Help me derive the Doppler effect ratio

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Homework Help Overview

The discussion revolves around deriving the expression for the ratio of time intervals, ΔTo/ΔTs, in the context of the Doppler effect. Participants are attempting to manipulate an equation involving these variables while refreshing their algebra skills.

Discussion Character

  • Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants share their attempts to isolate ΔTo/ΔTs from the given equation, expressing confusion over the recursive nature of the variables involved. There are questions about arithmetic operations and the proper application of algebraic principles.

Discussion Status

Some participants have provided guidance on factoring and rearranging terms, while others express uncertainty about their previous steps. The conversation reflects a collaborative effort to clarify the algebraic manipulation required to derive the desired expression.

Contextual Notes

Participants mention a lecture as a reference point, indicating that they are working within the constraints of a homework assignment and are focused on understanding rather than simply obtaining a solution.

Comscistudent
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Homework Statement


I'm watching this lecture and wanted to attempt to derive the expression for ΔTo/ΔTs

Homework Equations


ΔTo = ΔTs + VΔTo / U

The Attempt at a Solution


I have worked out that ΔTs = 1-V / U, however I'm stuck trying to get ΔTs / ΔTo because of the ΔTo on both sides of the original equation, it just turns into a recursive thing which I know must be wrong.

I'd appreciate any pointers that would help me get there myself rather than a solution. I'm only refreshing myself on algebra currently so I'm doubtless missing something simple
 
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Comscistudent said:

Homework Statement


I'm watching this lecture and wanted to attempt to derive the expression for ΔTo/ΔTs

Homework Equations


ΔTo = ΔTs + VΔTo / U

The Attempt at a Solution


I have worked out that ΔTs = 1-V / U
That's not what I get, starting from your relevant equation. What happened to ΔTo?
Please show us what you did.
Comscistudent said:
, however I'm stuck trying to get ΔTs / ΔTo because of the ΔTo on both sides of the original equation, it just turns into a recursive thing which I know must be wrong.

I'd appreciate any pointers that would help me get there myself rather than a solution. I'm only refreshing myself on algebra currently so I'm doubtless missing something simple
 
Sorry that was a mistake I forgot to divide the ΔTs by ΔTo. Now I've gotten this far but still end up with the variables on both sides

ΔTo = ΔTs + VΔTo / U
ΔTo = ΔTs / ΔTo + V / U
ΔTo / ΔTs = 1 / ΔTo + V/U / ΔTs
ΔTo / ΔTs = 1 / ΔTo + V / UΔTs
 
Comscistudent said:
Sorry that was a mistake I forgot to divide the ΔTs by ΔTo. Now I've gotten this far but still end up with the variables on both sides

ΔTo = ΔTs + VΔTo / U
ΔTo = ΔTs / ΔTo + V / U
The equation above is incorrect. You divided the right side by ΔTo, but not the left side. Any arithmetic operation you apply to one side of an equation, you must apply the same operation to the other side.
Instead, bring the terms with ΔTo to one side, and then factor ΔTo out.
Here's a start.
ΔTo - VΔTo / U = ΔTs
Now factor ΔTo from the two terms on the left side.

Comscistudent said:
ΔTo / ΔTs = 1 / ΔTo + V/U / ΔTs
ΔTo / ΔTs = 1 / ΔTo + V / UΔTs
 
Wow I'm stupid, thanks for putting up with me. I finally got there I think -

ΔTo - VΔTo / U = ΔTs
ΔTo( 1 - V / U ) = ΔTs
ΔTo / ΔTs ( 1 - V / U ) = 1
ΔTo / ΔTs = 1 / ( 1 - V / U )
ΔTo / ΔTs = (1 - V / U )-1
 
Comscistudent said:
Wow I'm stupid, thanks for putting up with me. I finally got there I think -

ΔTo - VΔTo / U = ΔTs
ΔTo( 1 - V / U ) = ΔTs
ΔTo / ΔTs ( 1 - V / U ) = 1
ΔTo / ΔTs = 1 / ( 1 - V / U )
ΔTo / ΔTs = (1 - V / U )-1
Since ##1 - \frac V U = \frac{U - V}{U}## you could go from the next to last line above, to this:
ΔTo / ΔTs = ##\frac U {U - V}##
 
That's even nicer, I wonder why they use the (1 - V / U)-1 in the lecture? Probably works out easier to apply or something, I'm sure I'll find out when I watch the rest of it.

Thanks again!
 

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