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Take the classical example of fast moving Car whizzing past you. You hear the .....oooooooOOooooooo..... sound.

But in that case most of the effect we are observing is the change in intensity of sound. The sound gets louder as car approaches and then fades away as car goes away.

A video showed that a 50mph speeding car, produces doppler effect of shifting frequency from 500Hz fo 530Hz.

Is this change in frequency what we hear mostly, or is it the changing loudness of sound that we mostly hear?

In other words, does the highly changing loudness cover the doppler effect of changing frequency?

Or, is the doppler effect distinctly detectable by our ears, despite the changing loudness effect?

Finally, is it possible to somehow observe the doppler effect experimentally (by human ears) without the changing intensity? How different (or better) will the effect feel(or sound) then?

Rohit Shekhawat
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The Doppler effect (i.e. the shift in frequency of the received sound) manifests itself as the change in the pitch of the sound, and is often easily noticeable to the human ear as it occurs in traffic, with or without the accompanying change in loudness. This is similar to distinguishing two different notes played on the same musical instrument, with different loudness. It is especially easy to do with the sirens of emergency vehicles which are loud enough to be heard for a long time as they approach and recede from you, and have distinct pitches you could "hum to". This might be harder to do with the engine noise of a typical car, but often still possible.

If a fire truck blaring its siren drives past you at a modest speed of 80 km/h = 22 m/s, using a speed of sound of 343 m/s, the ratio between the pitches heard as the truck is approaching and receding is $$\frac{f_a}{f_r}=\frac{343+22}{343-22}=1.138.$$ To put this in familiar terms, two semitones, which is the pitch increment between two of the closest black keys on a piano, or the pitch increment when moving two frets up on a guitar string, corresponds to a frequency ratio is $2^{2/12}=1.123$. This is less than the Doppler shift of the fire truck siren (or anything else passing by you at the same speed).

Puk
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  • Your equation corresponds to the case where both truck and me are moving towards each other at 22m/s. Am I right? – Rohit Shekhawat Mar 07 '23 at 08:58
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    @RohitShekhawat No. You are stationary. When the truck is approaching, $$f_a = \frac{343}{343-22}f_0.$$After the truck passes by you and starts to recede, $$f_r = \frac{343}{343+22}f_0.$$The equation in the answer is just the ratio of these two. – Puk Mar 07 '23 at 09:04
  • Oh sorry, I misinterpreted the meaning of 'fa' and 'fr'. You are very right for sure. – Rohit Shekhawat Mar 07 '23 at 09:13