Chapter 14, Problem 18P

A trumpet creates a sound intensity level of 1.15 × 10² dB at a distance of 1.00 m. (a) What is the sound intensity of a trumpet at this distance? (b) What is the sound intensity of five trumpets at this distance? (c) Find the sound intensity of five trumpets at the location of the first row of an audience, 8.00 m away, assuming, for simplicity, the sound energy propagates uniformly in all directions. (d) Calculate the decibel level of the five trumpets in the first row. (e) If the trumpets are being played in an outdoor auditorium, how far away, in theory, can their combined sound be heard? (f) In practice such a sound could not be heard once the listener was 2-3 km away. Why can’t the sound be heard at the distance found in part (e)? Hint: In a very quiet room the ambient sound intensity level is about 30 dB.

To determine

The sound intensity of a trumpet at a distance of 1m.

Answer to Problem 18P

Solution: The sound intensity of a trumpet at a distance of 1m is $0.316{\text{W/m}}^{\text{2}}$

Explanation of Solution

Given Info: The intensity level of sound of a trumpet is $1.15\times {10}^2\text{dB}$ at a distance of 1.00 m.

Formula to calculate the sound intensity of a trumpet is,

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

• $I_0$ is reference Intensity,
• $I$ is the intensity of a trumpet,

Simplify and rearrange the expression in terms of $I$.

$\begin{array}{c}\frac{\beta }{10}=\log \left(\frac{I}{I_0}\right)\\ {10}^{\frac{\beta }{10}}=\left(\frac{I}{I_0}\right)\\ I=I_0{10}^{\frac{\beta }{10}}\end{array}$

The value of reference intensity constant is $1.0\times {10}^{-12}{\text{W/m}}^{\text{2}}$

Substitute $1.15\times {10}^2\text{dB}$ for $\beta$ and $1.0\times {10}^{-12}{\text{W/m}}^{\text{2}}$ for $I_0$ in the above expression to get $I$.

$\begin{array}{c}I=\left(1.0\times {10}^{-12}{\text{W/m}}^{\text{2}}\right){10}^{\frac{1.15\times {10}^2}{10}}\\ =\left(1\times {10}^{-12}{\text{W/m}}^{\text{2}}\right)\times {10}^{11.5}\\ =0.316{\text{W/m}}^{\text{2}}\end{array}$

Conclusion:

The sound intensity of a trumpet at a distance of 1m is $0.316{\text{W/m}}^{\text{2}}$

To determine

The sound intensity of a five trumpet at a distance of 1m.

Answer to Problem 18P

Solution: The sound intensity of a trumpet at a distance of 1m is $1.58{\text{W/m}}^{\text{2}}$

Explanation of Solution

Given Info: The sound intensity of a trumpet at a distance of 1m is $0.316{\text{W/m}}^{\text{2}}$

Formula for intensity of five trumpets,

$I\text{'}=5I$

• $I^{\prime }$ is the intensity of sound for five trumpets.

Sound intensity of a trumpet at a distance of 1m from (a) is $0.316{\text{W/m}}^{\text{2}}$.

Substitute $0.316{\text{W/m}}^{\text{2}}$ for I in the above expression to get $I\text{'}$

$\begin{array}{c}I\text{'}=5\left(0.316{\text{W/m}}^{\text{2}}\right)\\ = 1.58{\text{W/m}}^{\text{2}}\end{array}$

Conclusion:

The sound intensity of a five trumpets at a distance of 1m is $1.58{\text{W/m}}^{\text{2}}$

To determine

The sound intensity of five trumpets at a location.

Answer to Problem 18P

Solution: The sound intensity of five trumpets located at the first row of an audience at a distance of 8m is $2.47\times {10}^{-2}{\text{W/m}}^{\text{2}}$

Explanation of Solution

Given Info: The distance of five trumpets from the first row of an audience is 8m.

Formula for ratio of sound intensity of trumpets,

$\frac{I_1}{I_2}=\frac{r_2{}^2}{r_1{}^2}$

• $I_1$ is the sound intensity at a distance of  1 m
• $I_2$ is the sound intensity at a distance of 8 m
• $r_1$ is the distance of the trumpet at 1 m.
• $r_2$ is the distance of a trumpet at a first row of an audience

Rearrange the expression in terms of $I_2$ .

