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.
The relationship between the frequency heard by the observer in motion and the original frequency of the sound is given by (Doppler effect) [tex]f'= \frac{v}{v-v_s}f [/tex] where f' is the frequency heard by the observer v is the speed of the wave (the speed of sound) [tex]v_s[/tex] 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 [tex]v_s=v( 1- \frac{f'}{f}) [/tex] and by plugging the data of the problem into the equation, we find [tex]v_s = (343 m/s)( 1- \frac{470 Hz}{450 Hz})=-15.2 m/s [/tex] so, the car is approaching the observer at 15.2 m/s.
