Chapter 3.5, Problem 49E

To determine

Find the percent decrease in the intensity level of the noise.

Answer to Problem 49E

The percent decrease in the intensity level of the noise $95\%$ .

Explanation of Solution

Given:

The noise level decreased from $93$ to $80$ decibels.

Calculation:

Intensity before the installation.

  $\begin{array}{l}\alpha =10\log \frac{I_b}{I_{\circ }}\\ \Rightarrow \frac{\alpha }{10}=\frac{10}{10}\log \frac{I_b}{I_{\circ }}\\ \Rightarrow \frac{\alpha }{10}=\log \frac{I_b}{I_{\circ }}\end{array}$

Taking base $10$ on both sides.

  $\begin{array}{l}\Rightarrow {10}^{\frac{\alpha }{10}}={10}^{\log \frac{I_b}{I_{\circ }}}\\ \Rightarrow {10}^{\frac{\alpha }{10}}=\frac{I_b}{I_{\circ }}\\ \Rightarrow I_{\circ }{10}^{\frac{\alpha }{10}}=I_b\\ \Rightarrow I_b=I_{\circ }{10}^{\frac{\alpha }{10}}\\ \Rightarrow I_b={10}^{-12}{10}^{\frac{93}{10}}\end{array}$

Intensity after the installation.

  $\begin{array}{l}\alpha =10\log \frac{I_a}{I_{\circ }}\\ \Rightarrow \frac{\alpha }{10}=\frac{10}{10}\log \frac{I_a}{I_{\circ }}\\ \Rightarrow \frac{\alpha }{10}=\log \frac{I_a}{I_{\circ }}\end{array}$

Taking base $10$ on both sides.

  $\begin{array}{l}\Rightarrow {10}^{\frac{\alpha }{10}}={10}^{\log \frac{I_a}{I_{\circ }}}\\ \Rightarrow {10}^{\frac{\alpha }{10}}=\frac{I_a}{I_{\circ }}\\ \Rightarrow I_{\circ }{10}^{\frac{\alpha }{10}}=I_a\\ \Rightarrow I_a=I_{\circ }{10}^{\frac{\alpha }{10}}\\ \Rightarrow I_a={10}^{-12}{10}^{\frac{80}{10}}\end{array}$

% change

  $\begin{array}{l}=\left|\frac{I_a-I_b}{I_b}\right|\cdot 100\\ =\left|\frac{{10}^{-12}{10}^{\frac{80}{10}}-{10}^{-12}{10}^{\frac{93}{10}}}{{10}^{-12}{10}^{\frac{93}{10}}}\right|\cdot 100\\ =\left|\frac{{10}^{\frac{80}{10}}-{10}^{\frac{93}{10}}}{{10}^{\frac{93}{10}}}\right|\cdot 100\\ =\left|{10}^{-\frac{13}{10}}-1\right|\cdot 100\\ =\left|0.05-1\right|\cdot 100\\ =0.95\cdot 100\\ =95\end{array}$

Hence the percent decrease in the intensity level of the noise $95\%$ .