Question

A hall of volume 5500m³ is found to have a reverberation time of 2.3s. The sound absorbing surface of the hall has an area of 750m². Calculate the average absorption coefficient.

Answer 1

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Answer 2

Given,

A hall with,

Volume,$V=5500m^{3}$

Reverberation time, $T=2.3s$

Sound absorbing surface area of hall, $A=750m^{2}$

To find,

The average absorption coefficient.

Solution,

We know that,

Reverberation time,

$T=\frac{0.158V}{A\alpha }$

Where,

$V=volume\\A=area\\\alpha =absorption constant$

So,

Putting the respective values we get,

$2.3s=\frac{0.158(5500m^{3} )}{750m^{2}.\alpha }$

⇒$2.3$×$750=\frac{869}{\alpha }$

⇒$1725=\frac{869}{\alpha }$

⇒$\alpha =\frac{869}{1725}$

⇒$\alpha =0.504$

Therefore, the average absorption constant is $\alpha =0.504$