The time of reverberation of a room $\mathrm{A}$ is one second. What will be the time (in seconds) of…
The time of reverberation of a room $\mathrm{A}$ is one second. What will be the time (in seconds) of reverberation of a room, having all the dimensions double of those of room $\mathrm{A}$ ?
1
2
4
$\frac{1}{2}$
Solution
Reverberation Time
$T=\frac{0 \cdot 61 V}{A S}$
Where $V=$ Volume of room in cubic metre
$A=$ Average coefficient of the room
$S=$ Total surface area of room
Then,
$\begin{aligned}
& T \propto \frac{V}{S} \\
& \text { or } \frac{T_1}{T_2}=\left(\frac{V_1}{V_2}\right)\left(\frac{S_2}{S_1}\right)=\left(\frac{V}{8 V}\right)\left(\frac{45}{5}\right) \\
& =\frac{1}{2} \\
& \therefore T_2=2 T_1=2 \times 1=2 \mathrm{sec} . \\
& \left(\because T_1=1 \mathrm{sec}\right) \\
&
\end{aligned}$
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