Chapter 12, Problem 23P

(a)

To determine

To Calculate: The ratio of highest to lowest intensities at given frequency.

Answer to Problem 23P

The ratio of highest to lowest intensity at the given frequency is ${10}^9$ .

Explanation of Solution

Given:

Intensity is, $f=100\text{Hz}$

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

Formula used:

Write the expression for intensity ratio.

  $\text{Intensity}\text{ratio}=\frac{I_{\text{highest}}}{I_{\text{lowest}}}$

Calculation:

The graph for sound level versus frequency is shown below.

  

The hearing threshold at frequency 100Hz is about $5\times {10}^{-9}\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.$ , observed from the given graph. The pain threshold is about $5\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.$ that is the highest frequency.

Therefore, the ratio of highest to lowest intensity is given as,

  $\frac{I_{\text{highest}}}{I_{\text{lowest}}}=\frac{5}{5\times {10}^{-9}}={10}^9$

Conclusion:

The ratio is ${10}^9$ .

(b)

To determine

To Calculate: The ratio of highest to lowest intensities at given frequency.

Answer to Problem 23P

The ratio of highest to lowest intensity at the given frequency is ${10}^{12}$ .

Explanation of Solution

Given:

Intensity is, $f=100\text{Hz}$

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

Formula used:

  $\text{Intensity}\text{ratio}=\frac{I_{highest}}{I_{lowest}}$

Calculation:

The graph for sound level versus frequency is shown below.

  

The hearing threshold at frequency 5000Hz is about ${10}^{-13}\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.$ , observed from the given graph. The pain threshold is about ${10}^{-1}\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.$ that is the highest frequency.

Therefore, the ratio of highest to lowest intensity is given as,

  $\frac{I_{highest}}{I_{lowest}}=\frac{{\text{10}}^{\text{-1}}\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.}{{\text{10}}^{\text{-13}}\raisebox{1ex}{$\text{W}$}\!\left/ \!\raisebox{-1ex}{${\text{m}}^{\text{2}}$}\right.}={10}^{12}$

Conclusion:

The ratio is ${10}^{12}$ .