Chapter 12, Problem 91GP

To determine

The factor by which the dispalacement amplitude varies.

Answer to Problem 91GP

  ${10}^6$

Explanation of Solution

Given:

The intensity at threshold frequency of the human ear, ${\mathrm{I}}_0={10}^{-12}{\text{W/m}}^{\text{2}}$

The intensity of threshold of pain, $\mathrm{I}=1{\text{W/m}}^{\text{2}}$

Formula used:

The intensity of the sound wave is given as:

  $\mathrm{I}=2{\mathrm{\pi }}^2\mathrm{\nu }\mathrm{\rho }{\mathrm{f}}^2{\mathrm{A}}^2$

Considering the product of constant parameters in the above expression as $\mathrm{k}$ .

Thus,

  $\mathrm{I}=\mathrm{k}{(\mathrm{f}\mathrm{A})}^2$

Here, $\mathrm{f}$ is the frequency of the sound wave.

  $\mathrm{A}$ is the amplitude of the sound wave.

  $\mathrm{\rho }$ is the density of the medium.

  $\mathrm{v}$ is the velocity of the sound wave.

Calculation:

Substitute the given values:

  $\begin{array}{l}$ I_0=k{(f{A}_0)}^2$$ I_f=k{(f{A}_f)}^2$\frac{{\mathrm{I}}_0}{{\mathrm{I}}_{\mathrm{f}}}=\frac{{\mathrm{A}}_0{}^2}{{\mathrm{A}}_{\mathrm{f}}{}^2}\\ \frac{{10}^{-12}}{1}={\left(\frac{{\mathrm{A}}_0}{{\mathrm{A}}_{\mathrm{f}}}\right)}^2\\ \left(\frac{{\mathrm{A}}_0}{{\mathrm{A}}_{\mathrm{f}}}\right)={10}^{-12/2}={10}^{-6}\\ \left(\frac{{\mathrm{A}}_{\mathrm{f}}}{{\mathrm{A}}_0}\right)={10}^6\end{array}$

Conclusion:

The amplitude will increase by a factor ${10}^6$ .