Chapter 12, Problem 63P

(a)

To determine

The maximum speed of an element of air in the sound wave.

Answer to Problem 63P

The maximum speed of an element of air in the sound wave is $6.7\times {10}^{-8}\text{ m/s}$ .

Explanation of Solution

Write the expression for the intensity of the sound.

$I=\frac{p_0^2}{2\rho v}$ (I)

Here, $I$ is the intensity of the sound, $p_0$ is the pressure amplitude, $\rho$ is the mass density of the medium and $v$ is the speed of the sound

Write the equation for $p_0$ .

$p_0=\omega v\rho s_0$ (II)

Here, $\omega$ is the angular frequency of the sound and $s_0$ is the amplitude of the sound wave

Put the above equation in equation (I).

$\begin{array}{c}I=\frac{{\left(\omega v\rho s_0\right)}^2}{2\rho v}\\ =\frac{{\left(\omega s_0\right)}^2\rho v}{2}\end{array}$ (II)

Write the equation for the maximum speed of the sound.

$v_m=\omega s_0$

Here, $v_m$ is the maximum speed of the sound wave

Put the above equation in equation (II) and rewrite it for $v_m$ .

$\begin{array}{c}I=\frac{v_m^2\rho v}{2}\\ v_m^2=\frac{2I}{\rho v}\\ v_m=\sqrt{\frac{2I}{\rho v}}\end{array}$ (III)

Conclusion:

Given that the intensity of the sound is $1.0\times {10}^{-12}{\text{ W/m}}^2$ , speed of the sound is $340\text{ m/s}$ and the density of the air is $1.3{\text{ kg/m}}^3$ .

Substitute $1.0\times {10}^{-12}{\text{ W/m}}^2$ for $I$ , $1.3{\text{ kg/m}}^3$ for $\rho$ and $340\text{ m/s}$ for $v$ in equation (III) to find $v_m$ .

$\begin{array}{c}{v}_m=\sqrt{\frac{2\left(1.0\times {10}^{-12}{\text{ W/m}}^2\right)}{\left(1.3{\text{ kg/m}}^3\right)\left(340\text{ m/s}\right)}}\\ =6.7\times {10}^{-8}\text{ m/s}\end{array}$

Therefore, the maximum speed of an element of air in the sound wave is $6.7\times {10}^{-8}\text{ m/s}$ .

(b)

To determine

The average kinetic energy of the eardrum.

Answer to Problem 63P

The average kinetic energy of the eardrum is $1\times {10}^{-19}\text{ J}$ .

Explanation of Solution

Write the equation for the average kinetic energy.

$K_{\text{av}}=\frac{K_m}{2}$ (IV)

Here, $K_{\text{av}}$ is the average kinetic energy and $K_m$ is the maximum kinetic energy

Write the equation for $K_m$ .

$K_m=\frac{1}{2}m{v}_m^2$

Here, $m$ is the mass of eardrum

Put the above equation in equation (IV).

$\begin{array}{c}{K}_{\text{av}}=\frac{1}{2}\left(\frac{1}{2}m{v}_m^2\right)\\ =\frac{1}{4}m{v}_m^2\end{array}$ (V)

Conclusion:

Given that the mass of the eardrum is $0.1\text{ g}$ .

Substitute $1.0\text{ kHz}$ for $f_s$ , $0.50$ for $v_o/v$ and $-0.50$ for $v_s/v$ in equation (II) to find $f_o$ .

$\begin{array}{c}{f}_o=\left(1.0\text{ kHz}\right)\frac{1-0.50}{1+0.50}\\ =330\text{ Hz}\end{array}$

Therefore, the observed frequency if the source and observer are moving away from each other is $330\text{ Hz}$ .

(c)

To determine

Comparison of the average kinetic energy of the eardrum due to collisions with air molecules in the absence of sound wave with result of part (b).

Answer to Problem 63P

The average kinetic energy of the eardrum with collisions with air molecules in the presence of sound wave at the threshold human hearing is of the order of ten times that it can be in the absence of a sound wave.

Explanation of Solution

Write the value of average kinetic energy of the eardrum due to collisions with air molecules in the absence of sound wave.

${K^{\prime }}_{\text{av}}={10}^{-20}\text{ J}$

Here, ${K^{\prime }}_{\text{av}}$ is the value of average kinetic energy of the eardrum due to collisions with air molecules in the absence of sound wave

Conclusion:

Comparison of $K_{\text{av}}$ and ${K^{\prime }}_{\text{av}}$ gives that $K_{\text{av}}$ is ten times ${K^{\prime }}_{\text{av}}$ and the ear is as sensitive it can be.

Therefore, the average kinetic energy of the eardrum with collisions with air molecules in the presence of sound wave at the threshold human hearing is of the order of ten times that it can be in the absence of a sound wave.