Chapter 3.3, Problem 82E

To determine

Tofind:the difference in loudness when two stereos of same intensity are played at same time compared to when just one stereo is played.

Answer to Problem 82E

  $3.01\text{ dB}$

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 $a\ne 1$ :

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

Let the intensity of the stereo be I .

So, when one stereo of intensity I is played, the loudness is,

  ${\beta }_1=10\left(\log I+12\right)$

When two stereos with same intensity are played, the new intensity is $2I$ .

Substitute $2I$ in place of $I$ in $\beta =10\left(\log I+12\right)$ ,

  ${\beta }_2=10\left(\log \left(2I\right)+12\right)$

Using the product property,

  ${\beta }_2=10\left(\log 2+\log I+12\right)$

Now find the difference,

  $\begin{array}{c}{\beta }_2-{\beta }_1=10\left(\log 2+\log I+12\right)-10\left(\log I+12\right)\\ =10\log 2+10\log I+120-10\log I-120\\ =10\log 2\\ \approx \text{3}\text{.01 dB}\end{array}$

Conclusion:

Hence, the difference in loudness is approximately 3.01 dB.