The Doppler shift is not the same for a moving observer as it is for a moving source. But show that, for $v_{S} << v$ and $v_{O} << v$ , the difference between the two effects becomes negligible.

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Solution

We know that the general expression of Doppler effect where observer and source both are moving is given by:
$f^{'} = f_{0} (\frac{v \pm v_{0}}{v \pm v_{s}})$
where $v_{0}$ and $v_{s}$ are the velocities of observer and source repectively.
We find that the moving observers affect the numerator and moving the source affects the denominator.
But in the case $v_{s} << v$ and $v_{o} << v$ the $v_{o}$ and $v_{s}$ terms can be neglected compared to $v$
So $f^{'} = f_{0} (\frac{v}{v})$
hence the difference between the two effects becomes negligible.

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The Doppler shift is not the same for a moving observer as it is for a moving source. But show that, for vS<<v and vO<<v, the difference between the two effects becomes negligible.

The Doppler shift is not the same for a moving observer as it is for a moving source. But show that, for $v_{S} << v$ and $v_{O} << v$ , the difference between the two effects becomes negligible.