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Second order differential equation.(Damped oscillation)

  1. Jun 11, 2013 #1
    Hi could do with a little help with this question please!

    The question
    A damped oscillation with no external forces can be modelled by the equation:

    [itex]\frac{d^2x}{dt^2}[/itex]+2[itex]\frac{dx}{dt}[/itex]+2x=0

    Where x mm is amplitude of the oscillation at time seconds. The initial amplitude of the oscillation is 3mm (i.e. when t=0) and the intial velocity is 5mm/s.
    Solve the equation for x.

    Ok! so far I have;

    [itex]\frac{d^2x}{dt^2}[/itex]+2[itex]\frac{dx}{dt}[/itex]+2x=0

    [itex] m^2+2m+2=0 [/itex]

    [itex]\frac{-b±\sqrt{b^2-4ac}}{2a}[/itex]

    [itex]\frac{-2±\sqrt{2^2-4x1x2}}{2x1}[/itex]

    [itex]\frac{-2±\sqrt{-4}}{2}[/itex]

    [itex]\frac{-2}{2}[/itex]±[itex]\frac{\sqrt{-4}}{2}[/itex]

    m=-1±j[itex]\frac{\sqrt{4}}{2}[/itex]

    Equating this with m=α±jβ

    Gives α=-1 β=1

    Substituting in to the general solution (complex roots)

    x=e-1t[Acos(t)+Bsin(t)]

    x=3 t=0

    3=e-1x0[Acos(0)+Bsin(0)]

    3=A

    This is the part we are unsure of!

    [itex]\frac{dx}{dt}[/itex]=e-1t[-Asin(t)+Bcos(t)]-1e-1t[Acos(t)+Bsin(t)]

    By the product rule;

    =e-1t[B-A]cos(t)-[A-B]sin(t)

    x=0 [itex]\frac{dx}{dt}[/itex]=5

    5=e0[B-A]cos(0)-[A-B]sin(0)

    5=[B-A]

    We know A=3 so therefore B=8

    x=e-1t[3cos(t)+8sin(t)]

    Is this correct? If not any help would be greatly appreciated.
     
  2. jcsd
  3. Jun 11, 2013 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    Yes, that is correct. I'm not sure why you had any question about it.
     
  4. Jun 11, 2013 #3
    Just wasn't sure. Followed it from a book from a similar question and don't fully understand it! The question was different enough to have doubts!
     
  5. Jun 11, 2013 #4
    when you have the solution, you can substitute it into your differential equation to see if you end up with 0.
     
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