From our discussion so far you can appreciate that the reflection qualities of a room will be a key determining factor in the reverb time. Consider the extreme example of perfectly reflecting walls (not really possible in practice!). Any sound emitted in such an enclosure would travel around forever adding to the background noise of the room. It would get noisy in such a room fast! The converse case is of a room with perfectly absorbing walls, i.e. no reflections. Again it is not possible to make a perfect example of such an enclosure in practice although one can come pretty close. Rooms with very absorbing walls mimic free field because there are no reflections. Such rooms are important for audio research purposes and are calledanechoicchambers.

Real enclosures fall somewhere between the extremes mentioned in the last paragraph. The material that lines the walls of the enclosure or is contained within the room is vital to determining the how long sounds will bounce around. To quantify how different materials respond to acoustic waves we define a quantity calledabsorption coefficient designated by the symbol a. If there is an acoustic intensity I incident on a surface and the amount of energy absorbed or transmitted on reflection from that surface is given by I_athen the absorption coefficient is defined by

a = l_a / l

Note that ifallthe incident sound intensity is absorbed (or transmitted) then a = 1. Similarly if all is the intensity is reflected then a = 0. These values are clearly the extremes we mentioned above and for real materials the value of a will lie somewhere between 0 and 1. Below is a table of experimentally determined values of a for a variety of different materials.

	125 Hz	250 Hz	500 Hz	1000 Hz	2000 Hz	4000 Hz
Concrete block (unpainted)	0.36	0.44	0.31	0.29	0.39	0.25
Window glass	0.35	0.25	0.18	0.12	0.07	0.04
Heavy Drapes	0.14	0.35	0.55	0.72	0.7	0.65
Terrazzo floor	0.01	0.01	0.02	0.02	0.02	0.02
Suspended acoustical tile	0.76	0.93	0.83	0.99	0.99	0.94

 

There is a more extensive table of absorption coefficient values in your text book on page 466. I have tried to give you a cross section of materials with fairly different acoustic properties.

From the table of values we can learn a couple of lessons. First, the value of a depends upon frequency. For example, in the case of window glass we see that a is larger at low frequencies than it is at high frequencies. This fact means that glass reflects high frequencies better than it reflects low frequencies. The second lesson gleaned is a fairly intuitive observation that hard surface such as Terrazzo are highly reflective absorbing little energy with each reflection whereas heavy drapes tend to absorb heavily. Suspended acoustical tile is very absorbing and is specifically designed to eat up a lot of acoustic energy and thus to reduce the build up of background noise in heavily populated places like offices.

Self Test Questions 1

1. What is the absorption coefficient a for an open window? [Hint: all of the sound is transmitted though an open window]

2. What are the units of absorption coefficient?

Calculating the Absorption of a Room

The absorption coefficient is only the first step in calculating the reverb time of a room. We know that a room may consist of many different materials on the walls, floor, and ceiling. How do we account for all these different acoustic materials? We define a quantity called theAbsorptionthat sums all of the surface areas of a material in a room multiplied by the appropriate absorption coefficient. Absorption is indicated by the letter A and mathematically defined by

A = S₁xa₁ + S₂xs₂ + xa₃ + . . .

where S₁is the surface area (in m²) occupied by the material with absorption coefficient a₁, and so on. The sum of terms goes on until all of the surface areas in the room are accounted for. The units of A are m²because we are adding surface areas with units m²multiplied by absorption coefficients a which are ratios with no units of their own.

An example, and a self test exercise are the best ways to understand how to calculate A.

Example

Consider a room with floor (and ceiling) of dimensions 5 m x 8 m and 3 m high walls. If the walls are painted concrete block, the floor terrazzo and the ceiling suspended acoustic tile, what is A at 500 Hz? The calculation is as follows

A = floor surface area x 0.02 + ceiling surface area x 0.83 + wall surface area x 0.31

have taken the a values from the table above. Now some simple geometry to get the surface areas and we have

A = (5 x 8) x 0.02 + (5 x 8) x 0.83 + (3 x 5 x 2 + 3 x 8 x 2) x 0.31 = 58.18 m²

One interpretation of the A value is to pretend that the walls are perfectly reflecting and that the A value represents the size of the open window cut in the enclosure to allow the sound to leak out. In this case the total internal surface area of the room is 158 m²and the opening A = 58.18 m².

Self Test Questions 2

Try implementing what we've learned by finding A for the following two enclosures.

1. A cubic room 7 m x 7 m x 7 m made out of unpainted concrete block at 500 Hz. (Gee sounds a bit like my office!)

2. Calculate A for the room in my example above but at 2000 Hz rather than 500 Hz. Is this room more or less damping than that calculated in the first question?

3. Calculate A at 500 Hz for a room with floor area 12 m x 10 m and 5 m high ceilings. The floor is carpeted (a₅₀₀= 0.14) the ceiling is plaster (a₅₀₀=0.06) and the walls are drywall (a₅₀₀= 0.05).

4. How would A change for the room in the previous question if suspended acoustic ceiling tile was installed?

5. What is A for a room with perfectly absorbing walls? [Hint: Ask yourself what is the absorption coefficient, a, for perfectly absorbing walls].

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