Physics: Frequency

Question

At very fast car races the sound of the engine as a car approaches is much different than when it passes you.

 

How fast would the car need to be going to have the change in frequency be one octave (frequency halved) difference? (assume velocity of sound in air is 343 m/s)

Answer 1

You will need to use the Doppler effect

1. For an object traveling towards a stationary observer: f_t = (v / (v-v_s))*f
2. For an object traveling away from a stationary observer: f_a = (v / (v+v_s))*f

Where f_t/a is the observed frequency, v is the velocity of sound, v_s is the velocity of the source, and f is the frequency of the source.

Due to the change in octave, the frequency observed when the car is moving away from the observer

is half the frequency observed when the car is moving towards the observer. This can be written as

f_a = (1/2)f_t

You will now want to substitute the Doppler effect equations into the octave relationship and solve for v_s. After solving for v_s, you will simply plug in your known variable.

If you need further help, feel free to ask.