Physics Asked on May 7, 2021
I have a few doubts about Doppler’s effect.
Edit: (Example where observer is moving to prove that wavelength is independent of observer’s velocity)
Please tell me if I am right in thinking so!
Everything seems to be right except the calculation of $λ$. that will be perceived by the observer is given as
$λ=frac{V}{f’}$ (all three are relative or apparent i.e. as perceived by the observer)
So instead of $V$, we write $V-U$ (since the observer is moving away the relative velocity will become slower, you’ve written $V+U$) this is the mistake you’ve made.
So no, it won’t be independent of observer’s velocity. I recommend seeing some of animation on Doppler effect for better visualisation.
Correct answer by Natru on May 7, 2021
The formula of Doppler Effect goes as follows:$$f_o=Bigl( frac{ vpm v_0}{vpm v_s}Bigl).f_s$$
Here ,
$f_o$=frequency measured by observer
$f_s$=frequency of source
Note:
When observer moves away from source - sign is used in numerator and if observer moves towards source + sign is used in numerator.
When source moves away from observer + sign is used in denominator and if source moves towards observer - sign is used in denominator.
Basically,
O.$rightarrow$ S.$rightarrow$ $f_o=Bigl( frac{ v+ v_0}{v+v_s}Bigl).f_s$
O.$rightarrow$ S.$leftarrow$ $f_o=Bigl( frac{ v+ v_0}{v- v_s}Bigl).f_s$
O.$leftarrow$ S.$rightarrow$ $f_o=Bigl( frac{ v- v_0}{v+v_s}Bigl).f_s$
O.$leftarrow$ S.$leftarrow$ $f_o=Bigl( frac{ v- v_0}{v-v_s}Bigl).f_s$
[Arrow indicates direction of velocity of observer(o) and source(s) and written beside it is the formula for that corresponding situation]
As you can see frequency measured by observer depends both on observer's velocity and also on source's velocity. So since speed of sound doesn't change we can say by $v=flambda$ that wavelength will also depend on both the velocities of source as well as observer.
By the way the formula you used would be valid for a stationery source i.e A SPECIAL CASE.
Answered by Möbius on May 7, 2021
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