I know the following:

Source is emitting frequency of 440Hz.

The source is moving towards me at 232 m/s.

The observed frequency is 880Hz

What is the speed of sound in this case, and how does the formula look like?

I`m not in any way a mathematician, so if you use us Vf or other symbols to explain a formula, please explain what the symbols represent too.

Cheers :)


closed as off-topic by Rob Jeffries, M. Enns, Michael Seifert, Jon Custer, SRS Aug 3 '17 at 18:48

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The Doppler shift equation is given as $$\frac{\Delta \lambda}{\lambda_s} = \frac{u}{v}$$ where $\Delta\lambda$ is the change in wavelength and $\lambda_s$ is the source wavelength, $u$ is the speed of the source and $v$ is the speed of sound.

To convert from the given frequency to wavelength, we need to use the equation $v = f\lambda$. Substituting this in gives us $$\frac{(\frac{v}{880} - \frac{v}{440})}{\frac{v}{440}} = \frac{232}{v}$$ Manipulating this mathematically we can work our way down to the equation $$v = \frac{232}{440(\frac{1}{880}-\frac{1}{440})}$$ From this point we can calculate $v$ to equal $-464~\rm ms^{-1}$

This answer comes out to be negative as in the question I have taken the speed of the source to be positive. This means that the change in distance between you and the source per second is taken as $232\rm m$ instead of $-232\rm m$, essentially flipping what is positive and negative in the answer. If you are taking the movement of the source towards you as positive movement then it is correct, however if you are taking the movement of the source towards you as negative then the correct answer is $+464 \rm ms^{-1}$

Hope this helps :)

  • $\begingroup$ You might want to explain why the velocity comes out negative - what that implies. $\endgroup$ – Floris Aug 3 '17 at 13:54
  • $\begingroup$ @Floris To my mind it comes out to be negative as I took the speed value of the source as positive, whereas it is moving towards us, which should make it negative. I can edit the post to make it the answer positive if that improves the answer. $\endgroup$ – CooperCape Aug 3 '17 at 13:59
  • $\begingroup$ No - you just have to explain what negative velocity means. Is the source moving towards you or away? Incidentally I have already upvoted... $\endgroup$ – Floris Aug 3 '17 at 14:01

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