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Simple harmonic oscillator

  1. Nov 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Show simple harmonic motion starting from Hooke's Law.


    3. The attempt at a solution

    [tex]F=-kx[/tex]

    [tex]=m\frac{d^2x}{dt^2}=-kx[/tex]

    [tex]\frac{1}{x}\frac{d^2x}{dt^2}=-\frac{k}{m}[/tex]

    [tex]=\frac{1}{x}\frac{d}{dt}\frac{dx}{dt}=-\frac{k}{m}[/tex]

    [tex]\int\int\frac{1}{x}d\left(dx\right)=-\frac{k}{m}\int dt\int dt[/tex]

    [tex]\int\frac{1}{x}dx=-\frac{k}{2m}t^2[/tex]

    [tex]\ln\left(x\right)=-\frac{k}{2m}t^2[/tex]

    [tex]x\left(t\right)=e^{-\frac{k}{2m}t^2}[/tex]

    But it should be:

    [tex]x\left(t\right)=e^{-i\sqrt{\frac{k}{2m}}t}[/tex]

    I thought I knew how to do this. :redface:
     
  2. jcsd
  3. Nov 6, 2009 #2

    rock.freak667

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    Homework Helper

    Well normally you could just put F=ma and then put it the form [itex]a= - \omega^2 x[/itex] and that would be sufficient to show SHM.


    Else to solve the DE, you would need to know that the equation y''+ky=0 has solutions y1=cos(kt) and y2=sin(kt)
     
  4. Nov 7, 2009 #3
    I want to solve the differential equation. But as you can see, somewhere I need to take the square root of the exponential argument.

    Where, in my steps, did I miss that?
     
  5. Nov 7, 2009 #4
    Nobody knows?
     
  6. Nov 7, 2009 #5
    I don't see how to use this to get to the right answer, but:

    [tex]\int\frac{1}{x}dx \neq \ln (x)[/tex]

    [tex]\int\frac{1}{x}dx = \ln |x| + C[/tex]

    So the derivation would continue:

    [tex]\ln |x| + C =-\frac{k}{2m}t^2 [/tex]

    [tex] |x| = e^{-\frac{k}{2m}t^2 + C} [/tex]

    Like I said, not sure how to use this, or if it helps. I could see squaring x, then taking the square root, in lieu of the absolute value signs, but, then, I don't see where the i comes out. Most of the derivations I've seen 'guess' at the solution of the 2nd order DE to be of the form [itex]a cos(\omega t)[/itex], and go from there.

    I'm going to look at this more tomorrow, I can't believe I don't know either.
     
  7. Nov 7, 2009 #6

    rock.freak667

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    Homework Helper


    [tex]m\frac{d^2x}{dt^2}=-kx \Rightarrow \frac{d^2x}{dt^2}+\frac{k}{m}x=0 [/tex]


    the auxiliary equation is r2+(k/m)=0 so r=±√(k/m)i

    when you have roots in the form r=λ±μi what is the general solution x(t) equal to?
     
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