# Shouldn't sound travel better through doors?

Sound travels much faster in solids than liquids and gases.

Then why do we hear a fainter sound from the other room if we close the door than open it?

As sound travels faster through solids...shouldn't we hear a much louder sound from the other room when the door is closed? (I know that loudness depends on amplitude but still...I don't get how a solid can make sound fainter)

• This is because a significant faction of the sound is reflected off the door. There needs to be impedance matching for there to be good transmission. Typically this concept is used in electronics but there is an acoustic analogue. See en.wikipedia.org/wiki/Impedance_matching#Acoustics – LasersMatter Sep 30 '16 at 17:15
• @LaserMatter i don't really know what impendance is...how can sound be absorbed? It doesn't really make sense to me...is the kinetic energy of the wave absorbed? – MartianCactus Sep 30 '16 at 17:56
• @Ruts i know but how can a solid decrease the sound wave's amplitude? – MartianCactus Sep 30 '16 at 17:56
• As Adam is explaining, it isn't that the door absorbs the majority of the sound it is that it is reflecting the sound. In general, when you have a wave (sound, electromagnetic,...) moving from one material or another there is a reflected and transmitted parts of the wave. Here is a video showing this youtube.com/watch?v=AjzBGC4tGTo . The wave the guy makes is like the sound in the room. The other half of the wave machine is like the door. If the impedance mismatch was greater there would be a larger reflection and less transmission. – LasersMatter Sep 30 '16 at 21:14

## 1 Answer

On the boundary of two substances with different propagation velocities, a wave traveling from A to B experiences both reflection and transmission. See the last animation on this site.

It can be shown that the reflection index (R) is the following $$R=\frac {Z_A-Z_B} {Z_A+Z_B}$$ where $Z=\rho v$ is the impedance of the materials, specifically $Z_\text {air}=420 \frac {\text {Pa}\cdot s} m$, $Z_\text {wood}=2.3\times 10^6 \frac {\text {Pa}\cdot s} m$, thus $R=-0.99963$. Moreover the sound exiting the door experiences the same degree of reflection.

• You should expand this answer to explain why, in the case of a door, more is reflected than transmitted. – Phiteros Sep 30 '16 at 17:39
• It is not a question of more being reflected than transmitted. The question is that part of it gets reflected, period. Never mind if it is 10% or 90% of the power that gets reflected, if some power gets reflected, then less power reaches the other side. Nevertheless, I hope this reply, or another, improves by adding some estimates of power loss for "common" doors – Rolazaro Azeveires Sep 30 '16 at 17:49
• Don't forget that the door has two surfaces that the sound has to successfully transmit through before it can be heard. – Sean E. Lake Sep 30 '16 at 18:06
• @James this is not true, if a ball bounces off a wall the transferred energy is less than when it sticks to the wall, my advice is to forget the metaphors and internalize the mathematical formalism, it may be a rougher route but if it really interests you and want to understand it then it is worth the time – Adam Sep 30 '16 at 20:23
• @James you are confusing energy and momentum. (But it took scientists a few hundred years to sort out the difference, so don't feel bad about making that mistake!) – alephzero Sep 30 '16 at 20:35