Chapter 12, Problem 16P

(a)

To determine

To Calculate: The power output of the loudspeaker.

Answer to Problem 16P

The power output of the speaker is $9.9\times {10}^2\text{W}$ .

Explanation of Solution

Given:

Intensity is, $I=130\text{dB}$

Hearing intensity, $I_0=1.0\times {10}^{-12}\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.$

The distance is $2.8\text{m}$

Formula used:

The intensity is given as,

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

Here, $I$ is sound intensity,

  $I_0$ is the hearing intensity.

The intensity is given as,

  $I=\frac{P}{A}$

Here, $P$ is the power,

  $A$ is the area of cross-section.

Calculation:

The power output corresponding to the given intensity is,

  $\begin{array}{l}\beta =130\text{dB}=10\log \frac{I_{2.8}}{I_0}\\ \Rightarrow 13=\log \frac{I_{2.8}}{I_0}\\ \Rightarrow I_{2.8}={10}^{13}{I}_0\\ =\left(1.0\times {10}^{-12}\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.\right)\times \left({10}^{13}\right)\\ =10\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.\end{array}$

So,

  $\begin{array}{l}P=IA=\left(10\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.\right)\times \left(4\pi {\left(2.8\right)}^2{\text{m}}^2\right)\\ P=\left(10\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.\right)\times \left(4\times 3.14\times {\left(2.8\right)}^2{\text{m}}^2\right)\\ =985\text{W}\end{array}$

Or,

  $P=9.9\times {10}^2\text{W}$

Conclusion:

The output power is $9.9\times {10}^2\text{W}$ .

(b)

To determine

To Calculate: The distance of the sound level.

Answer to Problem 16P

The sound level would be $2.8\times {10}^2\text{m}$ far away from the given value.

Explanation of Solution

Given:

The given value is $90\text{dB}$ .

Formula used:

The intensity is given as,

  $I=\frac{P}{A}$

Here, $P$ is the power,

  $A$ is the area of cross-section.

The intensity level is given as,

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

Here, $I$ is sound intensity,

  $I_0$ is the hearing intensity.

Calculation:

Write the expression for intensity level,

  $\begin{array}{l}\beta =90dB=10\log \frac{I}{I_0}\\ \Rightarrow 9=\log \frac{I}{I_0}\\ \Rightarrow I={10}^9{I}_0\\ =\left(1.0\times {10}^{-12}\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.\right)\times \left({10}^9\right)\\ =1.0\times {10}^{-3}\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.\end{array}$

And the distance in terms of the power output is,

  $\begin{array}{l}r=\sqrt{\frac{P}{4\pi I}}=\sqrt{\frac{985\text{W}}{4\times 3.14\times \left(1.0\times {10}^{-3}\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.\right)}}\\ =2.8\times {10}^2\text{m}\end{array}$

Conclusion:

The distance is $2.8\times {10}^2\text{m}$ .