Chapter 14, Problem 5CQ

Two cars are on the same straight road. Car A moves east at 55 mph and sounds its horn at 625 Hz. Rank from highest to lowest the frequency observed by Car B’s driver in each of these cases: Car B is moving east at 70 mph and is (a) behind and (b) in front of Car A. Car B is moving west at 70 mph and is (c) behind and (d) in front of Car A.

To determine

Rank from highest to lowest the frequency observed by Car B’s driver in each of these cases.

Answer to Problem 5CQ

Solution: $\underset{\_}{f_{O\left(d\right)}>f_{O\left(a\right)}>f_{O\left(b\right)}>f_{O\left(c\right)}}$.

Explanation of Solution

The frequency observed by the car B’s driver can be found using the concept of Doppler effect. Doppler effect is the change in frequency heard by an observer whenever there is relative motion between a source of sound and the observer is called Doppler erect.

Formula for Doppler effect is,

$f_O=f_S\left(\frac{v+v_O}{v-v_S}\right)$

• $f_O$ is the frequency observed by Car B’s driver.
• $f_S$ is the frequency of the source(horn).
• $v$ is the speed of air.
• $v_O$ is the speed of the observer Car B.
• $v_S$ is the speed of the source Car A.

A positive speed is substituted for $v_0$ when the observer moves towards the source and a negative speed is substituted for $v_0$ when the observer moves away from the source. Similarly, a positive speed is substituted for $v_S$ when the source moves towards the observer and a negative speed is substituted for $v_S$ when the source moves away from the observer.

Car A and Car B both moves in the same direction towards east. Here, the observer Car B is moving towards Car A and the source Car A is moving away from Car B. Hence, $v_O=70\text{mph}$, $v_S=-55\text{mph}$

Let $f_O{}_{\left(b\right)}$ frequency heard by the observer in scenario (a)

Substitute $625\text{Hz}$ for $f_S$, $343\text{m}/\text{s}$ for $v$ , $70\text{mph}$ for $v_O$ , $-\text{55}\text{mph}$ for $v_S$ to find $f_{O(a)}$.

$\begin{array}{c}{f}_{O(a)}=625\text{Hz}\left(\frac{343\text{m}/\text{s}+\left(70\text{mph}\times \frac{0.44704\text{m}/\text{s}}{1\text{mph}}\right)}{343\text{m}/\text{s}-\left(-55\text{mph}\times \frac{0.44704\text{m}/\text{s}}{1\text{mph}}\right)}\right)\\ =636\text{Hz}\end{array}$

Formula for Doppler effect is,

$f_O{}_{\left(b\right)}=f_S\left(\frac{v+v_O}{v-v_S}\right)$

• $f_O{}_{\left(b\right)}$ frequency heard by the observer in scenario (b)

Car A and Car B both moves in the same direction towards east. Car B is moving at greater speed than car A. So, Car B is moving away from Car A from behind. Here, the source Car A moves towards the observer and the observer Car B moves away from the source Car A, a negative sign is substituted to the speed of the car $v_O=-70\text{mph}$.

Substitute $625\text{Hz}$ for $f_S$, $343\text{m}/\text{s}$ for $v$ , $-70\text{mph}$ for $v_O$ , $\text{55}\text{mph}$ for $v_S$ to find $f_{O(b)}$.

$\begin{array}{c}{f}_{O(b)}=625\text{Hz}\left(\frac{343\text{m}/\text{s}-\left(70\text{mph}\times \frac{0.44704\text{m}/\text{s}}{1\text{mph}}\right)}{343\text{m}/\text{s}-\left(55\text{mph}\times \frac{0.44704\text{m}/\text{s}}{1\text{mph}}\right)}\right)\\ =612\text{Hz}\end{array}$

Formula for Doppler effect is,

$f_{O\left(c\right)}=f_S\left(\frac{v+v_O}{v-v_S}\right)$

• $f_O{}_{\left(c\right)}$ frequency heard by the observer in scenario (c)

Car A and Car B both moves in the opposite direction. The source Car A is moving away from the Car B and the observer Car B is moving away from the source Car A.

So, $v_S=-55\text{mph}$ $v_O=-\text{70}\text{mph}$

Substitute $625\text{Hz}$ for $f_S$, $343\text{m}/\text{s}$ for $v$ , $-70\text{mph}$ for $v_O$ , $-\text{55}\text{mph}$ for $v_S$ to find $f_{O\left(c\right)}$.

$\begin{array}{c}{f}_{O(c)}=625\text{Hz}\left(\frac{343\text{m}/\text{s}+\left(-70\text{mph}\times \frac{0.44704\text{m}/\text{s}}{1\text{mph}}\right)}{343\text{m}/\text{s}-\left(-55\text{mph}\times \frac{0.44704\text{m}/\text{s}}{1\text{mph}}\right)}\right)\\ =530\text{Hz}\end{array}$

Formula for Doppler effect is,

$f_{O\left(d\right)}=f_S\left(\frac{v+v_O}{v-v_S}\right)$

• $f_O{}_{\left(d\right)}$ frequency heard by the observer in scenario (d)

Car A and Car B both moves in the opposite direction. The source Car A is moving towards Car B and the observer Car B is moving towards the source Car A.

So, $v_S=55\text{mph}$ $v_O=\text{70}\text{mph}$

Substitute $625\text{Hz}$ for $f_S$, $343\text{m}/\text{s}$ for $v$ , $70\text{mph}$ for $v_O$ , $\text{55}\text{mph}$ for $v_S$ to find $f_{O\left(d\right)}$.

$\begin{array}{c}{f}_{O(d)}=625\text{Hz}\left(\frac{343\text{m}/\text{s}+\left(70\text{mph}\times \frac{0.44704\text{m}/\text{s}}{1\text{mph}}\right)}{343\text{m}/\text{s}-\left(55\text{mph}\times \frac{0.44704\text{m}/\text{s}}{1\text{mph}}\right)}\right)\\ =735\text{Hz}\end{array}$

Conclusion:

$\underset{\_}{f_{O\left(d\right)}>f_{O\left(a\right)}>f_{O\left(b\right)}>f_{O\left(c\right)}}$