Deriving Doppler Effect Frequency w/ Stationary Person & Moving Source

[annamal]

Can you derive the formula for frequency observed from doppler effect with stationary person and moving sound source away from the person like this:
$v_t = v + v_s$ where $v_t$ is the total velocity observed by stationary person from moving sound, v is velocity of sound and $v_s$ is velocity of sound source.
$$v = \lambda f$$
$$f_o = (v + v_s)/\lambda_o$$ where $f_o$ is frequency observed and $\lambda_o$ is wavelength observed

 

[Baluncore]

... from doppler effect with stationary person and moving sound source away from the person ...

If moving away, the wavelength will be longer so the frequency will be lower.
Maybe you need to change the sign to (v - vs).

 

[annamal]

If moving away, the wavelength will be longer so the frequency will be lower.
Maybe you need to change the sign to (v - vs).

I guess my question is not very clear. What I am asking is why the velocity of sound is constant regardless of whether the source be moving. Wouldn't the velocity of sound be added to the speed of source? I am likening it to a person running on a moving train. To a stationary viewer the velocity of runner = velocity of runner on train + velocity of moving train.

 

[PeroK]

What I am asking is why the velocity of sound is constant regardless of whether the source be moving. Wouldn't the velocity of sound be added to the speed of source?

Because sound is dependent only on the air. The speed is determined by the medium.

I am likening it to a person running on a moving train. To a stationary viewer the velocity of runner = velocity of runner on train + velocity of moving train.

If you put some air in a box and move the box, the then speed of a sound wave to an external observer IS the speed of the box plus the speed of sound in the box.