3.18 Sound pressure level in an interior sound field.
3.14.2 Sound Absorption
The sound absorption coefficient α of sound‐absorbing materials (curtains, drapes, carpets, clothes, fiberglass, acoustical foams, etc.), is defined as
(3.73)
Note that α also depends on the angle of incidence. The absorption coefficient of materials depends on frequency as well. Thicker materials absorb more sound energy (particularly important at low frequency). See Figure 3.19.
Figure 3.19 Sound absorption coefficient α of typical absorbing materials as a function of frequency.
If all the sound energy is absorbed, α = 1 (none reflected). If no sound energy is absorbed, α = 0:
If α = 1, the sound absorption is perfect (e.g. an open window).
The behavior of sound‐absorbing materials is described in more detail in Chapter 9 of this book.
3.14.3 Reverberation Time
In a reverberant space, the reverberation time TR is normally defined to be the time for the sound pressure level to drop by 60 dB when the sound source is cut off (see Figure 3.20). Different reverberation times are desired for different types of spaces (see Figure 3.21). The Sabine formula is often used, TR = T60 (for 60 dB):
where V is room volume (m3), c is the speed of sound (m/s), S is wall area (m2), and
(3.74)
where Si is ith wall area of absorption coefficient αi.
Figure 3.20 Measurement of reverberation time TR.
Figure 3.21 Examples of recommended reverberation times.
In practice, when the reverberation time is measured (see Figure 3.20), it is normal practice to ignore the first 5‐dB drop in sound pressure level and find the time between the 5‐dB and 35‐dB drops and multiply this time by 2 to obtain the reverberation time TR.
Example 3.13
A room has dimensions 5 × 6 × 10 m3. What is the reverberation time T60 if the floor (6 × 10 m) has absorbing material
Solution
We will assume that
Notice that the Sabine reverberation time formula T60 = 0.16 V/S
Example 3.14
A classroom has dimensions 4 × 6 × 10 m3 and a reverberation time of 1.5 seconds. (a) Determine the total sound absorption of the classroom; (b) if 35 students are in the classroom, and each is equivalent to 0.45 sabins (m2) sound absorption, what is the new reverberation time of the classroom?
Solution
1 the volume of the classroom is V = 240 m3. Therefore
2 The total sound absorption is now 25.8 + 35(0.45) = 41.55 sabins (m2). Then
Figure 3.22 Sound source in anechoic room.
3.15 Room Equation
If we have a diffuse sound field (the same sound energy at any point in the room) and the field is also reverberant (the sound waves may come from any direction, with equal probability), then the sound intensity striking the wall of the room is found by integrating the plane wave intensity over all angles θ, 0 < θ < 90°. This involves a weighting of each wave by cos θ, and the average intensity for the wall in a reverberant field becomes
(3.75)
Note
