(a) You are driving down the highway in your car when a police car sounding its siren overtakes you and passes you. If its frequency at rest is f0, is the frequency you hear while the car is catching up to you higher or lower than f0?(b) What about the frequency you hear after the car has passed you?

Doppler effect is the change in frequency of a (sound) wave in relation to an observer who is moving relative to the wave source. Based on this principle of Doppler effect we can determine the frequency of the police car that is catching up with you (the observer).

Formula for Doppler effect:

fL = ((v+vl)/(v-vs)) fs

fL refers to the frequency of sound which the observer hears (Hz, or 1/s)

v is the speed of sound in air (m/s)

vL refers to the velocity of the observer (m/s)

vs is the velocity of the source (police car) of the sound (m/s)

fs refers to the frequency of sound which the source emits (Hz, or 1/s)

a) As the police car is catching up to you while you're at rest, your velocity as the observer is zero. Hence we substitue vl with 0.

fL = ((v+0)/(v-vs)) fs = (v/(v-vs)) fs

The numerator v is larger than the denominator v -vs. Therefore the observed frequency is greater than the source frequency. Hence, The frequency you hear while the car is catching up to you is higher than f0.

b) Since the car has passed you and the source is moving away from you, vs is therefore negative. Substituting vl with 0 and vs with -vs.

fL = ((v+0)/(v-(-vs))) fs = (v/(v+vs)) fs

The numerator v is smaller than the denominator v -vs. Therefore the observed frequency is smaller than the source frequency. Hence, the frequency you hear after the car has passed you is less than f0.