Question

If the sound level in a room is increased from 50 dB to 60 dB, by what factor is the pressure amplitude increased?

Answer 1

Given:
Initial sound level ${\beta }_1$ = 50 dB
Final sound level ${\beta }_2$ = 60 dB
Constant reference intensity $I_0$ = 10^$-$12 W/m²
We can find initial intensity I₁ using:
${\beta }_1=10{\log }_{10}\left(\frac{I_1}{I_0}\right)$.
$\Rightarrow 50=10{\log }_{10}\left(\frac{I_1}{{10}^{-12}}\right)$
On solving, we get:
$I_1$ = 10^$-$7 W/m^2_.
Similarly,
${\beta }_2=10{\log }_{10}\left(\frac{I_2}{I_0}\right)$.
On substituting the values and solving, we get:
$I_2={10}^{-6}\mathrm{W}/{\mathrm{m}}^2$
As the intensity is proportional to the square of pressure amplitude (p),
we have:
$\frac{I_2}{I_1}={\left(\frac{p_2}{p_1}\right)}^2=\left(\frac{{10}^{-6}}{{10}^{-7}}\right)=10$
$$\begin{gathered} \therefore {\left(\frac{p_2}{p_1}\right)}^2=10 \\ \Rightarrow \frac{p_2}{p_1}=\sqrt{10} \end{gathered}$$
Hence, the pressure amplitude is increased by $\sqrt{10}$ factor.