Chapter 3.3, Problem 80E

To determine

Tofind:the difference in loudness between an average office with an intensity of $1.26\times {10}^{-7}$ watt per square meter and a broadcast studio with an intensity of $3.16\times {10}^{-10}$ watt per square meter.

Answer to Problem 80E

  $26\text{ dB}$

Explanation of Solution

Given:

The relationship between the number of decibels $\beta$ and the intensity of a sound I (in watts per square meter) is:

  $\beta =10\log \frac{I}{{10}^{-12}}=10\left(\log I+12\right)$

Concept Used:

Product property of logarithm states that for a positive real numbers a such that $a\ne 1$ :

  ${\log }_a\left(uv\right)={\log }_{a}u+{\log }_{a}v$

Power property of logarithm states that for a positive real numbers a such that $a\ne 1$ :

  ${\log }_a{u}^n=n{\log }_{a}u$

Quotient property of logarithm states that for a positive real numbers a such that $a\ne 1$ :

  ${\log }_a\left(\frac{u}{v}\right)={\log }_{a}u-{\log }_{a}v$

First find the number of decibels in average office with an intensity of $1.26\times {10}^{-7}$ watt per square meter

Substitute $I=1.26\times {10}^{-7}$ in $\beta =10\left(\log I+12\right)$ ,

  ${\beta }_{office}=10\left(\log \left(1.26\times {10}^{-7}\right)+12\right)$

Using product property,

  ${\beta }_{office}=10\left(\log \left(1.26\right)+\log \left({10}^{-7}\right)+12\right)$

Using the power property,

  ${\beta }_{office}=10\left(\log \left(1.26\right)-7\log \left(10\right)+12\right)$

Using ${\log }_{b}b=1$ ,

  $\begin{array}{c}{\beta }_{office}=10\left(\log \left(1.26\right)-7{\log }_{10}\left(10\right)+12\right)\\ =10\left(\log \left(1.26\right)-7+12\right)\\ =10\left(\log \left(1.26\right)+5\right)\\ =10\log \left(1.26\right)+50\end{array}$

Now, find the number of decibels in broadcast studio with an intensity of $3.16\times {10}^{-10}$ watt per square meter

Substitute $I=3.16\times {10}^{-10}$ in $\beta =10\left(\log I+12\right)$ ,

  ${\beta }_{studio}=10\left(\log \left(3.16\times {10}^{-10}\right)+12\right)$

Using product property,

  ${\beta }_{studio}=10\left(\log \left(3.16\right)+\log \left({10}^{-10}\right)+12\right)$

Using the power property,

  ${\beta }_{studio}=10\left(\log \left(3.16\right)-10\log \left(10\right)+12\right)$

Using ${\log }_{b}b=1$ ,

  $\begin{array}{c}{\beta }_{studio}=10\left(\log \left(3.16\right)-10{\log }_{10}\left(10\right)+12\right)\\ =10\left(\log \left(3.16\right)-10+12\right)\\ =10\left(\log \left(3.16\right)+2\right)\\ =10\log \left(3.16\right)+20\end{array}$

Now find the difference,

  $\begin{array}{c}{\beta }_{office}-{\beta }_{studio}=10\log \left(1.26\right)+50-\left(10\log \left(3.16\right)+20\right)\\ =10\log \left(1.26\right)+50-10\log \left(3.16\right)-20\\ =10\left[\log \left(1.26\right)-\log \left(3.16\right)\right]+30\end{array}$

Using quotient property,

  $\begin{array}{c}{\beta }_{office}-{\beta }_{studio}=10\left[\log \left(\frac{1.26}{3.16}\right)\right]+30\\ \approx 26\end{array}$

Hence, the difference in loudness is approximately 26 dB.