# Physically, why are high sound frequencies more easily absorbed than low sound frequencies?

If someone is playing music in a room, outside the room you generally hear a more pronounced reduction in high frequencies, compared to low.

I know that sound is most easily absorbed by a material when the absorbing material is $\frac{wavelength}{4}$ distance from a boundary.

Could anyone explain, in terms of boundary conditions (perhaps?) why a low frequency may be more likely to reflect from and travel through a material, than a high frequency?

Does the frequency at all effect any boundary conditions we apply to sound? I cant find much on-line but it makes sense in my head that since absorption is frequency Dependant, then the boundary conditions should reflect this.

Thanks

Actually, the correct answer involves the concept of acoustic impedance ($Z$), the idea that different media (materials) vary in their deformation due to acoustic pressure forces.

$$Z = \frac{acoustic \ \ pressure \ \ force}{acoustic \ \ particle \ \ velocity}$$

Acoustic waves move matter, so the material form of the system is very important when discussing impedance (or any acoustical topic, really.) So is the orientation of the object with respect to the pressure wave. Both affect the way acoustic energy travels.

This can make the actual analytical determination of frequency dependance quite complex.

Still, the music room example is relatively simple. Let's assume the wave coming from the instrument hits the (non-porous) concrete wall perpendicularly: And you're standing on the right-hand side.

Physical properties of the wall - generally it's mass, $m$, and stiffness, $\kappa$, determine which frequencies are transmitted (to you) and which frequencies are reflected back (to the music room.)

In the following section I'm framing the idea in the same way Matthew Schwartz does here, in §5 "Complex impedance" . The document contains more complete derivations, too.

The total acoustic impedance is given as the sum of a resistance term, a mass term and a stiffness term:

$$Z_{total} = R + Z_m + Z_{\kappa}$$

Generally (i.e., not specific to the form of the wall), we can write an equation for acoustic impedance in terms of frequency, $f$:

$$Z_{total}(f) = R + i \ 2 \pi f m \ \ - i \frac{\kappa}{2 \pi f}$$

We see that for high-frequency components of the incident wave the mass of the wall ($m$) causes higher values of impedance at these frequencies. The same is true for stiffness ($\kappa$).

What's important is actually the difference in $m$ and $\kappa$ between the two media. We can write an acoustic transmission coefficient as a function of $f$ as well:

$$T(f) = \frac{2 \ Z_{total, \ air}(f)}{Z_{total, \ wall}(f) + Z_{total, \ air}(f)}$$

(Note that when $Z_{wall} = Z_{air}$, $T = 1$ and all incident energy at that frequency transmitted. Much more likely is the case where $Z_{air} \ll Z_{wall}$, where we expect $T$ to become quite small.)

In this example, concrete is much more massive ($+ \Delta m$) and stiff ($+ \Delta \kappa$) than air and so it lets less high frequency energy through due to both factors. However, you can imagine a situation where two adjacent media have opposing changes in each property, for example ice and water. Ice is less massive (dense) than water ($- \Delta m$), but also much stiffer ($+ \Delta \kappa$). In such a case we might expect a central frequency to be the most effectively transmitted.

• Can you please explain to me how the higher frequencies experience higher impedance than lower frequencies when they hit the wall? I dont understand equations,could you show me in simple way how can I calculate effective wall impedance for given frequency,mass and stiffness? Mar 12, 2018 at 12:54
• @wavscientist I understand that equations aren't really your preferred way of thinking about physical systems. But consider again the one I've included above: $Z_{total}(f) = \ i \ 2 \pi f m \ \ + \ -i \ \frac{\kappa}{2 \pi f}$ Hopefully you can see from it that increasing the mass $m$ of the wall, or stiffness $\kappa$ will increase the impedance. Increasing frequency also increases the impedance (in two ways, by increasing the mass term, and by reducing the stiffness term.) Mar 13, 2018 at 5:04
• But how much does it increase,that is what I would like to know.I will give you three scenarios and you tell me how much does impedance increase please.... 1. Frequency doubles,mass and stiffness remains same..... 2. Stiffness doubles,frequencu and mass remains same.... 3. Mass doubles,frequency and stiffness remains same.... Would doubling of either frequency,mass or stiffness double effective impedance,or does one of these quadruple impedance? Mar 13, 2018 at 14:45
• @wavscientist would you consider asking this as a new question? I'd be happy to make some plots if those are more helpful to you. The short answer is that it really depends on what the initial values are - the way the equation works you can't reliably expect the same scaling principle at all different frequencies. Mar 13, 2018 at 16:23
• I made new question physics.stackexchange.com/questions/392012/… Mar 13, 2018 at 21:48

The amount of absorption depends on the frequency of the sound. A high frequency sound has many cycles in a second, and the particles in the medium are therefore vibrating very rapidly. Just as when you rub your hands together very rapidly, this produces more heat than if you rub your hands together slowly. Since the molecules get their energy to vibrate from the sound wave, the sound wave will run out of energy sooner when it is a high frequency sound. This means that, under the same conditions, a high frequency sound won't travel as far as a low frequency sound.