Question

People with good hearing can perceive sounds as low in level as$-8.00\mathrm{d}\mathrm{B}$at a frequency of$3000\mathrm{H}\mathrm{z}$. What is the intensity of this sound in watts per meter squared?

Short answer

The intensity is $1.58\times {10}^{-13}\mathrm{W}/{\mathrm{m}}^2$.

Step-by-step solution

Given data

The intensity level is $d=-8.00\mathrm{d}\mathrm{B}$.

The frequency is $f=3000\mathrm{H}\mathrm{z}$.

Definition of sound intensity

At low frequencies, loudness requires more intensity to perceive equal loudness, whereas the mid-frequency range requires less intensity to perceive the same.

Calculation of the intensity of the first sound

The intensity of the sound is,

$d=10\log \frac{I}{{10}^{-12}}$

So,

$\begin{array}{rl}-8.00& =10\log \frac{I}{{10}^{-12}}\\ -0.8& =\log \frac{I}{{10}^{-12}}\\ {10}^{-0.8}& =\frac{I}{{10}^{-12}}\\ I& =1.58\times {10}^{-13}\mathrm{W}/{\mathrm{m}}^2\end{array}$

The intensity of this sound in watts per meter squared is $1.58\times {10}^{-13}\mathrm{W}/{\mathrm{m}}^2$.