The first formula I found in a video from Khan Academy:

$$ f_0 = f_s\times \frac{v}{v\pm v_s} $$ $f_0$: Observed frequency (Hz)

$f_s$: Source frequency (Hz)

$v$: Wave velocity (m s$^{-1}$)

$v_s$: Source velocity (m s$^{-1}$)

And second formula is from my physics class:

$$ f_b = f(1 + v_b/c) $$

I just don't know the difference and its bothering me. Any help would be nice

  • $\begingroup$ The look the same to me. $\endgroup$
    – M. Enns
    Nov 19, 2017 at 22:53
  • $\begingroup$ @M.Enns how so? I am just curious $\endgroup$
    – user204946
    Nov 20, 2017 at 5:21

1 Answer 1


You haven't spelled out what $f_b$ and $v_b$ represent in the second equation so I can't give you a definitive answer but if $f_b$ is the frequency of the source and $v_b$ is the speed of the source away from the observer then the two equations are the same. $c$ is customarily used for wave speed, in this case with respect to the medium (air).

Let's rearrange the second equation to solve for $f$ which I assume is the observed frequency.

$$f_b = f(1 + v_b/c)$$ $$f_b = f\frac{c+v_b}{c}$$ $$f_b\frac{c}{c+v_b}=f$$

So the second equation is the same as the first equation with the only difference being the $\pm$ symbol in the denominator of the first equation. The idea is that if the source is moving away from the observer you add and if it's moving towards the observer you subtract. If you are using second equation you would need some sort of sign convention i.e. making $v_b$ positive for moving away and negative for moving towards the observer.


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