Chapter 13, Problem 40P

(a)

To determine

The maximum linear speed of the heart wall.

Answer to Problem 40P

The maximum linear speed of the heart wall is $\underset{\_}{0.0217\text{m}/\text{s}}$

Explanation of Solution

Write the expression for the maximum speed.

    $v_{\max }=\omega A$        (II)

Here, $v_{\text{max}}$ is the maximum speed, $\omega$ is the angular velocity, $A$ is the amplitude.

Write the expression for the angular velocity.

    $\omega =2\pi f$        (III)

Here, $f$ is the frequency of the wave.

Conclusion:

Substitute $115\text{beats per minute}$ for $f$ in equation (III) to find $\omega$.

    $\begin{array}{c}\omega =2\pi \left(\frac{115{\text{min}}^{-1}}{60.0\text{s}/\text{min}}\right)\\ =12.0\text{rad}/\text{s}\end{array}$

Substitute $12.0\text{rad}/\text{s}$ for $\omega$ and $1.80\text{mm}$ for $A$ in equation (III) to find $v_{\text{max}}$.

    $\begin{array}{c}{v}_{\text{max}}=\left(12.0\text{rad}/\text{s}\right)\left(1.80\text{mm}\times \frac{1\text{m}}{{10}^3\text{mm}}\right)\\ =0.0217\text{m}/\text{s}\end{array}$

Therefore, the maximum linear speed of the heart wall is $\underset{\_}{0.0217\text{m}/\text{s}}$

(b)

To determine

The maximum change in frequency between the sound that arrives at the wall of the baby’s heart and the sound emitted by the source.

Answer to Problem 40P

The maximum change in frequency between the sound that arrives at the wall of the baby’s heart and the sound emitted by the source is $\underset{\_}{28.9\text{Hz}}$.

Explanation of Solution

The frequency of the wave when the heart approaches the source at maximum velocity is given by,

    $f^{\prime }=f\left(\frac{v+v_w}{v}\right)$        (III)

Here, $f^{\prime }$ is the frequency of the wave when the heart approaches the source at maximum velocity, $v$, $v_w$ is the speed of the heart wall.

The maximum change in frequency is,

    $\Delta f_{\text{max}}=f^{\prime }-f$        (IV)

Use equation (III) in (IV),

    $\begin{array}{c}\Delta f_{\text{max}}=f\left(\frac{v+v_w}{v}\right)-f\\ =f\left(\frac{v_w}{v}\right)\end{array}$        (V)

Conclusion:

Substitute $2000000\text{Hz}$ for $f$, $0.0217\text{m}/\text{s}$ for $v_w$ and $1.50\text{km}/\text{s}$ for $v$ in equation (V) to find $\Delta f_{\text{max}}$.

    $\begin{array}{c}\Delta f_{\text{max}}=\left(2000000\text{Hz}\right)\left(\frac{0.0217\text{m}/\text{s}}{1.50\text{km}/\text{s}\times \frac{1\text{m}}{{10}^{-3}\text{km}}}\right)\\ =28.9\text{Hz}\end{array}$

Therefore, The maximum change in frequency between the sound that arrives at the wall of the baby’s heart and the sound emitted by the source is $\underset{\_}{28.9\text{Hz}}$.

(c)

To determine

The maximum change in frequency between the reflected sound received by the detector and that emitted by the source.

Answer to Problem 40P

The maximum change in frequency between the reflected sound received by the detector and that emitted by the source is $\underset{\_}{57.9\text{Hz}}$.

Explanation of Solution

The frequency of the wave when the heart approaches the source at maximum velocity is,

    ${f^{\prime }}^{\prime }=f^{\prime }\left(\frac{v}{v-v_w}\right)$        (VI)

Use equation (VI) for the change in frequency when the heart wall is a moving source.

    $\Delta {f^{\prime }}^{\prime }=f^{\prime }\left(\frac{v}{v-v_s}\right)-f$        (VII)

Here, $v_s$ is the speed of the source.

Use equation (III) in equation (VII),

    $\Delta {f^{\prime }}^{\prime }=f\left(\frac{v+v_w}{v}\right)\left(\frac{v}{v-v_s}\right)-f$        (VIII)

Solve equation (VIII) for $\Delta {f^{\prime }}^{\prime }$,

    $\begin{array}{c}\Delta {f^{\prime }}^{\prime }=f\left[\frac{v\left(v+v_w\right)}{v\left(v-v_s\right)}-\left(\frac{v\left(v-v_s\right)}{v\left(v-v_s\right)}\right)\right]\\ =f\left(\frac{v_0+v}{v-v_s}\right)\end{array}$        (IX)

Conclusion:

The velocity of the source and the observer in expression (IX) refers to the movement of the heart wall, and the velocity of the sound wave is much greater than those velocities,

    $\Delta {f^{\prime }}^{\prime }=f\left(\frac{2v_s}{v}\right)$        (X)

Substitute $2.00\times {10}^6\text{Hz}$ for $f$, $0.0217\text{m}/\text{s}$ for $v_s$ and $1.50\text{km}/\text{s}$ for $v$ in equation (X) to find $\Delta {f^{\prime }}^{\prime }$.

    $\begin{array}{c}\Delta f^{″}=\left(2.00\times {10}^6\text{Hz}\right)\left(\frac{2\times 0.0217\text{m}/\text{s}}{1.50\text{km}/\text{s}\times \frac{1\text{m}}{{10}^{-3}\text{km}}}\right)\\ =57.9\text{Hz}\end{array}$

Therefore, the maximum change in frequency between the reflected sound received by the detector and that emitted by the source is $\underset{\_}{57.9\text{Hz}}$.