Chapter 22, Problem 55E

To determine

The intensity level at $1000\text{m}$.

Answer to Problem 55E

The intensity level at $1000\text{m}$ is $\underset{\_}{49.53\text{dB}}$.

Explanation of Solution

Write the expression for intensity level of sound.

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

Here, $\beta$ is the intensity level of sound, $I_0$ is the lowest audible intensity and $I_1$ is the intensity of sound at $3000\text{m}$ distance.

Given that the intensity level of sound is $40\text{dB}$ at a distance of $3000\text{m}$.

Apply the above condition in equation (I), and rearrange to find $I_1$.

    $\begin{array}{c}40\text{dB}=10{\log }_{10}\left(\frac{I_1}{I_0}\right)\\ 4={\log }_{10}\left(\frac{I_1}{I_0}\right)\\ {10}^4=\left(\frac{I_1}{I_0}\right)\\ I_1={10}^4\times I_0\end{array}$        (II)

Since the lowest audible intensity is equal to ${10}^{-12}{\text{Wm}}^{-2}$, equation (II) is written as

    $\begin{array}{c}{I}_1=\left({10}^4\right)\left({10}^{-12}{\text{Wm}}^{-2}\right)\\ ={10}^{-8}{\text{Wm}}^{-2}\end{array}$

Write the relation between acoustic power in terms of intensity.

    $P=I_1\left(4\pi r^2\right)$        (III)

Here, $P$ is the acoustic power at $3000\text{m}$ distance, and $r$ is the distance from the source.

Since the intensity of sound at $3000\text{m}$ distance is ${10}^{-8}{\text{Wm}}^{-2}$.

Use the above condition in equation (III).

    $\begin{array}{c}P=\left({10}^{-8}{\text{Wm}}^{-2}\right)\left(4\pi {\left(3000\text{m}\right)}^2\right)\\ =1.13\text{W}\end{array}$        (IV)

Write the relation between acoustic power in terms of intensity.

    $P=I^{\prime }\left(4\pi r^2\right)$        (V)

Here, $P$ is the acoustic power at $1000\text{m}$ distance, $I^{\prime }$ is the intensity at $1000\text{m}$, and $r$ is the distance from the source

Rearrange the above equation to find $I^{\prime }$.

    $I^{\prime }=\frac{P}{4\pi r^2}$

Since the acoustic power at $1000\text{m}$ distance is $1.13\text{W}$, and the distance from the source is $1000\text{m}$ the above equation is written as

    $\begin{array}{c}{I}^{\prime }=\frac{1.13\text{W}}{4\pi {\left(1000\text{m}\right)}^2}\\ =8.99\times {10}^{-8}{\text{Wm}}^{-2}\end{array}$        (VI)

Write the expression for intensity level of sound at a distance of $1000\text{m}$.

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

Here, ${\beta }^{\prime }$ is the intensity level of sound, $I_0$ is the lowest audible intensity and $I^{\prime }$ is the intensity of sound at a distance of $1000\text{m}$.

Conclusion:

Substitute $8.99\times {10}^{-8}{\text{Wm}}^{-2}$ for $I^{\prime }$, and ${10}^{-12}{\text{Wm}}^{-2}$ for $I_0$ in equation (VII), to find ${\beta }^{\prime }$.

    $\begin{array}{c}{\beta }^{\prime }=10{\log }_{10}\left(\frac{8.99\times {10}^{-8}{\text{Wm}}^{-2}}{{10}^{-12}{\text{Wm}}^{-2}}\right)\\ =49.53\text{dB}\end{array}$

Therefore, the intensity level at $1000\text{m}$ is $\underset{\_}{49.53\text{dB}}$.