Chapter 10, Problem 40P

(a)

To determine

Whether the maximum velocity and the greatest acceleration of the eardrum are for high-frequency sounds or low-frequency sounds for a given vibration amplitude.

Answer to Problem 40P

The maximum velocity and the greatest acceleration of the eardrum are for high-frequency sounds for a given vibration amplitude.

Explanation of Solution

Write the expression for the maximum velocity.

$v_m=\omega A$ (I)

Here, $v_m$ is the maximum velocity, $\omega$ is the angular frequency and $A$ is the vibration amplitude

Write the expression for $\omega$.

$\omega =2\pi f$ (II)

Here, $f$ is the frequency

Put equation (II) in equation (I).

$v_m=2\pi fA$ (III)

Equation (III) implies that $v_m\propto f$.

Write the expression for the maximum acceleration.

$a_m={\omega }^{2}A$ (IV)

Here, $a_m$ is the maximum acceleration

Put equation (II) in equation (IV).

$\begin{array}{c}{a}_m={\left(2\pi f\right)}^{2}A\\ =4{\pi }^2{f}^{2}A\end{array}$ (V)

Equation (V) implies that $a_m\propto f^2$.

Conclusion:

Since maximum velocity is proportional to frequency and the maximum acceleration is proportional to square of the frequency, maximum velocity and acceleration will occur for high frequency sounds.

Therefore, the maximum velocity and the greatest acceleration of the eardrum are for high-frequency sounds for a given vibration amplitude.

(b)

To determine

The maximum velocity and acceleration of the eardrum for vibrations of amplitude $1.0\times {10}^{-8}\text{ m}$ at frequency of $20.0\text{ Hz}$.

Answer to Problem 40P

The maximum velocity of the eardrum for vibrations of amplitude $1.0\times {10}^{-8}\text{ m}$ at frequency of $20.0\text{ Hz}$ is $1.3\times {10}^{-6}\text{ m/s}$ and the maximum acceleration is $1.6\times {10}^{-4}{\text{ m/s}}^2$.

Explanation of Solution

Equation (III) can be used to find the maximum velocity and equation (V) can be used to find the maximum acceleration.

Conclusion:

Substitute $20.0\text{ Hz}$ for $f$ and $1.0\times {10}^{-8}\text{ m}$ for $A$ in equation (III) to find $v_m$.

$\begin{array}{c}{v}_m=2\pi \left(20.0\text{ Hz}\right)\left(1.0\times {10}^{-8}\text{ m}\right)\\ =1.3\times {10}^{-6}\text{ m/s}\end{array}$

Substitute $20.0\text{ Hz}$ for $f$ and $1.0\times {10}^{-8}\text{ m}$ for $A$ in equation (V) to find $a_m$.

$\begin{array}{c}{a}_m=4{\pi }^2{\left(20.0\text{ Hz}\right)}^2\left(1.0\times {10}^{-8}\text{ m}\right)\\ =1.6\times {10}^{-4}{\text{ m/s}}^2\end{array}$

Therefore, the maximum velocity of the eardrum for vibrations of amplitude $1.0\times {10}^{-8}\text{ m}$ at frequency of $20.0\text{ Hz}$ is $1.3\times {10}^{-6}\text{ m/s}$ and the maximum acceleration is $1.6\times {10}^{-4}{\text{ m/s}}^2$.

(c)

To determine

The maximum velocity and acceleration of the eardrum for vibrations of amplitude $1.0\times {10}^{-8}\text{ m}$ at frequency of $20.0\text{ kHz}$.

Answer to Problem 40P

The maximum velocity of the eardrum for vibrations of amplitude $1.0\times {10}^{-8}\text{ m}$ at frequency of $20.0\text{ kHz}$ is $0.0013\text{ m/s}$ and the maximum acceleration is $160{\text{ m/s}}^2$.

Explanation of Solution

Equation (III) can be used to find the maximum velocity and equation (V) can be used to find the maximum acceleration.

Conclusion:

Substitute $20.0\text{ kHz}$ for $f$ and $1.0\times {10}^{-8}\text{ m}$ for $A$ in equation (III) to find $v_m$.

$\begin{array}{c}{v}_m=2\pi \left(20.0\text{ kHz}\right)\left(1.0\times {10}^{-8}\text{ m}\right)\\ =2\pi \left(20.0\times {10}^3\text{ Hz}\right)\left(1.0\times {10}^{-8}\text{ m}\right)\\ =0.0013\text{ m/s}\end{array}$

Substitute $20.0\text{ kHz}$ for $f$ and $1.0\times {10}^{-8}\text{ m}$ for $A$ in equation (V) to find $a_m$.

$\begin{array}{c}{a}_m=4{\pi }^2{\left(20.0\text{ kHz}\right)}^2\left(1.0\times {10}^{-8}\text{ m}\right)\\ =4{\pi }^2{\left(20.0\times {10}^3\text{ Hz}\right)}^2\left(1.0\times {10}^{-8}\text{ m}\right)\\ =160{\text{ m/s}}^2\end{array}$

Therefore, the maximum velocity of the eardrum for vibrations of amplitude $1.0\times {10}^{-8}\text{ m}$ at frequency of $20.0\text{ kHz}$ is $0.0013\text{ m/s}$ and the maximum acceleration is $160{\text{ m/s}}^2$.