Chapter 13, Problem 71P

The Doppler equation presented in the text is valid when the motion between the observer and the source occurs on a straight line so that the source and observer are moving either directly toward or directly away from each other. If this restriction is relaxed, one must use the more general Doppler equation

$f\prime =\left(\frac{v+v_o\cos {\theta }_o}{v-v_s\cos {\theta }_s}\right)f$

where θ_o and θ_s are defined in figure P13.7la. Use the preceding equation to solve the following problem. A train moves at a constant speed of v = 25.0 m/s toward the intersection shown in Figure P13.71b. A car is stopped near the crossing, 30.0 m from the tracks. The train’s horn emits a frequency of 500 Hz when the train is 40.0 m from the intersection. (a) What is the frequency heard by the passengers in the car? (b) If the train emits this sound continuously and the car is stationary at this position long before the train arrives until long after it leaves, what range of frequencies do passengers in the car hear? (c) Suppose the car is foolishly trying to beat the train to the intersection and is traveling at 40.0 m/s toward the tracks. When the car is 30.0 m from the tracks and the train is 40.0 m from the intersection, what is the frequency heard by the passengers in the car now?

To determine

The frequency heard by the passengers in the car.

Answer to Problem 71P

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

Explanation of Solution

The Doppler equation is given by,

    $f^{\prime }=\left(\frac{v+v_0\cos {\theta }_0}{v-v_s\cos {\theta }_s}\right)f$        (I)

Here, $f^{\prime }$ is the new frequency, $v$ is the speed , $v_0$ is the motion of the observer, $v_s$ is the motion of the source.

In this case the value of $v_0$ is $0\text{m}/\text{s}$, equation (I) changes to,

    $f^{\prime }=\frac{v}{v-v_s\cos {\theta }_s}f$        (II)

Conclusion:

Since the train is $40.0\text{m}$ from the intersection and the car is $30.0\text{m}$ from the intersection,

    $\text{cos}{\theta }_s=\frac{4}{5}$

Substitute $343\text{m}/\text{s}$ for $v$, $25.0\text{m}/\text{s}$ for $v_S$ and $0.800$ for $\cos {\theta }_S$ and $500\text{Hz}$ for $f$ in equation (II) to find $f^{\prime }$.

    $\begin{array}{c}{f}^{\prime }=\frac{343\text{m}/\text{s}}{343\text{m}/\text{s}-0.800\left(25.0\text{m}/\text{s}\right)}\left(500\text{Hz}\right)\\ =531\text{Hz}\end{array}$

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

To determine

The range of frequencies that heard by the passenger.

Answer to Problem 71P

The range of frequencies that heard by the passenger is from $\underset{\_}{539\text{Hz}}$ to $\underset{\_}{466\text{Hz}}$.

Explanation of Solution

The value of ${\theta }_s$ varies from $0°$ to $180°$. frequency is given by,

    ${f^{\prime }}_{\text{max}}=\frac{v}{v-v_S\cos 0°}f$        (III)

    ${f^{\prime }}_{\text{min}}=\frac{v}{v-v_S\cos 180°}f$        (IV)

Conclusion:

Substitute $343\text{m}/\text{s}$ for $v$, $25.0\text{m}/\text{s}$ for $v_S$ $500\text{Hz}$ for $f$ in equation (III) to find ${f^{\prime }}_{\text{max}}$.

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

Substitute $343\text{m}/\text{s}$ for $v$, $25.0\text{m}/\text{s}$ for $v_S$ $500\text{Hz}$ for $f$ in equation (IV) to find ${f^{\prime }}_{\text{min}}$.

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

Therefore, The range of frequencies that heard by the passenger is from $\underset{\_}{539\text{Hz}}$ to $\underset{\_}{466\text{Hz}}$.

To determine

The frequency heard by the passengers in the car when the car is $30.0\text{m}$ from the tracks and the train is $40.0\text{m}$ from the intersection.

Answer to Problem 71P

The frequency heard by the passengers in the car when the car is $30.0\text{m}$ from the tracks and the train is $40.0\text{m}$ from the intersection is $\underset{\_}{568\text{Hz}}$.

Explanation of Solution

Since the rain is $40.0\text{m}$ from the intersection and the car is $30.0\text{m}$ from the intersection is

    $\cos {\theta }_0=\frac{3}{5}$

Substitute $343\text{m}/\text{s}$ for $v$, $25.0\text{m}/\text{s}$ for $v_S$, $40.0\text{m}/\text{s}$ for $v_0$ $500\text{Hz}$ for $f$ in equation (III) to find ${f^{\prime }}_{\text{max}}$.

    $f^{\prime }=\frac{343\text{m}/\text{s}}{}$ $\begin{array}{c}{f}^{\prime }=\frac{343\text{m}/\text{s}+0.600\left(40.0\text{m}/\text{s}\right)}{343\text{m}/\text{s}+0.800\left(25.0\text{m}/\text{s}\right)}\left(500\text{Hz}\right)\\ =568\text{Hz}\end{array}$

Conclusion:

Therefore, The frequency heard by the passengers in the car when the car is $30.0\text{m}$ from the tracks and the train is $40.0\text{m}$ from the intersection is $\underset{\_}{568\text{Hz}}$.