Chapter 3.5, Problem 50E

To determine

To find: the percent decrease in the intensity of the noise as a result of the installation of these materials.

Answer to Problem 50E

  $97.49\%$

Explanation of Solution

Given:

The number of decibels $\beta$ of a sound with an intensity of I watts per square meter is given by:

  $\beta =10\log \left(\frac{I}{I_0}\right)$.........(1)

Where $I_0={10}^{-12}{\text{ watt per m}}^2$

Also, it is given that due to the installation of a muffler, the noise level of an engine decreased from 88 to 72 decibels.

Formula used:

1. $a^m\times a^n=a^{m+n}$

2. $a^{{\log }_{a}x}=x$

First find the intensity $I_B$ before the installation of muffler

Substitute $\beta =88$ in the above model.

  $\begin{array}{c}88=10\log \left(\frac{I_B}{{10}^{-12}}\right)\\ \frac{88}{10}=\log \left(\frac{I_B}{{10}^{-12}}\right)\\ 8.8=\log \left(\frac{I_B}{{10}^{-12}}\right)\end{array}$

Changing the above expression into exponential form,

  $\begin{array}{c}{10}^{8.8}=\frac{I_B}{{10}^{-12}}\\ I_B={10}^{8.8}\times {10}^{-12}\\ I_B={10}^{8.8+\left(-12\right)}\\ I_B={10}^{-3.2}\end{array}$

Now find the intensity $I_A$ after the installation of muffler

Substitute $\beta =72$ in the above model.

  $\begin{array}{c}72=10\log \left(\frac{I_A}{{10}^{-12}}\right)\\ \frac{72}{10}=\log \left(\frac{I_A}{{10}^{-12}}\right)\\ 7.2=\log \left(\frac{I_A}{{10}^{-12}}\right)\end{array}$

Changing the above expression into exponential form,

  $\begin{array}{c}{10}^{7.2}=\frac{I_A}{{10}^{-12}}\\ I_A={10}^{7.2}\times {10}^{-12}\\ I_A={10}^{7.2+\left(-12\right)}\\ I_A={10}^{-4.8}\end{array}$

The percent decrease in the intensity of the noise is:

  $\begin{array}{c}\%\text{ change}=\left|\frac{I_A-I_B}{I_B}\right|\times 100\%\\ =\left|\frac{{10}^{-4.8}-{10}^{-3.2}}{{10}^{-3.2}}\right|\times 100\%\\ =97.49\%\end{array}$