$I_2=\frac{I_1{r}_1{}^2}{r_2{}^2}$

Substitute 1 m for $r_1$, 8 m for $r_2$  and $1.58{\text{W/m}}^{\text{2}}$ for $I_1$ in the above expression to get $I_2$.

$\begin{array}{c}{I}_2=\frac{\left(1.58{\text{W/m}}^{\text{2}}\right){\left(1\text{m}\right)}^2}{{\left(8\text{m}\right)}^2}\\ =2.47\times {10}^{-2}{\text{W/m}}^{\text{2}}\end{array}$

Conclusion:

Sound intensity of five trumpets located at the first row of an audience at a distance of 8m is $2.47\times {10}^{-2}{\text{W/m}}^{\text{2}}$

To determine

The sound intensity level for first five trumpets

Answer to Problem 18P

Solution: The sound intensity level for first five trumpets is 104 dB

Explanation of Solution

Given Info: Sound intensity of a trumpet at a distance of 8 m from (c) is $2.47\times {10}^{-2}{\text{W/m}}^{\text{2}}$ and intensity constant is $1\times {10}^{-12}{\text{W/m}}^{\text{2}}$

Formula for intensity level of a sound,

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

• $I_0$ is intensity constant
• $I$ is the intensity of a sound

Substitute $2.47\times {10}^{-2}{\text{W/m}}^{\text{2}}$ for $I$ and $1\times {10}^{-12}{\text{W/m}}^{\text{2}}$ for $I_0$ in the above expression to get $\beta$

$\begin{array}{c}\beta =10\log \left(\frac{2.47\times {10}^{-2}{\text{W/m}}^{\text{2}}}{1\times {10}^{-12}{\text{W/m}}^{\text{2}}}\right)\\ =104\text{dB}\end{array}$

Conclusion:

The sound intensity level for first five trumpets is 104 dB

To determine

Distance heard in the outdoor auditorium

Answer to Problem 18P

Solution: Distance heard in the outdoor auditorium is $1.26\times {10}^6\text{m}$

Explanation of Solution

Given Info: distance of five trumpets located at the first row of an audience is 8m, intensity constant is from (a) $1\times {10}^{-12}{\text{W/m}}^{\text{2}}$ and Sound intensity of a trumpet at a distance of 8 m from (c) is $2.47\times {10}^{-2}{\text{W/m}}^{\text{2}}$

Formula for ratio of sound intensity of trumpets,

$\frac{I_1}{I_2}=\frac{r_2{}^2}{r_1{}^2}$

• $I_1$ is the reference intensity.
• $I_2$ is the sound intensity of a trumpet at a distance of  8 m
• $r_1$ is the distance of the trumpet heard in the outdoor auditorium
• $r_2$ is the distance of a trumpet from a first row of an audience 8 m.

Rearrange the expression in terms of  $r_1$.

$r_1=r_2\sqrt{\frac{I_2}{I_1}}$

Substitute 8 m for $r_2$, $1\times {10}^{-12}{\text{W/m}}^{\text{2}}$ for $I_1$ and $2.47\times {10}^{-2}{\text{W/m}}^{\text{2}}$ for $I_2$  in the above expression to get $r_1$.

$\begin{array}{c}{r}_1=8\text{m}\sqrt{\frac{2.47\times {10}^{-2}{\text{W/m}}^{\text{2}}}{1\times {10}^{-12}{\text{W/m}}^{\text{2}}}}\\ =1.26\times {10}^6\text{m}\end{array}$

Conclusion:

Distance heard in the outdoor auditorium is $1.26\times {10}^6\text{m}$

To determine

Why the sound can be heard at a large distance.

Answer to Problem 18P

Therefore, the sound intensity of wave is much lower than the ambient noise intensity at the distance found in (e)

Explanation of Solution

The proportionality expression relating the intensity $I=\frac{1}{r^2}$.

As you can notice from the above equation, the intensity falls rapidly with increase in distance.

As the distance increases the intensity of the sound decreases and finally it becomes much lower than the ambient noise.

Conclusion:

Therefore, the sound intensity of wave is much lower than the ambient noise intensity at the distance found in (e)