Chapter 17, Problem 71CP

(a)

To determine

The frequency heard by passenger in the car.

Answer to Problem 71CP

The frequency heard by passenger in the car is $\underset{\_}{531\text{Hz}}$.

Explanation of Solution

Write the expression for the frequency of sound heard by the passengers in the car.

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

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

The car stopped and not moving hence the speed of the observer will be zero.

Rewrite equation (I) by substituting $0$ for ${\upsilon }_o$.

  $f^{\prime }=f\left(\frac{\upsilon }{\upsilon -{\upsilon }_s\cos {\theta }_s}\right)$                                                                                              (II)

The train is $40\text{m}$ from the intersection, and car is $30\text{m}$ from the intersection.

Hence using hypotenuse theorem the distance between train and the car is.

  $h^2=\sqrt{a^2+b^2}$                                                                                                         (III)

Here, $h$ is the hypotenuse, $a$ is the altitude, and $b$ is the base length.

Write the expression for $\cos {\theta }_s$.

  $\cos {\theta }_s=\frac{b}{h}$                                                                                                               (IV)

Conclusion:

Substitute, $40\text{m}$ for $b$, and $30\text{m}$ for $a$ in equation (III).

  $\begin{array}{c}{h}^2=\sqrt{{\left(40\text{m}\right)}^2+{\left(30\text{m}\right)}^2}\\ h=50\text{m}\end{array}$

Substitute, $40\text{m}$ for $b$, and $50\text{m}$ for $h$ in equation (IV).

  $\begin{array}{c}\cos {\theta }_s=\frac{40\text{m}}{50\text{m}}\\ =\frac{4}{5}\end{array}$

Substitute, $343\text{m}/\text{s}$ for $\upsilon$, $\frac{4}{5}$ for $\cos {\theta }_s$, $500\text{Hz}$ for $f$, and $25.0\text{m}/\text{s}$ for ${\upsilon }_s$ in equation (II).

  $\begin{array}{c}{f}^{\prime }=\left(500\text{Hz}\right)\left(\frac{343\text{m}/\text{s}}{343\text{m}/\text{s}-\left(25.0\text{m}/\text{s}\right)\left(4/5\right)}\right)\\ =531\text{Hz}\end{array}$

Therefore, the frequency heard by passenger in the car is $\underset{\_}{531\text{Hz}}$.

(b)

To determine

The range of frequencies heard by the passengers in the car.

Answer to Problem 71CP

The range of frequencies heard by the passengers in the car is in between $\underset{\_}{466\text{Hz-536}\text{Hz}}$.

Explanation of Solution

As train approaches and departs, the angle ${\theta }_s$ varies from $0°$ to $180°$. The maximum frequency heard by the passengers when ${\theta }_s$ equal to $0°$, and minimum when ${\theta }_s$ equal to $180°$.

Write the expression for maximum frequency.

  ${f^{\prime }}_{\text{max}}=f\frac{\upsilon }{\upsilon -{\upsilon }_s\cos 0°}$                                                                                              (V)

Write the expression for maximum frequency.

  ${f^{\prime }}_{\text{min}}=f\frac{\upsilon }{\upsilon -{\upsilon }_s\cos 180°}$                                                                                          (VI)

Conclusion:

Substitute, $343\text{m}/\text{s}$ for $\upsilon$, $0$ for $\cos {\theta }_s$, $500\text{Hz}$ for $f$, and $25.0\text{m}/\text{s}$ for ${\upsilon }_s$ in equation (V).

  $\begin{array}{c}{f}^{\prime }=\left(500\text{Hz}\right)\left(\frac{343\text{m}/\text{s}}{343\text{m}/\text{s}-\left(25.0\text{m}/\text{s}\right)\left(1\right)}\right)\\ =539\text{Hz}\end{array}$

Substitute, $343\text{m}/\text{s}$ for $\upsilon$, $180°$ for $\cos {\theta }_s$, $500\text{Hz}$ for $f$, and $25.0\text{m}/\text{s}$ for ${\upsilon }_s$ in equation (VI).

  $\begin{array}{c}{f}^{\prime }=\left(500\text{Hz}\right)\left(\frac{343\text{m}/\text{s}}{343\text{m}/\text{s}-\left(25.0\text{m}/\text{s}\right)\left(-1\right)}\right)\\ =466\text{Hz}\end{array}$

Therefore, the range of frequencies heard by the passengers in the car is in between $\underset{\_}{466\text{Hz-536}\text{Hz}}$.

(b)

To determine

The frequency heard by the passengers in car when car is trying to beat the train.

Answer to Problem 71CP

The frequency heard by the passengers in car when car is trying to beat the train is $\underset{\_}{568\text{Hz}}$.

Explanation of Solution

Write the Doppler equation.

  $f^{\prime }=f\frac{\upsilon +{\upsilon }_0\cos {\theta }_o}{\upsilon -{\upsilon }_s\cos {\theta }_s}$                                                                                               (VII)

Let ${\theta }_o$ be the angle made by distance from intersection to car and distance between train and car.

Write the expression for $\cos {\theta }_o$.

  $\cos {\theta }_o=\frac{a}{h}$                                                                                                             (VIII)

Conclusion:

Substitute, $30\text{m}$ for $b$, and $50\text{m}$ for $h$ in equation (VIII).

  $\begin{array}{c}\cos {\theta }_s=\frac{30\text{m}}{50\text{m}}\\ =\frac{3}{5}\end{array}$

Substitute, $343\text{m}/\text{s}$ for $\upsilon$, $\frac{4}{5}$ for $\cos {\theta }_s$, $\frac{3}{5}$ for $\cos {\theta }_s$, $40\text{m}/\text{s}$ for ${\upsilon }_o$, $500\text{Hz}$ for $f$, and $25.0\text{m}/\text{s}$ for ${\upsilon }_s$ in equation (VII).

  $\begin{array}{c}{f}^{\prime }=\left(500\text{Hz}\right)\left(\frac{343\text{m}/\text{s+}40\text{m}/\text{s}\left(3/5\right)}{343\text{m}/\text{s}-\left(25.0\text{m}/\text{s}\right)\left(4/5\right)}\right)\\ =568\text{Hz}\end{array}$

Therefore, the frequency heard by the passengers in car when car is trying to beat the train is $\underset{\_}{568\text{Hz}}$.