I was recently considering how to justify the formula relating to the Doppler effect for sound waves to a group of eleventh grade students who are likely encountering it for the first time. The formula in question is

$$f'=f \frac{v-v_o}{v+v_s}, $$

where $v$ is the speed of the sound wave, $f$ is the frequency of the wave emitted by a source moving at speed $v_s$ through a medium, and $f'$ is the frequency heard by an observer moving away from the source at speed $v_o$ through the medium.

These students will have no experience with results from special relativity, so I'm not considering any effects of special relativity. However, the students will be familiar with the concept of relative velocity as well as the formula $v = f \lambda.$

I thought about justifying the formula in the following manner:

  Once the sound wave has been emitted, the distance in space between two successive crests (i.e. the wavelength) has a fixed value, say $\lambda^*$. From the source's reference frame, the frequency of the sound wave travelling towards the observer is $f$ and it is moving away with speed $v + v_s$. We can then write \begin{equation} v+v_s = f\lambda^*. \end{equation}

  On the other hand, in the observer's reference frame the frequency of the sound wave is $f'$ and it is moving towards them with speed $v-v_o$. Hence \begin{equation} v-v_o = f'\lambda^*. \end{equation}

  Combining these two equations gives the desired result.

A number of textbooks and other sources give a derivation involving a change in the effective wavelength, even though the derivation above seems simpler. For this reason, I'm not convinced that this derivation is correct. Is this approach valid? If I have made an error or oversight, I would also like to understand why the correct result still emerges.

  • $\begingroup$ "From the source's reference frame, the frequency of the sound wave travelling towards the observer is ๐‘“ and it is moving away with speed ๐‘ฃ+๐‘ฃ๐‘ ." But what relevance does the velocity in the source's frame of reference have to the observed frequency? How does your argument work? $\endgroup$ – Philip Wood Feb 21 '19 at 19:13
  • $\begingroup$ @PhilipWood My understanding is that the velocity of the source through the medium will have no impact on the frequency heard by someone in the source's reference frame, since such an observer will be at rest with respect to the source regardless of the speed of their reference frame. The relevance of this velocity is that it can be related to $\lambda^*$, which is agreed upon by both observers. $\endgroup$ – Zac Feb 22 '19 at 0:43
  • $\begingroup$ Agree with first sentence of your comment. What I'm not clear about is "The relevance of this velocity [velocity of sound relative to source?] is that it can be related to ๐œ†โˆ—" I agree that it $is$ related to $\lambda*,$ but I can't quite see why it works. It seems easier to me to stay in the frame in which the medium is stationary and say that wavelengths between source and observer (assumed stationary for the time being) are expanded in the ratio $\frac{v+v_s}{v}.$ Incidentally there is a time-of-flight method of deriving the formula that doesn't bring in wavelength at all. $\endgroup$ – Philip Wood Feb 22 '19 at 11:47
  • $\begingroup$ Yes, I should have mentioned I was talking about the velocity of the sound relative to the source. For me, the key idea is that observers in each frame agree on the distance between successive crests, and both can apply the equation $v=f \lambda$. After thinking about it and considering the principle of relativity, I'm more convinced of this derivation. In terms of actually explaining the phenomenon (why the pitch of a sound changes) in an intuitive way, I don't think this has much value. The way that you have just described does a much better job of that. Thanks for your help! $\endgroup$ – Zac Feb 23 '19 at 12:48

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