Question

A stationary police car emits a sound of frequency 1200 Hz that bounces off a car on the highway and returns with a frequency of 1250 Hz. The police car is right next to the highway, so the moving car is traveling directly toward or away from it. (a) How fast was the moving car going? Was it moving toward or away from the police car? (b) What frequency would the police car have received if it had been traveling toward the other car at 20.0 m/s?

Short answer

(a) Speed of car is 7.02m/s and it is moving towards the police officer.

(b) Frequency received by the police car is 1404Hz

Step-by-step solution

Step 1:

Given data:

As by the doppler effect:

$f_L=\frac{v+v_L}{v+v_S}{f}_S$

$f_L=$Frequency observed by the listener

$v=$speed of sound

$v_L=$speed of listener

$f_S=$frequency of source

$v_S=$speed of the source of sound

And here _{$v_L$} is positive as the velocity of the listener from Listener to Source and _{$v_S$} is positive as the velocity of the source from Listener to Source and velocity is negative.

Step 2:

(a) Consider the car as if it is moving away, and if the car has a negative speed, it is moving in the opposite direction.

$v_{car}\to$velocity of moving car

Thus, two doppler shifts:

First case: $v_L\to$it is negative as the listener is moving away from the source.

$$\begin{gathered} f_L=\frac{v+v_L}{v+v_S}{f}_S \\ =\frac{v-v_{Car}}{v+0}\left(1200\mathrm{Hz}\right) \end{gathered}$$

Second case: source is the moving car and the police are the listener

$f_s\to$first shifted frequency, $v_L\to 0,v_S\to$it is positive as the source is moving away from the listener.

role="math" localid="1668071173728" $$\begin{gathered} f_L=\frac{v+v_L}{v+v_S}{f}_S=\frac{v+0}{v+v_{Car}}\times \frac{v-v_{Car}}{v+0}\left(1200\mathrm{Hz}\right)=1250\mathrm{Hz} \\ \frac{v\left(v-v_{Car}\right)}{v\left(v+v_{Car}\right)}=\frac{1250\mathrm{Hz}}{1200\mathrm{Hz}}=\frac{25}{24},\to solve\mathrm{for}{v}_{Car} \\ 25v+25v_{Car}=24v-24v_{Car} \\ 49v_{Car}=-v \\ So{v}_{Car}=-7.02\mathrm{m}/\mathrm{s} \end{gathered}$$

Hence, the speed of the car is 7.02m/s and it is moving toward the police officer.

Step 3:

(b) As $v_{car}\to$velocity of the moving car

Thus, two doppler shifts:

First case: $v_L\to$it is positive as the listener is moving towards from the source,$v_s=v_{police}\to$it is negative as moving towards the listener.

$$\begin{gathered} f_L=\frac{v+v_L}{v+v_S}{f}_S \\ =\frac{v+7.02m/s}{v-20m/s}\left(1200\mathrm{Hz}\right) \\ =1300.07\mathrm{Hz} \end{gathered}$$

Second case: source is the moving car and police is the listener

$f_S=1300.07Hz\to$first shifted frequency, $v_L=20m/s\to$it is positive as it is moving towards the source, $v_S\to$it is negative as the source is moving towards the listener.

$$\begin{gathered} f_L=\frac{v+v_L}{v+v_S}{f}_S \\ =\frac{v+v_{police}}{v-v_{Car}}\left(1300\mathrm{Hz}\right) \\ =\frac{344\mathrm{m}/\mathrm{s}+20\mathrm{m}/\mathrm{s}}{344\mathrm{m}/\mathrm{s}-7.02}\left(1300\mathrm{Hz}\right) \\ =1404\mathrm{Hz} \end{gathered}$$

Hence, the frequency received by the police car is 1404Hz.