Chapter 12, Problem 15P

(a)

To determine

The estimated sound level for amplifier A and amplifier B.

Answer to Problem 15P

The estimated sound level for the both cases are $122\text{dB}$ and $114\text{dB}$ .

Explanation of Solution

Given:

The power of amplifier A is $P_A=250\text{W}$ .

The power of amplifier B is $P_B=40\text{W}$ .

The distance between amplifier and the loudspeaker is $d=\text{3}.\text{5}\text{m}$ .

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

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,

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

Here, $I$ is sound intensity,

  $I_0$ is the hearing intensity.

Calculation:

Assume that the sound spreads spherically outward, then the intensity of amplifier is A.

  $\begin{array}{l}{I}_A=\frac{P}{A}=\frac{250\text{W}}{4\pi {\left(3.5\right)}^2}\\ =\frac{250\text{W}}{4\times 3.14\times {\left(3.5\right)}^2}\\ =1.6\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.\\ \end{array}$

Calculate the intensity level for the Amplifier A.

  $\begin{array}{l}{\beta }_A=10\log \frac{I}{I_0}\\ =10\times \log \left(\frac{1.6\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.}{1.0\times {10}^{-12}\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.}\right)\\ =122\text{dB}\end{array}$

Similarly, for the case of $40\text{W}$ power,

  $\begin{array}{l}{I}_B=\frac{P}{A}=\frac{40\text{W}}{4\pi {\left(3.5\right)}^2}\\ =\frac{40\text{W}}{4\times 3.14\times {\left(3.5\right)}^2}\\ =0.26\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.\end{array}$

And,

  $\begin{array}{l}{\beta }_B=10\log \frac{I}{I_0}\\ =10\times \log \left(\frac{0.26\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.}{1.0\times {10}^{-12}\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.}\right)\\ =114\text{dB}\end{array}$

Conclusion:

The sound level for the respective cases are, $122\text{dB}$ and $114\text{dB}$ .

(b)

To determine

To Check: The expensive amplifier sound loud.

Answer to Problem 15P

No, the expensive amplifier will not sound twice as loud as the cheaper one.

Explanation of Solution

Introduction:

The loudness of the sound is compared on the basis of the intensity of the sound. If the intensity of sound is higher that indicates that amplifier produces a louder sound.

The intensities need to differ by the factor of 10 for a sound to be perceived as twice as loud as another. Therefore, they differ from each other by a factor of $\frac{1.6}{0.26}\approx 6$ .

Conclusion:

Hence, the expensive amplifier will not sound loud.