Chapter 23, Problem 56SP

One sound has an intensity level of 75.0 dB, while a second has an intensity level of 72.0 dB. What is the intensity level when the two sounds are combined?

To determine

The intensity level of the sound if one sound with the intensity level of $75.0\text{ dB}$ and a second sound with the intensity level of $72.0\text{ dB}$ are combined.

Answer to Problem 56SP

Solution:

$76.8\text{ dB}$

Explanation of Solution

Given data:

The intensity level of the first sound is $75.0\text{ dB}$.

The intensity level of the second sound is $72.0\text{ dB}$.

Formula used:

The expression for theintensity level of sound is written as

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

Here, $\beta$ is the intensity level of the sound, $I$ is the intensity of the sound at any point, and $I_0$ is the standard intensity of the sound.

The standard intensity of the sound is ${10}^{-12}{\text{ W/m}}^2$.

Explanation:

Write the expression for the intensity level of the first sound:

${\beta }_1=10{\log }_{10}\left(\frac{I_1}{I_0}\right)$

Here, $I_1$ is the intensity of the first sound.

Substitute $75.0\text{ dB}$ for ${\beta }_1$ and ${10}^{-12}{\text{ W/m}}^2$ for $I_0$

$\begin{array}{c}75.0\text{ dB}=10{\log }_{10}\left(\frac{I_1}{{10}^{-12}{\text{ W/m}}^2}\right)\\ 7.5={\log }_{10}\left(\frac{I_1}{{10}^{-12}{\text{ W/m}}^2}\right)\\ I_1=\left({10}^{-12}{\text{ W/m}}^2\right)\left[{\text{antilog}}_{10}\left(7.5\right)\right]\\ =3.2\times {10}^{-5}{\text{ W/m}}^2\end{array}$

Write the expression for the intensity level of thesecond sound:

${\beta }_2=10{\log }_{10}\left(\frac{I_2}{I_0}\right)$

Here, $I_2$ is the intensity of the second sound.

Substitute $72.0\text{ dB}$ for ${\beta }_2$ and ${10}^{-12}{\text{ W/m}}^2$ for $I_0$

$\begin{array}{c}72.0\text{ dB}=10{\log }_{10}\left(\frac{I_2}{{10}^{-12}{\text{ W/m}}^2}\right)\\ 7.2={\log }_{10}\left(\frac{I_2}{{10}^{-12}{\text{ W/m}}^2}\right)\\ I_2=\left({10}^{-12}{\text{ W/m}}^2\right)\left[{\text{antilog}}_{10}\left(7.2\right)\right]\\ =1.6\times {10}^{-5}{\text{ W/m}}^2\end{array}$

Write the expression for total intensity for the combination of both intensities:

$I_t=I_1+I_2$

Substitute $3.2\times {10}^{-5}{\text{ W/m}}^2$ for $I_1$ and $1.6\times {10}^{-5}{\text{ W/m}}^2$ for $I_2$

$\begin{array}{c}{I}_t=\left(3.2\times {10}^{-5}{\text{ W/m}}^2\right)+\left(1.6\times {10}^{-5}{\text{ W/m}}^2\right)\\ =4.8\times {10}^{-5}{\text{ W/m}}^2\end{array}$

Again, write the expression for intensity level of sound for the combination as

$\beta \text{'}=10{\log }_{10}\left(\frac{I_t}{I_0}\right)$

Here, $\beta \text{'}$ is the total intensity level of sound due to combination and $I_t$ is the total intensity.

Substitute $4.8\times {10}^{-5}{\text{ W/m}}^2$ for $I_t$ and $1.6\times {10}^{-5}{\text{ W/m}}^2$ for $I_0$

$\begin{array}{c}\beta \text{'}=10{\log }_{10}\left(\frac{4.8\times {10}^{-5}{\text{ W/m}}^2}{{10}^{-12}{\text{ W/m}}^2}\right)\\ =10{\log }_{10}\left(4.8\times {10}^7\right)\\ =76.8\text{ dB}\end{array}$

Conclusion:

Therefore, the intensity level of sound when the given two sounds are combined is $76.8\text{ dB}$.