I have been asked to solve this exercise.

An environment of $V=1500$ $m^3$, it is considered a reverberated field acoustic pressure level of $70$ $dB$ when an acoustic source of $80$ $dB$ is activated. Compute the total absorption (in $m^2$) in order to obtain a reverberation time of $4$ seconds.

Let me show you my idea: I have to use Sabine's formula $$ T_{60} = \beta \frac{V}{A} \text{ with } \beta=0.16$$

The surface $A$ should be computed as

$$A = \sum_i a_i S_i $$

where $a_i$ is the absorption coefficient of the $i$-th surface $S_i$.

Can someone help me to find the missing surfaces and coefficients because I can't understand in which way I must find them?

Thanks in advance.

up vote 2 down vote accepted

Sabin formula for Reverberation Time:

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You know the target RT and the volume of the room. What you are looking for is the total absorption A, S is just a surface area. Technically, the unit of A is called Sabine, not squared meters. So rearrange the equation and solve it:

enter image description here

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