Chapter 8.6, Problem 22PPS

(a)

To determine

To find: How many times minimum intensity of sound detectable by the human ear with the old muffler.

Answer to Problem 22PPS

The sound heard was $316227766$ times the minimum intensity of detectable by the human ear with old muffler.

Explanation of Solution

Given information:

The given noise level of engine dropped from $85$ decibels to $73$ decibels, and $m$ is $1$ .

Calculation:

Consider the noise level of the engine using old muffler is,

  $L_1=85$ decibels.

And the noise level of the engine using new muffler is,

  $L_2=73$ decibels.

If $m=1$ and the intensity of the sound id $I_1$ times of the sound detectable by the human ear with old muffler.

Thus, the old muffler is,

  $L=10\log \frac{I}{m}$ .

Substitute the value of $I$ and $m$ .

  $\begin{array}{c}10\log \frac{I_1}{1}=85\\ \log I_1=8.5\\ I_1={10}^{8.5}\\ \approx 316227766\end{array}$

Therefore, the sound heard was $316227766$ times the minimum intensity of detectable by the human ear with old muffler.

(b)

To determine

To find: How many times minimum intensity of sound detectable by the human ear with the new muffler..

Answer to Problem 22PPS

The sound heard was $19952623.15$ times and $93.69\%$ of intensity decreased after using new muffler.

.

Explanation of Solution

Given information:

The given noise level of engine dropped from $85$ decibels to $73$ decibels, and $m$ is $1$ .

Calculation:

Consider $m=1$ and the intensity of the sound id $I_2$ times of the sound detectable by the human ear with new muffler.

Thus, the new muffler is,

  $L=10\log \frac{I}{m}$ .

Substitute the value of $I$ , and $m$ .

  $\begin{array}{c}10\log \frac{I_2}{1}=73\\ \log I_2=7.3\\ I_2={10}^{7.3}\\ \approx 19952623.12\end{array}$

The percentage of intensity decreased using new muffler is,

  $\begin{array}{l}=\frac{{10}^{8.5}-{10}^{7.3}}{{10}^{8.5}}\times 100\%\\ =93.69\%\end{array}$

Therefore, the sound heard was $19952623.15$ times and $93.69\%$ of intensity decreased after using new muffler.