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Doppler effect : what happens when velocities of source and observer are comparable to velocity of sound

My text says that Doppler effect is applicable only when the velocities of source and observer are less as compared to the velocity of sound , whereas in a reference book (waves by D.C.Pandey ) I came across a problem in which they actually used Doppler effect when source velocity was equal to that of sound.

Please guide me to the correct answer.

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    $\begingroup$ As you might have noticed while using a whip, it produces a load bang. This is because it exceeds the speed of sound. And in the moment when the speed of a whip is comparable to the speed of sound, the sound amplifies. Though I'm also curious to how long that would amplify the sound. $\endgroup$ – MaDrung Feb 6 '17 at 8:47
  • $\begingroup$ Well so in terms of quantititative analysis would that mean that the loud bang - is independent of the actual frequency or rather independent of the Doppler effect ,I would seek more explanation on this topic $\endgroup$ – Aditya Sher Feb 6 '17 at 8:52
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    $\begingroup$ If the source is moving and the observer is stationary then the relevant equation is $f_{\rm observer} = \dfrac {v_{\rm sound}}{v_{\rm sound} - v_{\rm source}}f_{\rm source}$. I do not think that your could a better explanation than in this link which has some nice animations acs.psu.edu/drussell/Demos/doppler/doppler.html $\endgroup$ – Farcher Feb 6 '17 at 10:15
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Doppler effect is not valid only when the source velocity is greater than that of sound. In other situations, it is valid. When the source velocity approaches the speed of sound, the subsequent waves come closer and closer together and the wavelength approaches 0. At the speed of sound, this is what happens- the source gives off a wave which travels at the speed of sound and the source follows the wave as it is moving at the same speed. This means that the next wave that it gives off is along the first wave and subsequent waves all combine and get bunched up together, moving together. So instead of even rise and fall of air pressure, you get an abrupt dramatic increase then fall in air pressure, creating a shock wave, which is the sonic boom.

Resnick Halliday (Walker version) has a good diagram for this phenomenon, in the waves chapter. Try to get hold of a copy( actual or online).

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  • $\begingroup$ I'm having troubles with picturing how if you have a wave, can you push more stuff into that wave (to create even more pressure) if the matter is already compressed in that wave as it is. That would create a backwards push that would limit how loud a bang you could get from it. I bet my thinking is flawed, but I would really like some help with it. $\endgroup$ – MaDrung Feb 6 '17 at 16:09
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    $\begingroup$ Like, I said Resnick Halliday Walker version has a very good diagram that explains this. If I could figure how to post it here, I would. You can achieve more pressure as the waves gets bunched up together( this will not create a backward push). But the loud bang is due to the huge pressure fluctuation . Sound does not depend upon the pressure but upon the fluctuation. $\endgroup$ – TheFool Feb 6 '17 at 16:14
  • $\begingroup$ @The fool what would be the case when the source moves away with velocity v , so does it imply that frequency will become zero?, And thankyou for the answer. $\endgroup$ – Aditya Sher Feb 6 '17 at 16:20
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    $\begingroup$ When the source moves away with velocity of sound, the Doppler effect equation simply gives you half the frequency( As the distance between two successive waves i.e wavelength, doubles). P.S- You are welcome. Please do not forget to close the question by accepting the answer if it satisfies you. $\endgroup$ – TheFool Feb 6 '17 at 16:25

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