Chapter 22, Problem 60E

(a)

To determine

The upper limit of hearing of elephants and rats.

Answer to Problem 60E

The upper limit of hearing of elephants is $4642.37\text{Hz}$; the upper limit of hearing of rats is $1.71\times {10}^5\text{Hz}$.

Explanation of Solution

Write the expression for separation between ears.

  $r\propto m^{3/8}$

Express the above relation in equation form.

  $r=k{m}^{3/8}$        (I)

Rearrange the above expression in terms of $k$.

  $k=\frac{r}{m^{3/8}}$        (II)

Here, $r$ is separation between ears, $m$ is mass of species and $k$ is proportionality constant.

Write the expression for wavelength of sound.

  $\lambda =\frac{c}{f}$

Here, $c$ is speed of sound, $\lambda$ is wavelength of sound and $f$ is frequency of sound.

The separation between ears must be equal to the wavelength of sound that a human ear perceives; it means $r=\lambda$.

Substitute $r$ for $\lambda$ in above expression.

  $r=\frac{c}{f}$        (III)

Rearrange the above expression in terms of $f$.

  $f=\frac{c}{r}$        (IV)

Conclusion:

For humans:

Substitute $344\text{m}/\text{s}$ for $c$ and $19000\text{Hz}$ for $f$ in equation (III).

  $\begin{array}{c}r=\frac{344\text{m}/\text{s}}{19000\text{Hz}}\\ =1.81\times {10}^{-2}\text{m}\end{array}$

Substitute $1.81\times {10}^{-2}\text{m}$ for $r$ and $70\text{kg}$ for $m$ in equation (II).

  $\begin{array}{c}k=\frac{\left(1.81\times {10}^{-2}\text{m}\right)}{{\left(70\text{kg}\right)}^{3/8}}\\ =3.68\times {10}^{-3}\text{m}\cdot {\left(\text{kg}\right)}^{-3/8}\end{array}$

For elephants:

Substitute $3.68\times {10}^{-3}\text{m}\cdot {\left(\text{kg}\right)}^{-3/8}$ for $k$ and $3\times {10}^3\text{kg}$ for $m$ in equation (I).

  $\begin{array}{c}r=\left(3.68\times {10}^{-3}\text{m}\cdot {\left(\text{kg}\right)}^{-3/8}\right){\left(3\times {10}^3\text{kg}\right)}^{3/8}\\ =\left(3.68\times {10}^{-3}\text{m}\cdot {\left(\text{kg}\right)}^{-3/8}\right)\left(20.13{\left(\text{kg}\right)}^{3/8}\right)\\ =7.41\times {10}^{-2}\text{m}\end{array}$

Substitute $7.41\times {10}^{-2}\text{m}$ for $r$ and $344\text{m}/\text{s}$ for $c$ in equation (IV).

  $\begin{array}{c}f=\frac{344\text{m}/\text{s}}{7.41\times {10}^{-2}\text{m}}\\ =4642.37\text{Hz}\end{array}$

For rats:

Substitute $3.68\times {10}^{-3}\text{m}\cdot {\left(\text{kg}\right)}^{-3/8}$ for $k$ and $2\times {10}^{-1}\text{kg}$ for $m$ in equation (I).

  $\begin{array}{c}r=\left(3.68\times {10}^{-3}\text{m}\cdot {\left(\text{kg}\right)}^{-3/8}\right){\left(2\times {10}^{-1}\text{kg}\right)}^{3/8}\\ =\left(3.68\times {10}^{-3}\text{m}\cdot {\left(\text{kg}\right)}^{-3/8}\right)\left(0.5469{\left(\text{kg}\right)}^{3/8}\right)\\ =0.201\times {10}^{-2}\text{m}\end{array}$

Substitute $0.201\times {10}^{-2}\text{m}$ for $r$ and $344\text{m}/\text{s}$ for $c$ in equation (IV).

  $\begin{array}{c}f=\frac{344\text{m}/\text{s}}{0.201\times {10}^{-2}\text{m}}\\ =1.71\times {10}^5\text{Hz}\end{array}$

Thus, the upper limit of hearing of elephants is $4642.37\text{Hz}$; the upper limit of hearing of rats is $1.71\times {10}^5\text{Hz}$.

(b)

To determine

The experimental scaling exponent for upper limit frequency versus mass.

Answer to Problem 60E

The experimental scaling exponent for upper limit frequency versus mass is  $-0.20$.

Explanation of Solution

Write the expression for upper hearing limit.

  $f\propto m^n$

Here, $f$ is upper limit frequency, $m$ is mass of mammal and $n$ is the power corresponding to mass.

Express the above relation in ratio for two mammals.

       $\frac{f_1}{f_2}={\left(\frac{m_1}{m_2}\right)}^n$

Here, subscript1 is used for first mammal and subscript2 is used for second mammals.

Apply logarithm on both sides in above expression.

  $\log \left(\frac{f_1}{f_2}\right)=\log {\left(\frac{m_1}{m_2}\right)}^n$

Rearrange the above expression in terms of $n$.

  $n=\frac{\log \left(f_1/f_2\right)}{\log \left(m_1/m_2\right)}$        (V)

Conclusion:

Substitute $11,000\text{Hz}$ for $f_1$ , $72,000\text{Hz}$ for $f_2$ , $3\times {10}^3\text{kg}$ for $m_1$ and $2\times {10}^{-1}\text{kg}$ for $m_2$ in equation (V).

  $\begin{array}{c}n=\frac{\log \left(11,000\text{Hz}/72,000\text{Hz}\right)}{\log \left(3\times {10}^3\text{kg}/2\times {10}^{-1}\text{kg}\right)}\\ =\frac{\log \left(11/72\right)}{\log \left(1.5\times {10}^4\right)}\\ =-\frac{0.816}{4.176}\\ \approx -0.20\end{array}$

Thus, experimental scaling exponent for upper limit frequency versus mass is $-0.20$.