Sound absorption

Room acoustics describes how sound behaves in a space. That means the listener and the sound source are in the same room. If the room has nearly no sound absorbing surfaces (wall, roof and floor), the sound will bounce between the surfaces and it takes a long time before the sound dies out. The listener in this kind of room will then have a problem registering the speaker because he hears both the direct sound and repeated reflected sound waves.

If the surfaces instead are covered with sound absorbing material, the reflected sound will decrease much quicker and the listener will only hear the direct sound. Also, the general sound level in the room will decrease.

Room acoustics sound absorption

A material's sound absorbing properties are expressed by the sound absorption coefficient, α, (alpha), as a function of the frequency. Alpha (α) ranges from 0 to 1.00 (from total reflection to total absorption).

The sound absorbers can be divided into three main categories

  • Porous absorbents 
  • Resonance absorbents
  • Single absorbers 


Pours absorbers

A good example of a porous sound absorbent is stone wool. When the sound wave penetrates the mineral wool, the sound energy through friction is changed into heat.

The material thickness has a great impact on the material's sound absorbing qualities. High frequencies (above 500 Hz) are easier to handle with 30–50 mm stone wool thicknesses. More challenging are the sounds in frequencies below 500 Hz. Here we need thicker stone wool slabs to create better sound absorption. Material thickness can also be compensated for with air space behind an acoustic ceiling or wall panel to improve low frequency performance.

For these sound absorbents, it is very important not to put an airtight layer on the surface, such as a vapour barrier or paint as this will reduce the sound absorbing properties significantly. You can see the effect of the airtight layer in the picture below (the dotted line):

Sound absorption curves for stone wool insulation

Here you can find some practical absorption coefficients for certain materials:

Octave band (Hz) 125 250 500 1000 2000 4000
Concrete

0.02

0.02 

0.02 

0.02 

0.03 

0.04 

Gypsum board on stud 

0.2

0.15

0.1

0.08

0.05

0.05

Windows

0.35

0.25

0.18

0.12

0.07

0.04

50-mm mineral wool slab*

0.2

0.65

1.0

1.0

1.0

1.0

100-mm mineral wool slab*

0.45

0.9

1.0

1.0

1.0

1.0

* with solid backing

Resonance absorbers

Resonant absorbers consist of a mechanical or acoustical oscillation system. One case of this is membrane absorbers – for example, a solid plate with a tight air space behind. The absorption reaches its maximum at the resonance frequency. If the cavity is filled with a pours material such as stone wool, the sound absorption over the frequency range is broadened.

Single absorbers

In this category, you will find objects such as tables, chairs, people, etc. The absorption for these is normally given as m2 in Sabine’s formula per object.

The reverberation time of a room characterises how long acoustic energy remains in it. It is usually defined as the time for the acoustic intensity to decrease by a factor of one million (60 dB).

Since a reasonably loud clap is about 100 dB (SPL) and a whisper is about 40 dB, you can easily estimate the reverberation time for a room by clapping and listening to how long you can still hear some remaining sound from the clap. This assumes that the room is not particularly unusual in its dimensions and that it is reasonably quiet.

Single absorbers

In a small room or hall (volume <1000 m3) where the sound field is diffused and the average absorption is less than 0.3, an empirical formula called the Sabine formula can be used to calculate the reverberation time:

 

RT = 0.16 x V / A


T = reverberation time, s
V = volume of the room, m3
A = (Σ surface area (S) x α) = absorption area of the room, m2

Absorption area of the room A is the sum of each surface area S multiplied by its absorption coefficient α.
For example, if the desired reverberation time in a classroom is 0.8 seconds and the dimensions of the classroom are 6 x 10 x 3 m and the intension is to use 45 mof absorbing ceiling material, what then is the required absorption coefficient for the product? 

Answer: A = 0.16 x V/T = 0.16 x 180/0.8 = 36 m2 x α = 36/45 = 0.8


The optimum reverberation time for a space depends on the size, materials and type of room. Every object placed within the enclosure can also affect this reverberation time, including people and their belongings.

Rooms for speech require a shorter reverberation time than for music. A longer reverberation time can make it difficult to understand speech. If, on the other hand, the reverberation time is too short, tonal balance and loudness may suffer.

Noise reduction in large industrial premises

In industrial halls with a volume exceeding approximately 1000 m3, the height is normally much less than both the length and the width of the hall. In this case, the height and the furnishing density have a considerable influence on the sound field. In such a hall, the sound field is generally not diffuse and it is therefore not useful to calculate the reverberation time by using the Sabine formula.