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Hooke's Law and differential equations

  1. Feb 17, 2007 #1
    1. The problem statement, all variables and given/known data
    [tex]-kx=m\frac{d^2x}{dt^2}[/tex]

    I don't know how to solve differential equations, can someone show me how to do it, with this example.
     
  2. jcsd
  3. Feb 17, 2007 #2

    cristo

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    This is an example of a second order differential equation; let's rewrite it as [tex]\frac{d^2x}{dt^2}+\frac{k}{m}x=0[/tex] This will have a general solution of the form x=Aeat. Plug this into the equation and we obtain the equation a2+k/m=0, which gives us the values for a; namely [itex]a=\pm \sqrt{-\frac{k}{m}}=\pm i\sqrt{\frac{k}{m}}[/itex]. Letting [itex]\sqrt{k/m}=\omega[/itex] we obtain [itex]x(t)=Ae^{i\omega t}+Be^{-i\omega t}[/itex]. Now, recognising that this is the form of a sum of cosine and sine functions, we obtain the general solution [itex]x(t)=C\cos(\omega t)+D\sin(\omega t)[/itex]

    I hope you follow that; however, if you have not studied differential equations, then I would suggest first learning about first order differential equations, and then moving onto second order equations (although, I suspect you want this solution for a specific purpose).

    [edit: I didn't notice that this was in the homework forum, and so a full solution shouldn't be given; however, I suspect from previous posts that you may be self-learning the subject, and so this is not a homework question]
     
    Last edited: Feb 18, 2007
  4. Feb 17, 2007 #3
    Why do you need both solutions?
     
  5. Feb 18, 2007 #4

    cristo

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    Since both [itex]e^{i\omega t}[/itex] and [itex]e^{-i\omega t}[/itex] are solutions to the equation, then the general solution will be a linear combination of the two, with the constants determined by the boundary conditions.
     
  6. Feb 18, 2007 #5
    Thanks cristo. How can you pass from e^(something) to cosine and sine functions?
     
  7. Feb 18, 2007 #6

    cristo

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    Use the definition of complex exponential, namely [itex]e^{\pm i\theta}=\cos\theta \pm i\sin\theta[/itex]. With a bit of rearrangement, we find that [tex]\cos\theta =\frac{1}{2}(e^{i\theta}+e^{-i\theta})[/tex] and [tex]\sin\theta=\frac{1}{2i}(e^{i\theta}-e^{-i\theta})[/tex]
     
  8. Feb 18, 2007 #7

    robphy

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    While the method described above is very useful and practical, don't forget what the equation is asking:

    Can you determine which functions x(t) are proportional to minus their second derivatives?

    You don't need a [crash] course in differential equations to answer that question.
     
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