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):

equation

where V is room volume (m3), c is the speed of sound (m/s), S is wall area (m2), and images is the angle‐averaged wall absorption coefficient, or

(3.74)equation

where Si is ith wall area of absorption coefficient αi.

Graph depicts the measurement of reverberation time TR.
Figure 3.20 Measurement of reverberation time TR.
Graph depicts a few examples of recommended reverberation times.
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 images = 0.5 placed on it?

SOLUTION

We will assume that images = 0 on the other surfaces (that are made of hard painted concrete.)

equation

Notice that the Sabine reverberation time formula T60 = 0.16 V/S images still predicts a reverberation time as images → 1, which does not agree with the physical world. This is approximately the case of an anechoic room (see Figure 3.22). Some improved formulas have been devised by Eyring and Millington‐Sette that overcome this problem. Sabine’s formula is acceptable, provided images ≤ 0.5.

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. Thereforeequation
  2. The total sound absorption is now 25.8 + 35(0.45) = 41.55 sabins (m2). Thenequation
Schematic illustration of sound source in anechoic room.
Figure 3.22 Sound source in anechoic room.

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