Question

A car approaching a stationary observer emits 450. hz from its horn. if the observer detects a frequency pf 470. hz, how fast is the car moving? the speed of sound is 343 m/s.

Answer 1

The relationship between the frequency heard by the observer in motion and the original frequency of the sound is given by (Doppler effect)
$f'= \frac{v}{v-v_s}f$
where
f' is the frequency heard by the observer
v is the speed of the wave (the speed of sound)
$v_s$ is the speed of the source relative to the observer (= the speed of the car), and it is negative when the source is approaching the observer
f is the original frequency of the sound

By re-arranging the formula, we get
$v_s=v( 1- \frac{f'}{f})$
and by plugging the data of the problem into the equation, we find
$v_s = (343 m/s)( 1- \frac{470 Hz}{450 Hz})=-15.2 m/s$
so, the car is approaching the observer at 15.2 m/s.