Question

What is the intensity in watts per meter squared of 80.9 dB sound?

Answer 1

We are asked to determine the intensity of 80.9 dB of a sound wave. To do that we will use the following formula:

$I=I_0(10^{\frac{\beta}{10}})$

Where:

$\begin{gathered} I=\text{ intesity} \\ \beta=\text{ decibells. } \end{gathered}$

The reference value is the following:

$I_0=1\times10^{-12}\frac{W}{m^2}$

Substituting the values:

$I=(1\times10^{-12}\frac{W}{m^2})(10^{\frac{80.9}{10}})$

Solving the operation we get:

$I=1.2\times10{}^{-4}\frac{W}{m^2}$

Therefore, the intensity is 1.2 x 10 ^-4 W/m^2.