Chapter 22, Problem 57E

(a)

To determine

The intensity in watts per square.

Answer to Problem 57E

The intensity in watts per square is $\underset{\_}{{10}^{-8}{\text{Wm}}^{-2}}$.

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 $10\text{m}$ distance.

Given that the intensity level of sound is $40\text{dB}$.

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

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

Conclusion:

Therefore, the intensity in watts per square is $\underset{\_}{{10}^{-8}{\text{Wm}}^{-2}}$.

(b)

To determine

The intensity level in decibels $3\text{m}$ from the radio.

Answer to Problem 57E

The intensity level in decibels $3\text{m}$ from the radio is $\underset{\_}{50.45\text{dB}}$.

Explanation of Solution

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 $10\text{m}$ distance, and $r$ is the distance from the source.

From subpart(a), the intensity of sound at $10\text{m}$ distance is ${10}^{-8}{\text{Wm}}^{-2}$.

Use the above condition in equation (III).

    $\begin{array}{c}P=\left({10}^{-8}{\text{Wm}}^{-2}\right)\left(4\pi {\left(10\text{m}\right)}^2\right)\\ =1.256\times {10}^{-5}\text{W}\end{array}$        (IV)

Write the expression for intensity of sound at a distance of $3\text{m}$.

    $I^{\prime }=\frac{P}{4\pi {\left(3\text{m}\right)}^2}$

Here, $I^{\prime }$ is the intensity of sound at a distance of $3\text{m}$.

Use equation (IV) in the above equation.

    $\begin{array}{c}{I}^{\prime }=\frac{1.256\times {10}^{-5}\text{W}}{4\pi {\left(3\text{m}\right)}^2}\\ =1.11\times {10}^{-7}{\text{Wm}}^{-2}\end{array}$

Write the expression for intensity level of sound at a distance of $3\text{m}$.

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

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 $3\text{m}$.

Conclusion:

Substitute $1.11\times {10}^{-7}{\text{Wm}}^{-2}$ for $I^{\prime }$, and ${10}^{-12}{\text{Wm}}^{-2}$ for $I_0$ in equation (IV), to find ${\beta }^{\prime }$.

    $\begin{array}{c}{\beta }^{\prime }=10{\log }_{10}\left(\frac{1.11\times {10}^{-7}{\text{Wm}}^{-2}}{{10}^{-12}{\text{Wm}}^{-2}}\right)\\ =50.45\text{dB}\end{array}$

Therefore, the intensity level in decibels $3\text{m}$ from the radio is $\underset{\_}{50.45\text{dB}}$.

(c)

To determine

The output power of the radio in watts that is spread uniformly from a forward hemisphere.

Answer to Problem 57E

The output power of the radio in watts that is spread uniformly from a forward hemisphere is $\underset{\_}{6.28\times {10}^{-6}\text{W}}$.

Explanation of Solution

Write the expression for acoustic power from a hemisphere.

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

Here, $P^{\prime }$ is the acoustic power from a hemisphere.

Conclusion:

Substitute $1.11\times {10}^{-7}{\text{Wm}}^{-2}$ for $I^{\prime }$, and $3\text{m}$ for $r$ in equation (V), to find $P^{\prime }$.

    $\begin{array}{c}{P}^{\prime }=\left(1.11\times {10}^{-7}{\text{Wm}}^{-2}\right)\left(2\pi {\left(3\text{m}\right)}^2\right)\\ =6.273\times {10}^{-6}\text{W}\\ \approx 6.28\times {10}^{-6}\text{W}\end{array}$

Therefore, The output power of the radio in watts that is spread uniformly from a forward hemisphere is $\underset{\_}{6.28\times {10}^{-6}\text{W}}$.