this is not always possible one could use the reflection principles described above to direct sound energy away from the adjacent rooms while at the same time trying to convert as much energy as possible to thermal. The result usually is "sandwich" multi-layer setups that induce reflections at each layer interface (see the last image above) with some of the layers being absorptive to allow for conversion losses. Combining this strategy with as much vibration isolation (rubber and springs are the norm here) as possible good results can be achieved.

2. Putting it all together

Now we do have enough knowledge to describe what is going on when you place an absorptive panel, either close to or on the wall.

Sound reaches the panel and since the impedance of the panel is neither $0$ nor infinite - $\infty$ - part of the incident energy is reflected with some phase shift. The rest of the energy denoted $E_{t}$ and equal $1 - E_{r}$ propagates into the panel. Part of this energy is converted to thermal energy and the rest reaches the interface of the panel and the next material. Let's assume that this is the wall and we have a case similar to that of the last picture above. Part of the remaining energy will be reflected into the panel due to the impedance mismatch between the panel and the wall and the rest will go into the wall (in there some of it may reflect at this interface and be re-radiated into the room but we'll assume this won't happen for simplicity).

Now, the portion of the energy reflected into the panel will go through some additional losses (conversion to thermal energy) and will reach the first interface between the panel and air but from the side of the panel this time. The same process will take place with a part of the energy being re-radiated into the room and part of it going back into the panel. This is depicted in the image mentioned before except for the transmission through the wall (it is assumed completely reflective).

Two important things to mention here.

  1. The absorptive material/panel allows sound to pass through. Otherwise, it wouldn't be able to incur losses (conversion to thermal energy).
  2. The absorptive material does not convert all the acoustic energy to thermal. This leads to some energy reaching the wall, thus allowing for sound to be transmitted to adjacent rooms through it, as well as some of the remaining energy to return to the room. The latter constitutes a reflection.

Related to 1: Think of the situation where you speak towards an absorptive panel with someone on the other side of it. Do you think they'll hear nothing? Of course not; they will hear your voice but with some frequencies (as the attenuation is frequency dependent) attenuated.

An additional remark related to 2: Sound absorption as approached from the "Room Acoustics" viewpoint is a statistical treatment of sound. This is a direct consequence of the fact that not all energy is converted to thermal inside an absorptive panel and some energy "escapes" back into the room.

Thus, one cannot completely "kill" a reflection with the placement of a single panel at the position of incidence. The use of absorption in control rooms of recording studios at the positions where the first reflections reach the sweet spot is not invalidated here but it is instructive to say that what is achieved is the reduction of reflected energy reaching the ears and not complete absorption of the reflected energy.

I hope this clarifies things related to absorption, insulation, transmission and reflections.

  • $\begingroup$ What do you mean."Now, the portion of the energy reflected into the panel will go through some additional losses (conversion to thermal energy) and will reach the first interface between panel and air but from the side of the panel this time." What do you mean by the side of the panel? $\endgroup$
    – John Smith
    Commented Apr 6, 2023 at 22:49
  • $\begingroup$ Also from my understanding of reading this, using a material with a lot of mass is best to block sound from traveling through a wall into another room, but sound-absorbing foam would quiet the sound that does get transmitted through the wall since some of its energy gets absorbed? $\endgroup$
    – John Smith
    Commented Apr 6, 2023 at 22:59
  • $\begingroup$ @JohnSmith, what I mean is that the reflected wave will be travelling inside the panel towards the interface of the panel with the air (since it is reflected off of the wall at the back side of the panel). Thus, the interface of panel-air will be reached again (this is the first interface found by the incident wave) but now the incident wave (that reflected off the wall) will be on the side of the panel, not the air. In the second image above, this is the reflected wave (arrow) in the second layer moving towards the first interface. If this is not clear please let me know to try and fix it. $\endgroup$
    – ZaellixA
    Commented Apr 7, 2023 at 7:21
  • $\begingroup$ @JohnSmith, regarding the second comment: When you have a lot of mass (what is also very important is the surface density of the material) you will manage to reflect the sound down to very low frequencies and induce some attenuation (conversion to thermal energy) in the process (inside the material). On the other hand, "foamy" absorbers are more effective at transforming the acoustic energy to thermal inside of them, so sound has to go through them to manage to reduce its energy. As you may have already understood, the latter is not a very effective sound insulator ;). $\endgroup$
    – ZaellixA
    Commented Apr 7, 2023 at 7:24
  • $\begingroup$ Please let me know if these are not very intuitive and I'll do my best to make clarifications. $\endgroup$
    – ZaellixA
    Commented Apr 7, 2023 at 7:24

First of all, neither of these are absolutely true. The material may do a good job limiting reflections, but there will be some echoes.

The sound waves have energy. To get rid of all the sound, you have to dissipate all the energy. This can be difficult to do, especially in a small material.

All the energy that isn't absorbed will go somewhere else, either reflected or transmitted. A regular wall has a very strong boundary between the air and the wall. The strong impedance mismatch between the two makes a lot of the energy reflect from the wall. It is possible to have material that can ease that transition. That allows more energy to flow through the wall than would happen otherwise.

If I put thick foam on my walls, wouldn't that make the sound that travels into the next room quieter than if I hadn't put the foam on the walls?

It really depends on the material and the design. Some foam may do a good job of absorbing the energy. Such foam could reduce the total energy transmitted through the wall. I'm pretty sure if you put up a huge slab of foam, say 10cm thick, you're going to attenuate a lot of sound.

But you could design a thin piece that did not absorb much, while still reducing reflected energy. It is quite possible for such material to make it louder in the next room.

See also: https://physics.stackexchange.com/a/538496/55662 https://physics.stackexchange.com/a/368200/55662

  • $\begingroup$ How could you design a piece of foam that does not absorb sound much, but greatly reduces reflected energy? $\endgroup$
    – John Smith
    Commented Apr 4, 2023 at 23:28
  • $\begingroup$ Simple, the thinner the foam wall is, the less it will absorb. So if you make a wall consisting of 0 mm of foam (and nothing else) it will absorb nothing and also reflect nothing. In effect - the perfect non-reflacting wall is the absence of wall. $\endgroup$
    – Penguino
    Commented Apr 4, 2023 at 23:30
  • $\begingroup$ It needs to reduce the impedance mismatch between air and the wall. Something with a speed of sound midway between the two, but thin to reduce absorption would do it. I don't know what material would do the job. $\endgroup$
    – BowlOfRed
    Commented Apr 4, 2023 at 23:31
  • $\begingroup$ @Penguino, I think the assumption is that whatever material you have is added to the wall. You don't get to remove the wall entirely. :-) $\endgroup$
    – BowlOfRed
    Commented Apr 4, 2023 at 23:32
  • 1
    $\begingroup$ Exactly. This is what the bones in your ear are doing. They are an impedance ladder between the eardrum and the cochlea. $\endgroup$
    – BowlOfRed
    Commented Apr 5, 2023 at 0:21

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