Question

The faintest sound that can be heard has a pressure amplitude of about $4\times {10}^{-5}{\text{N/m}}^2$ and the loudest that can be heard without pain has a pressure amplitude of about $2.8{\text{N/m}}^2$. Determine the intensity of the faintest sound (both in ${\text{W/m}}^2$ and in $\text{dB}$).
[Assume, air density is $1.3{\text{kg/m}}^3$ and velocity of sound is $330\text{m/s}$ and take ${\log }_{10}1.86=0.269$]

• A. $2\times {10}^{-12}{\text{W/m}}^2,1\text{dB}$
• B. $1.86\times {10}^{-12}{\text{W/m}}^2,1.6\text{dB}$
• C. $1.86\times {10}^{-12}{\text{W/m}}^2,2.7\text{dB}$
• D. $2\times {10}^{-12}{\text{W/m}}^2,2.7\text{dB}$

Answer 1

The correct option is C $1.86\times {10}^{-12}{\text{W/m}}^2,2.7\text{dB}$

We know that,
Relation between intensity and pressure of wave is given by
$I=\frac{(\Delta P{)}_{max}^2}{2\rho v}$

For finest sound, substituting the data given in the question we get,

$I=\frac{(4\times {10}^{-5}{)}^2}{2\times 1.3\times 330}=1.86\times {10}^{-12}{\text{W/m}}^2$
From the relation between loudness and intensity of a sound wave,
$\beta =10{\log }_{10}\left(\frac{I}{I_0}\right)$
From the data obtained,

$\beta =10{\log }_{10}\left(\frac{1.86\times {10}^{-12}}{{10}^{-12}}\right)=2.69\approx 2.7\text{dB}$