In class we were explained that if the source moves toward the stationary observer(adsbygoogle = window.adsbygoogle || []).push({});

[tex] f' = f_{0} \frac{v}{v-v_{s}} [/tex]

where f0 is then aturla frequency of the wave, v is hte speed of the wave, and vs is the speed of the source

and [tex] f' = f_{0} \frac{v_{rel}+v}{v} [/tex]

where vrel is the speed of the observer

a)SHow that the result are thesame if the source and observer speed are much less than the wave speed

Now i can easilky say that [tex] v - v_{s} \neq 0 [/tex] and similarly [tex] v_{rel} + v \neq 0 [/tex] but is that really enough??

It is worth 5 marks in thsi assignment

b) FInd the difference between theclassical Doppler shift and the relativitistic Dopper shift up to the second order terms in (v/c)^2

now i know the relativistic expression is

[tex] f' = f_{0} \sqrt{\frac{1+\beta}{1-\beta}} [/tex]

but how do i expand this. I do know how to expand [tex] \frac{1}{1-x} [/tex]

but in this case i can get v / v-c and c/v-c how would that work??

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# Homework Help: Doppler Effect aturla frequency

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