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Verifying a solution to Damped SHM

  1. Jan 25, 2016 #1

    RJLiberator

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    1. The problem statement, all variables and given/known data
    Verify that Ae^(-βt)cos(ωt) is a possible solution to the equation:

    d^2(x)/dt^2+ϒdx/dt+(ω_0)^2*x = 0

    and find β and ω in terms of ϒ and ω_0.

    2. Relevant equations
    N/a, trig identities I suppose.

    3. The attempt at a solution
    I think this is simply a 'plug and chug' type equation, but I'm having alll sorts of difficulty canceling things.

    I first calculated the first and second derivative of the given possible solution.

    First derivative = Ae^(-βt)*(-ωsin(ωt))+(-Aβe^(-βt)cos(ωt))
    Second derivative = -Ae^(-βt)ω^2cos(ωt)+Aβe^(-βt)ωsin(ωt)+Aβ^2e^(-βt)cos(ωt)+Aβe^(-βt)ωsin(ωt)

    I then plugged them into their respective spots into the equation and tried to simplify.
    I factored and divided both sides by Ae^(-βt).

    I am at:
    2βωsin(ωt)+β^2cos(ωt)-ϒ(ωsin(ωt)+βcos(ωt))+(ω_0)^2cos(ωt)-ω^2cos(ωt) = 0

    The problem I am having is visualizing how the terms can cancel.
    How does a term with ϒ cancel with terms with ω and terms with (ω_0)^2

    I have to be missing something.
     
  2. jcsd
  3. Jan 25, 2016 #2

    TSny

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    You a free to choose values for β and ω so that you have a solution.
     
  4. Jan 25, 2016 #3

    RJLiberator

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    Ah, that is the trick I was overlooking. I do recall that from my notes.


    We let ω = 0 such that we are left with

    β^2- ϒβ+(ω_0)^2 = 0

    Now we let β = ϒ/2 and we KNOW that (ω_0)^2 = ω^2+ϒ^2/4 but since ω^2 = 0 we are left with the following from the above

    ϒ^2/4-ϒ^2/2+ϒ^2/4 which indeed equals 0.

    So, in short, choose:
    ω = 0
    β = ϒ/2
    and know
    (ω_0)^2 = ϒ^2/4 + ω^2

    and the first part is solved!
     
  5. Jan 25, 2016 #4

    RJLiberator

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    The second part comes directly from the first
    β = ϒ/2
    (ω_0) = sqrt(ϒ^2/4 + ω^2)
     
  6. Jan 25, 2016 #5

    TSny

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    You had shown the relation that must be satisfied:
    2βωsin(ωt)+β^2cos(ωt)-ϒ(ωsin(ωt)+βcos(ωt))+(ω_0)^2cos(ωt)-ω^2cos(ωt) = 0

    This must be satisfied for all values of t. The only way this can happen is if the overall coefficient of sin(ωt) is zero and the overall coefficient of cos(ωt) is zero.

    This will tell you the values required for both β and ω.
     
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