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Homework Help: Making something 1% larger than something

  1. Sep 3, 2016 #1


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    1. The problem statement, all variables and given/known data

    In the Michelson-Morley experiment
    a) What would v/c have to be to make tACA 1% larger than tABA?

    2. Relevant equations

    [tex] t_{ACA} =\frac{2l}{c}\frac{1}{1 - v^2/c^2} [/tex]

    [tex] t_{ABA} =\frac{2l}{c}\frac{1}{\sqrt{1 - v^2/c^2}} [/tex]

    3. The attempt at a solution

    The way I set up my equation is as follows.

    I let x = v/c

    [tex] t_{ACA} - \frac{1}{100}t_{ACA} = t_{ABA} [/tex]

    I plug in the equations, omitting 2l/c as it cancels, into wolfram alpha and it finds a value of x = sqrt(199)/100.

    I then plug it into the calculations to check and find

    [tex] t_{ACA} = 1.01989, t_{ABA} = 1.009899, \frac{1}{100}t_{ACA} = 0.010198 [/tex]

    That is rather close to what I am looking for.

    Is this the correct way to solve this type of mathematical problem?
  2. jcsd
  3. Sep 3, 2016 #2
    Let's say I have 2 barrels of apples - one labeled TABA and one labeled TACA.
    If TACA has 100 apples and TABA has 99 apples, is that the same as if TACA has 101 apples and TABA has 100 apples?
  4. Sep 3, 2016 #3


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    No, if Taca has 100 apples and Taba has 99 apples then 99.99 would be 1% larger than Taba.

    If Taca has 101 apples and Taba has 100 apples then 101 would be 1% larger than 100 and indeed Taca is correct.
  5. Sep 3, 2016 #4
    I guess I would write it like this:
    If TACA = 100 and TABA = 99, then TACA/TABA = 1.0101 = 101.01%
    If TACA = 101 and TABA = 100, then TACA/TABA = 1.0100 = 101.00%, making TACA 1% larger than TABA
  6. Sep 3, 2016 #5

    Ray Vickson

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    Yes, but an even easier way would be to take ##\sqrt{1-v^2/c^2} = y##, so your equation becomes ##1/y^2 = 1.01/y## after cancelling the ##2l/c## as you did. Thus , ##y = 100/101##.

    You really should AVOID using tools like Wolfram Alpha at this stage of your learning experience; doing much of the work by hand and then using computer assistance at the end (primarily for doing arithmetic) is the very best way to learn the material.
  7. Sep 3, 2016 #6


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    Ah, that makes sense.

    I should have made an easier substitution instead of trying to do the calculation in one big wolfram alpha enhanced step.
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