Chapter 4.6, Problem 42E

To determine

Find the ratio of the intensity of the rock music to that of the power mower.

Answer to Problem 42E

  $\boxed{1:0.0398}$

Explanation of Solution

Given information:

The noise from a power mower was measured at $106dB$ . The noise level at a rock concert was measured at $120dB$ . Find the ratio of the intensity of the rock music to that of the power mower.

Calculation:

Consider the intensity of power mower was measured at $106dB$ .

  $\begin{array}{l}B=10\log \frac{I}{I_0}=10\log \frac{I}{{10}^{-12}}\\ 106=10\log \frac{I}{{10}^{-12}}\Rightarrow 10.6=\log \frac{I}{{10}^{-12}}\\ 10.6=\log I+12\log 10\\ \log I=10.6-12\log 10\\ =-1.4\\ I={10}^{-1.4}\\ I=0.0398W/m^2\end{array}$

Now the noise level at a rock concert was measured at $120dB$ .

  $\begin{array}{l}B=10\log \frac{I}{I_0}=10\log \frac{I}{{10}^{-12}}\\ 120=10\log \frac{I}{{10}^{-12}}\Rightarrow 12=\log \frac{I}{{10}^{-12}}\\ 12=\log I+12\log 10\\ \log I=12-12\log 10\\ =0\\ I={10}^0\\ I=1W/m^2\end{array}$

Therefore, the ratio of the intensity of the rock music to that of the power mower $\boxed{1:0.0398}$

Hence, the solution is, $\boxed{1:0.0398}$