Question

For a person with normal hearing, the faintest sound that can be heard at a frequency of 400 Hz has a pressure amplitude of about $6.0\times {10}^{-10}$ Pa. Calculate the corresponding intensity and sound intensity level at ${20}^{\circ }C$. (Assume $\nu =330m/s$ and ${\rho }_{air}=1.29kg/m^3$)

Answer 1

The sound pressure $p$ and sound intensity $I$ are related by the equation,

$I=\frac{p^2}{Z}$

where $Z$ is the acoustic impedance of the medium in which the sound wave propagates.

The acoustic impedance of air is ${20}^{0}C$. That is,

$Z=416.9Ns/m^3=416.9Pa/m$

Hence $I=\frac{(6\times {10}^{-10}{)}^2}{416.9}=4.2\times {10}^{-11}W/m^2$

Sound intensity level in dB is defined as

$L=10lo{g}_{10}.\frac{I}{I_0}$

with $I_0=1.0\times {10}^{-12}W/m^2$

Hence $L=10lo{g}_{10}\left(\frac{4.2\times {10}^{-11}}{1.0\times {10}^{-12}}\right)=6.23dB$