0

I am a high school student and recently learnt about the doppler effect. My teacher showed me two different formulas for the doppler effect with moving observer and moving source. Why are there different formulas because isn't movement relative, and it shouldn't matter whether the source or the observer is moving?

The formulas:

Moving source: $$f' = f \frac{v}{v+u_{s}}$$

Where $v$ is the speed of sound, $u_{s}$ is the speed of the source, $f$ is the frequency of the sound wave, and $f'$ is the frequency heard by the observer.

Moving observer: $$f' = f \frac{v+u_{o}}{v}$$

Qmechanic
  • 201,751
ved jain
  • 3
  • 2

1 Answers1

4

"Why are there different formulas because isn't movement relative, and it shouldn't matter whether the source or the observer is moving?"

Your reasoning applies to light but not to sound. This is because, in the case of sound, there is a third thing involved: the medium. It isn't just a case of source and observer. To understand this properly there's no substitute for actually deriving the equations for the two cases.

You might like the approach below. It treats the moving source and moving observer cases in a very similar fashion, without using the idea of wavelength. [It is also very easy to adapt to give the relativistic Doppler formula for light.]

For a given medium under given conditions, sound travels at a fixed speed, $v$, relative to that medium.

Suppose that the source frequency is $f_s$, so its period is $\tau_s = \frac1{f_s}$. But $\tau_s$ is the time between two successive wavefronts of the same phase (two maxima of pressure, perhaps) leaving the source.

If the observer is stationary in the medium but the source is moving away from the observer at speed $v_s$, then the extra distance the second wavefront has to travel is $u_s \tau_ s$, so the time, $\tau_o$ between the wavefronts reaching the observer is $$\tau_o=\tau_s+\frac {u_s \tau_s}v.$$ So we have $$\tau_o =\tau_s(1+u_s /v)\ \ \ \ \text{that is}\ \ \ \ f_o=\frac 1 {1+\frac{u_s}v}f_s $$

On the other hand, if the source is stationary in the medium and the observer is moving away from the source at speed $u_o$, then the extra distance the second wavefront has to travel is now $u_o \tau_o$, so the time, $\tau_o$, between the wavefronts reaching the observer is $$\tau_o=\tau_s+\frac {u_o \tau_o}v.$$ So we have $$\tau_o =\frac{\tau_s}{1-u_o /v}\ \ \ \ \text{that is}\ \ \ f_o=\left(1-\frac{u_o}v\right)f_s $$

[Make $u_s$ negative if the source is moving towards the observer, or $u_o$ negative if the observer is moving towards the source.]

Philip Wood
  • 35,641
  • Sometimes I write in MathJaX ${}o$ instead of ${}{0}$ to indicate that it is the observer. +1 – Sebastiano Sep 13 '21 at 12:16
  • 1
    @Sebastiano Hello and thank you. I typed lower case o (and not zero) in MathJax. But I was too lazy to make the subscripts non-italic. What do you think about 'ob' and 'so' instead of 'o' and 's'? – Philip Wood Sep 13 '21 at 12:40
  • Philips I have understood immediately :-)....'o'=observer, and 's' it is the source..It is easy and it is more clear almost for me. :-). – Sebastiano Sep 13 '21 at 21:58