A car traveling 35.0 m/s overtakes another car going only 22.0 m/s. When the faster car is still behind the slower one, it sounds a horn of frequency 1500.0Hz. What is the frequency heard by the driver of the slower car?
For observer and source both moving: fobserver=fsource [(1+ or - vobserver/v)/(1+ or -vsource/v)] v= velocity of sound in air signs in numerator: + if observer moving toward source and - if away from source signs in denominator: - if source moving toward observer and + if away from observer
Reference https://www.physicsforums.com/threads/doppler-effect-source-observer-both-moving.625483/
To find the frequency heard by the driver of the slower car, we need to consider the concept of the Doppler effect. The Doppler effect refers to the change in frequency or wavelength of a wave as observed by an observer moving relative to the source of the wave.
In this scenario, the faster car is approaching the slower car from behind. When this happens, the sound waves emitted by the faster car will be "compressed" or have a higher frequency as observed by the driver of the slower car.
To calculate the frequency heard by the driver of the slower car, we can use the following formula:
f' = (v + v₀) / (v - v₀) * f₀
where:
f' is the frequency observed by the driver of the slower car
v is the velocity of sound (assumed constant at 343 m/s)
v₀ is the velocity of the slower car (22.0 m/s)
f₀ is the frequency emitted by the faster car (1500.0 Hz)
Now we can substitute the values and calculate the frequency observed by the driver of the slower car:
f' = (343 + 22.0) / (343 - 22.0) * 1500.0
f' = 365.6 / 321 * 1500.0
f' ≈ 1704 Hz
Therefore, the frequency heard by the driver of the slower car is approximately 1704 Hz.