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Heavily Damped Oscillator Equation

  1. Feb 3, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Screen Shot 2016-02-03 at 6.19.37 PM.png

    2. Relevant equations


    3. The attempt at a solution
    I'm not really sure how to go about this.

    They tell me the hint, and to use the simplification. I assume when they say (1+y)^n they take y to be in general, anything. It was confusing at first to see a y in the format when no y was present in any of the previous equations.

    So, we know ϒ^2/4 >> w_0^2
    So, could we say α^2 = ϒ^2/4 -w_0^2 can turn into α^2 = ϒ^2/4 => α = ϒ/2.
    But this is not of the form α = C(1+y)^n.

    So, we know ϒ = b/m, but that doesn't seem to help too much as α = b/2m is still not in that form.

    Any hints on how I can start this off on the right way? I have a lot of equations in front of me and I feel with a good start up hint ill be off to the races.
     
  2. jcsd
  3. Feb 4, 2016 #2

    Orodruin

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    Read the given hint. It is there for a reason. You have only kept the leading term and not the first order correction.

    I also disagree with the problem, I find it extremely intuitive that an overdamped system decays slower.
     
  4. Feb 4, 2016 #3

    RJLiberator

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    α^2=(ϒ^2/4)-w_0^2 and we want it in the form α = C(1+y)^n

    I'm not quite sure what you mean with this quote.

    The only way I can think of it is if we immediately take a square root of both sides.

    α = sqrt(ϒ^2/4-w_0^2)

    This is somewhat like what the hint desires, but I feel like I won't get anywhere with this (As ive tried some calculations with this).

    I also did as well.
     
  5. Feb 4, 2016 #4

    Orodruin

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    Rewrite the square root as an exponent ....
     
  6. Feb 4, 2016 #5

    RJLiberator

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    Ah, so you are suggesting there is more here.

    α = sqrt(ϒ^2/4-w_0^2)
    α = [ϒ^2/4-w_0^2]^(1/2)
    In this case, C = 1
    But, we need it to be in the form (1+y)^n
    So, perhaps let's try to factor out a ϒ^2/4 from both terms
    So we get
    α = ϒ/2[1-4*w_0^2/ϒ^2]^(1/2)

    Now it is appearing to be in correct form.
    So,
    α = ϒ/2[1-(1/2)*4*w_0^2/ϒ^2]

    Simplifying we get
    α = [ϒ/2-w_0^2/ϒ]


    Oh my god.
    I just solved it.

    I love math. I also love you.
    Why oh why did this take me hours.
     
  7. Feb 4, 2016 #6

    Orodruin

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    Maybe this particular type of series expansion is not fundamental in your research subject - it is in mine. :rolleyes:
     
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