Chapter 4.5, Problem 62E

Solution

i.

To calculate: The decibel level of dog’s sound.

  $80\text{ decibels}$

Given:

The intensity of the dog’s sound $={10}^{-4}W/m^2$

Formula Used:

Use the function $L\left(I\right)=10\log \left(\frac{I}{I_{\circ }}\right)$ to compute the decibel level where $I_{\circ }={10}^{-12}W/m^2$.

Calculation:

Substitute known values in $L\left(I\right)=10\log \left(\frac{I}{I_{\circ }}\right)$.

  $\begin{array}{l}L\left({10}^{-4}\right)=10\log \left(\frac{{10}^{-4}}{{10}^{-12}}\right)\\ L\left({10}^{-4}\right)=10\log {10}^8\\ L\left({10}^{-4}\right)=80\text{ decibels}\end{array}$

Conclusion:

The dog’s sound decibel level is $80\text{ decibels}$.

ii.

To calculate: The decibel level of sound produced by the ambulance siren.

  $120\text{ decibels}$

Given:

The intensity of the siren’s sound $={10}^{0}W/m^2$.

Formula Used:

Use the function $L\left(I\right)=10\log \left(\frac{I}{I_{\circ }}\right)$ to compute the decibel level where $I_{\circ }={10}^{-12}W/m^2$.

Calculation:

Substitute known values in $L\left(I\right)=10\log \left(\frac{I}{I_{\circ }}\right)$.

  $\begin{array}{l}L\left({10}^0\right)=10\log \left(\frac{{10}^0}{{10}^{-12}}\right)\\ L\left({10}^0\right)=10\log {10}^{12}\\ L\left({10}^0\right)=120\text{ decibels}\end{array}$

Conclusion:

The siren’s sound decibel level is $120\text{ decibels}$.

iii.

To calculate: The decibel level of Bee’s sound.

  $\text{55 decibels}$

Given:

The intensity of the Bee’s sound $={10}^{-6.5}W/m^2$.

Formula Used:

Use the function $L\left(I\right)=10\log \left(\frac{I}{I_{\circ }}\right)$ to compute the decibel level where $I_{\circ }={10}^{-12}W/m^2$.

Calculation:

Substitute known values in $L\left(I\right)=10\log \left(\frac{I}{I_{\circ }}\right)$.

  $\begin{array}{l}L\left({10}^{-6.5}\right)=10\log \left(\frac{{10}^{-6.5}}{{10}^{-12}}\right)\\ L\left({10}^{-6.5}\right)=10\log {10}^{5.5}\\ L\left({10}^{-6.5}\right)=55\text{ decibels}\end{array}$

Conclusion:

The Bee’s sound decibel level is $\text{55 decibels}$.