Chapter 12, Problem 14P

(a)

To determine

To Calculate: The amount of energy absorbed by the eardrum.

Answer to Problem 14P

The power output of the sound from speaking a person in normal conversation is $9\times {10}^{-6}\text{W}$ .

Explanation of Solution

Given:

Intensity is, $I=50\text{dB}$

Area is, $A=5.0\times {10}^{-5}{\text{m}}^{\text{2}}$

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

Formula used:

The sound intensity level is given as,

  $\beta =10\log \frac{I}{I_0}$

Here, $I_0$ hearing intensity,

  $I$ is the intensity generated by normal conversation.

Calculation:

The power of the sound wave is equal to the absorbed amount of energy;

So, it is given as,

  $\begin{array}{l}50\text{dB}=10\log \frac{I}{I_0}\\ \Rightarrow 5=\log \frac{I}{I_0}\\ \Rightarrow I=I_0\times {10}^5\\ =\left(1.0\times {10}^{-12}\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.\right)\times \left({10}^5\right)\\ =1.0\times {10}^{-7}\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.\end{array}$

The power output is given as,

  $\begin{array}{l}P=IA=\left(1.0\times {10}^{-7}\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.\right)\times \left(5.0\times {10}^{-5}{\text{m}}^{\text{2}}\right)\\ P=5.0\times {10}^{-12}\text{W}\end{array}$

Conclusion:

The energy per second is $5.0\times {10}^{-12}\text{W}$ .

(b)

To determine

To Calculate: The time period taken by the eardrum.

Answer to Problem 14P

The time is taken by the eardrum to absorbed the total energy $1.0\text{J}$ is $6.3\times {10}^3\text{year}$ .

Explanation of Solution

Given:

Total energy is, $E_{\text{total}}=1.0\text{J}$

Calculation:

  $5.0\times {10}^{-12}\text{W}$ energy is absorbed by the eardrum in one second, which means that,

In $3.16\times {10}^7\text{s}$ , the amount of absorbed energy is $1.0\text{J}$ . So, the time taken by the eardrum to absorb the total energy $5.0\times {10}^{-12}\text{W}$ is,

  $\begin{array}{l}t=\left(1.0\text{J}\right)\times \left(\frac{1\text{s}}{5.0\times {10}^{-12}\text{J}}\right)\times \left(\frac{1\text{year}}{3.16\times {10}^7\text{s}}\right)\\ =6.3\times {10}^3\text{year}\end{array}$

Conclusion:

The time is $6.3\times {10}^3\text{year}$ .