Doppler Effect aturla frequency

[stunner5000pt]

In class we were explained that if the source moves toward the stationary observer

$$f' = f_{0} \frac{v}{v-v_{s}}$$
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 $$f' = f_{0} \frac{v_{rel}+v}{v}$$
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 $$v - v_{s} \neq 0$$ and similarly $$v_{rel} + v \neq 0$$ 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

$$f' = f_{0} \sqrt{\frac{1+\beta}{1-\beta}}$$

but how do i expand this. I do know how to expand $$\frac{1}{1-x}$$
but in this case i can get v / v-c and c/v-c how would that work??

 

[Davorak]

In class we were explained that if the source moves toward the stationary observer

$$f' = f_{0} \frac{v}{v-v_{s}}$$
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 $$f' = f_{0} \frac{v_{rel}+v}{v}$$
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 $$v - v_{s} \neq 0$$ and similarly $$v_{rel} + v \neq 0$$ but is that really enough??

From your question it seems like you are trying to pove:
$$f' = f_{0} \frac{v}{v-v_{s}} =f_{0} \frac{v_{rel}+v}{v}$$
$$\frac{1}{1-x} = 1 + x \ \ {When\ x\ll1}$$
so we have
$$f' = f_{0} \frac{v}{v-v_{s}} = f_{0} \frac{1}{1-\frac{v_{s}}{v}} \cong f_{0} ( 1 + \frac{v_{s}}{v} ) = f_{0} \frac{v_{rel}+v}{v} =f_{0} (\frac{v_{rel}}{v}+1)$$
Since $$v_{s} = v_{rel}$$

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

$$f' = f_{0} \sqrt{\frac{1+\beta}{1-\beta}}$$

but how do i expand this. I do know how to expand $$\frac{1}{1-x}$$
but in this case i can get v / v-c and c/v-c how would that work??

hmm I would use the binomial Expansion. I am lazy and did not want to write out the binomial Expansion so here is a link:
http://hyperphysics.phy-astr.gsu.edu/hbase/alg3.html

 

[dextercioby]

a)No...U should try to put the expressions in a way in which u can take the relevant limit $v<<v_{s}$

b)Use this trick:
$$\frac{1+\beta}{1-\beta}=1+\frac{2\beta}{1-\beta}=1+\frac{1}{something}$$

and now to take out terms in \beta^{2} outta the square root...

Daniel.

 

[HallsofIvy]

In class we were explained that if the source moves toward the stationary observer

$$f' = f_{0} \frac{v}{v-v_{s}}$$
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 $$f' = f_{0} \frac{v_{rel}+v}{v}$$
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 $$v - v_{s} \neq 0$$ and similarly $$v_{rel} + v \neq 0$$ but is that really enough??

Wouldn't be enough for me! Saying they are not 0 doesn't say anything about the relationship of v- v_s and v_r- v. What is the relationship between v_s and v_r?

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

$$f' = f_{0} \sqrt{\frac{1+\beta}{1-\beta}}$$

but how do i expand this. I do know how to expand $$\frac{1}{1-x}$$
but in this case i can get v / v-c and c/v-c how would that work??

What is &beta;? You should know how to expand $\sqrt{1- x^2}$ in a power series. Once you have expanded it in a power series, replace &beta; by its expression in v and c.