Chapter 17, Problem 43P

(a)

To determine

The maximum linear speed of the heart wall.

Answer to Problem 43P

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

Explanation of Solution

Write the expression for maximum linear speed in terms of angular frequency.

  ${\upsilon }_{\max }=\omega A$                                                                                                                   (I)

Here, $\omega$ is the angular frequency, $A$ is the amplitude.

Write the expression for angular frequency.

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

Here, $f$ is the frequency.

Conclusion:

Substitute, $115{\text{min}}^{-1}$ for $f$ in equation (II).

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

Substitute, $12.0\text{rad}/\text{s}$ for $\omega$, and $1.80\times {10}^{-3}\text{m}$ for $A$ in equation (I).

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

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 43P

The maximum change in frequency is $\underset{\_}{28.9\text{Hz}}$.

Explanation of Solution

Write the expression for shift in frequency.

  $f^{\prime }=f\frac{\left(\upsilon +{\upsilon }_o\right)}{\left(\upsilon -{\upsilon }_s\right)}$                                                                                                          (I)

Here, $f$ is the frequency of the source, $\upsilon$ is the speed of the sound, ${\upsilon }_o$ is the speed of the observer, and ${\upsilon }_s$ is the speed of the source.

The source is at rest, and the speed of the source is equal to zero.

Substitute, $0$ for ${\upsilon }_s$ in equation (I).

  $f^{\prime }=f\frac{\left(\upsilon +{\upsilon }_o\right)}{\left(\upsilon \right)}$                                                                                                        (II)

Let the change in frequency between the sound that arrives at the wall of the baby’s heart and the sound emitted by the source be $\Delta f^{\prime }$, it is the difference of the shift in frequency and the original frequency of the source.

  $\begin{array}{c}\Delta f^{\prime }=f\frac{\left(\upsilon +{\upsilon }_o\right)}{\left(\upsilon \right)}-f\\ =\frac{\upsilon f+{\upsilon }_{o}f-\upsilon f}{\upsilon }\\ =f\frac{{\upsilon }_o}{\upsilon }\end{array}$                                                                                               (II)

Conclusion:

Substitute, $2000000\text{Hz}$ for $f$, $0.0217\text{m}/\text{s}$ for ${\upsilon }_o$, and $1500\text{m}/\text{s}$ for $\upsilon$ in equation (II).

  $\begin{array}{c}\Delta f^{\prime }=\left(2000000\text{Hz}\right)\frac{0.0217\text{m}/\text{s}}{1500\text{m}/\text{s}}\\ =28.9\text{Hz}\end{array}$

Therefore, the maximum change in frequency 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 43P

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

Let $\Delta f^{″}$ is the maximum change in frequency between the reflected sound received by the detector and that emitted by the source.

In this case the speed of the observer is zero.

The expression for the change in frequency.

  $f^{″}=f^{\prime }\left(\frac{\upsilon }{\upsilon -{\upsilon }_s}\right)$                                                                                                     (III)

The expression for $\Delta f^{″}$.

  $\begin{array}{c}\Delta f^{″}=f^{″}-f\\ =f^{\prime }\left(\frac{\upsilon }{\upsilon -{\upsilon }_s}\right)-f\end{array}$                                                                                             (IV)

Substitute, equation (II) in (IV).

  $\begin{array}{c}\Delta f^{″}=f\left(\frac{\upsilon +{\upsilon }_o}{\upsilon }\right)\left(\frac{\upsilon }{\upsilon -{\upsilon }_s}\right)-f\\ =f\left(\frac{\upsilon \left(\upsilon +{\upsilon }_o\right)-\upsilon \left(\upsilon -{\upsilon }_s\right)}{\upsilon \left(\upsilon -{\upsilon }_s\right)}\right)\\ =f\left[\frac{{\upsilon }_s+{\upsilon }_o}{\upsilon -{\upsilon }_s}\right]\end{array}$                                                                              (V)

The velocity of both source and observer are same, and the speed of sound is greater than the speed of the source. Hence equation (V) becomes.

  $\Delta f^{″}=f\left[\frac{2{\upsilon }_s}{\upsilon }\right]$                                                                                                        (VI)

Conclusion:

Substitute, $2.00\times {10}^6\text{Hz}$ for $f$, $0.0434\text{m}/\text{s}$ for ${\upsilon }_s$, $1500\text{m}/\text{s}$ for $\upsilon$ in equation (VI).

  $\begin{array}{c}\Delta f^{″}=2.00\times {10}^6\text{Hz}\left[\frac{2\times 0.0434\text{m}/\text{s}}{1500\text{m}/\text{s}}\right]\\ =57.9\text{Hz}\end{array}$

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