The complete question is:
You are driving along a highway at 35.0 m/s when you hear the siren of a police car approaching you from behind at constant speed and you perceive the frequency as 1340 Hz. You are relieved that he is in pursuit of a different driver when he continues past you, but now you perceive the frequency as 1300 Hz. What is the speed of the police car? The speed of sound in air is 343m/s.
Answer:
V_s = 30 m/s
Step-by-step explanation:
The change in frequency observation occur due to doppler effect is given by the equation;
f_o = [(V ± V_o)/(V ∓ V_s)]f_s
Where;
f_o is observed frequency
f_source is frequency of the source
V is speed of sound
V_o is velocity of the observer
V_s is velocity of the source
Now, When the police is coming to you , you hear a higher frequency and thus, we'll use the positive sign on the numerator and negative sign on denominator.
Thus,
f_o = [(V + V_o)/(V - V_s)]f_s
Plugging in relevant values, we have;
1340 = [(343 + 35)/(343 - V_s)]f_s
1340 = [(378)/(343 - V_s)]f_s - - (eq1)
when the police is passing you , you hear a lesser frequency, and thus, we'll use the negative sign on the numerator and positive sign on denominator. thus;
f_o = [(V - V_o)/(V + V_s)]f_s
Plugging in the relevant values to get;
1300 = [(343 - 35)/(343 + V_s)]f_s
1300 = [(308)/(343 + V_s)]f_s - - eq2
Divide eq2 by eq1 with f_s canceling out to give
1340/1300 = [(378)/(343 - V_s)]/[(308)/(343 + V_s)]
V_s = 30 m/s