Chapter 17, Problem 74PQ

(a)

To determine

The intensity of the sound.

Answer to Problem 74PQ

The intensity of the sound is $\underset{\_}{13.8{\text{W/m}}^{\text{2}}}$.

Explanation of Solution

Write the equation of intensity of sound.

  $I=\frac{P}{A}$                                                                                                                 (I)

Here, $I$ is the intensity, $P$ is the power and $A$ is the area.

Write the expression for the power.

  $P=\frac{E}{t}$                                                                                                              (II)

Here, $E$ is the energy and $t$ is the time.

Rewrite the expression for the intensity by using equation (II).

  $I=\frac{E}{At}$                                                                                                             (III)

Conclusion:

Substitute $275\text{J}$ for $E$, $5.00{\text{m}}^{\text{2}}$ for $A$ and $4.00\text{s}$ for $t$ in equation (III) to find $I$.

  $\begin{array}{c}I=\frac{\left(275\text{J}\right)}{\left(5.00{\text{m}}^{\text{2}}\right)\left(4.00\text{s}\right)}\\ =\frac{\left(275\text{J}\right)}{\left(20.0{\text{m}}^{\text{2}}\text{/s}\right)}\\ =13.8{\text{W/m}}^{\text{2}}\end{array}$

Thus, the intensity of the sound is $\underset{\_}{13.8{\text{W/m}}^{\text{2}}}$.

(b)

To determine

The amount of sound level.

Answer to Problem 74PQ

The amount of sound level is $\underset{\_}{131\text{dB}}$.

Explanation of Solution

Write the equation of sound level.

  $\beta =10\log \left(\frac{I}{I_o}\right)$                                                                                               (IV)

Here, $\beta$ is the sound level, $I$ is the intensity and $I_o$ is the initial intensity.

Conclusion:

Substitute $13.8{\text{W/m}}^{\text{2}}$ for $I$ and ${\text{10}}^{-12}{\text{W/m}}^{\text{2}}$ for $I_o$ in equation (IV) to find $\beta$.

  $\begin{array}{c}\beta =10\log \left(\frac{\left(13.8{\text{W/m}}^{\text{2}}\right)}{\left({\text{10}}^{-12}{\text{W/m}}^{\text{2}}\right)}\right)\\ =10\log \left(13.8\times {10}^{12}\right)\\ =131\text{dB}\end{array}$

Thus, the amount of sound level is $\underset{\_}{131\text{dB}}$.

(c)

To determine

The intensity of the sound.

Answer to Problem 74PQ

The intensity of the sound is $\underset{\_}{3.44{\text{W/m}}^{\text{2}}}$.

Explanation of Solution

Write the equation of intensity of sound.

  $I=\frac{P}{A}$                                                                                                                 (I)

Here, $I$ is the intensity, $P$ is the power and $A$ is the area.

Write the expression for the power.

  $P=\frac{E}{t}$                                                                                                              (II)

Here, $E$ is the energy and $t$ is the time.

Rewrite the expression for the intensity by using equation (II).

  $I=\frac{E}{At}$                                                                                                            (III)

Conclusion:

Substitute $275\text{J}$ for $E$, $20.00{\text{m}}^{\text{2}}$ for $A$ and $4.00\text{s}$ for $t$ in equation (III) to find $I$.

  $\begin{array}{c}I=\frac{\left(275\text{J}\right)}{\left(20.00{\text{m}}^{\text{2}}\right)\left(4.00\text{s}\right)}\\ =\frac{\left(275\text{J}\right)}{\left(80.00{\text{m}}^{\text{2}}\text{/s}\right)}\\ =3.44{\text{W/m}}^{\text{2}}\end{array}$

Thus, the intensity of the sound is $\underset{\_}{3.44{\text{W/m}}^{\text{2}}}$.

(d)

To determine

The amount of sound level.

Answer to Problem 74PQ

The amount of sound level is $\underset{\_}{125\text{dB}}$.

Explanation of Solution

Write the equation of sound level.

  $\beta =10\log \left(\frac{I}{I_o}\right)$                                                                                               (IV)

Here, $\beta$ is the sound level, $I$ is the intensity and $I_o$ is the initial intensity.

Conclusion:

Substitute $3.44{\text{W/m}}^{\text{2}}$ for $I$ and ${\text{10}}^{-12}{\text{W/m}}^{\text{2}}$ for $I_o$ in equation (IV) to find $\beta$.

  $\begin{array}{c}\beta =10\log \left(\frac{\left(3.44{\text{W/m}}^{\text{2}}\right)}{\left({\text{10}}^{-12}{\text{W/m}}^{\text{2}}\right)}\right)\\ =10\log \left(3.44\times {10}^{12}\right)\\ =125\text{dB}\end{array}$

Thus, the amount of sound level is $\underset{\_}{125\text{dB}}$.