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

${\beta }_1=50dB,{\beta }_2=60dB,$

$I_1={10}^{-7}W/m^2,I_2={10}^{-6}W/m^2$

[becuase $\beta =10lo{g}_{10}\left(\frac{I}{I_0}\right)$, where $I_0={10}^{-12}W/m^2]$

Again, $\frac{I_2}{I_1}={\left(\frac{p_2}{p_1}\right)}^2=\left(\frac{{10}^{-6}}{{10}^{-7}}\right)=10$

(where p =pressure amplitude)

$\therefore {\left(\frac{p_2}{p_1}\right)}^2=10$

$\Rightarrow \frac{P_2}{P_1}=\sqrt{10}$