Chapter 3.5, Problem 49E

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 49E

  $95\%$

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 noise suppression materials, the noise level in an auditorium decreased from 93 to 80 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 noise suppression materials

Substitute $\beta =93$ in the above model.

  $\begin{array}{c}93=10\log \left(\frac{I_B}{{10}^{-12}}\right)\\ \frac{93}{10}=\log \left(\frac{I_B}{{10}^{-12}}\right)\\ 9.3=\log \left(\frac{I_B}{{10}^{-12}}\right)\end{array}$

Changing the above expression into exponential form,

  $\begin{array}{c}{10}^{9.3}=\frac{I_B}{{10}^{-12}}\\ I_B={10}^{9.3}\times {10}^{-12}\\ I_B={10}^{9.3+\left(-12\right)}\\ I_B={10}^{-2.7}\end{array}$

Now find the intensity $I_A$ after the installation of noise suppression materials

Substitute $\beta =80$ in the above model.

  $\begin{array}{c}80=10\log \left(\frac{I_A}{{10}^{-12}}\right)\\ \frac{80}{10}=\log \left(\frac{I_A}{{10}^{-12}}\right)\\ 8=\log \left(\frac{I_A}{{10}^{-12}}\right)\end{array}$

Changing the above expression into exponential form,

  $\begin{array}{c}{10}^8=\frac{I_A}{{10}^{-12}}\\ I_A={10}^8\times {10}^{-12}\\ I_A={10}^{8+\left(-12\right)}\\ I_A={10}^{-4}\end{array}$

The percent decrease in the intensity of the noise is:

  $\begin{array}{c}\%\text{ change}=\left|\frac{I_A-I_B}{I_B}\right|\times 100\%\\ =\left|\frac{{10}^{-4}-{10}^{-2.7}}{{10}^{-2.7}}\right|\times 100\%\\ =94.99\%\\ \approx 95\%\end{array}$