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Hooke's Law function

  1. Mar 27, 2005 #1
    I dont know anything about diff eq but:

    [tex] F = ma = -kx(t) [/tex]

    [tex] a = \frac{d^2(x)}{dt^2} [/tex]

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

    So we need a function whos second derivative is the same as the function itself.

    I know hooke's law says the function is [tex] cos(\omega t) [/tex] but I dont see why [tex] e^x [/tex] doesnt satisfy the original condition.

    Can anyone shed some light?
  2. jcsd
  3. Mar 27, 2005 #2


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    It does,but with a complex exponent.

    [tex] e^{i\omega t},e^{-i\omega t} [/tex]

    are the 2 independent solutions which form a basis in the solution space.

  4. Mar 27, 2005 #3
    It's because of the negative sign on the left side. This is why the complex exponent is required as dexter indicated (differentiate twice to get a factor of [itex](\pm i)^2 = -1[/itex] in front). The complex exponential solutions [itex]e^{i\omega t}[/itex] and [itex]e^{-i\omega t}[/itex] are equivalent to the usual [itex]\cos{\omega t}[/itex] and [itex]\sin{\omega t}[/itex], and in fact are more useful in some situations (e.g. barrier penetration in quantum mechanics).
    Last edited: Mar 27, 2005
  5. Mar 27, 2005 #4
    So you technically could use [tex] e^{i\omega t} [/tex] ? I was thinking you could just associate the negative with the constant k/m, but you cant because mass isnt negative and a spring coefficient is the same, right?

    A more physics question, where does the [tex] \omega t [/tex] come from, instead of just t?
  6. Mar 27, 2005 #5


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    Nope.U need to adjust the complex exponentials & constants of integration,as to cancel away all imaginary parts of the final solution.Remember that this is a physics problem whose solutions have physical meaning,namely length which must be real...

    As for [itex] \omega t[/itex],it's simple:check the units in SI...That product is adimensional (the radian & the steradian are not standard units).That [itex] \omega [/itex] is the angular frequency of oscillation...

  7. Mar 27, 2005 #6
    The signs of [itex]k[/itex] and [itex]m[/itex] are immaterial. If they are negative or positive in one place, then they are in any other as well. You can't "associate" the negative sign with them.

    The solution to the ODE

    [tex] \frac{d^2x}{dt^2} = -\alpha x(t)[/tex]

    is always

    [tex] x(t) = Ae^{-i\omega t} + Be^{i\omega t}[/tex]

    where [itex] \omega = \sqrt{\alpha}[/itex]. If [itex] \alpha[/itex] happens to be negative then taking the square root will give you another [itex]i[/itex] and you'll get real exponents once simplified.

    Alternatively you could group constants and just solve

    [tex] \frac{d^2x}{dt^2} = \beta x(t)[/tex]

    where [itex]\beta = -\alpha[/itex]. This would have the solution

    [tex] x(t) = Ae^{\gamma t} + Be^{-\gamma t}[/tex]

    where [itex]\gamma = \sqrt{\beta} = \sqrt{-\alpha} = i\sqrt{\alpha} = i \omega[/itex] so you get the same result as before.

    In the case of your particular equation, you just have [itex] \alpha = \frac{k}{m}[/itex] so you get the familiar

    [tex] \omega = \sqrt{\frac{k}{m}}[/tex]

    As to your question, mathematically, the reason that you need the [itex] \omega t[/itex] instead of just [itex]t[/itex] is that otherwise your "solutions" simply won't satisfy the DE.

    Take [itex]x(t) = \cos{t}[/itex]. Then

    [tex] \frac{dx}{dt} = -\sin{t} \Longrightarrow \frac{d^2x}{dt^2} = -\cos{t} = -x(t) \neq -\frac{k}{m}x(t)[/itex]

    on the other hand taking [itex]x(t) = \cos{\omega t}[/itex] gives

    [tex] \frac{dx}{dt} = -\omega \sin{\omega t} \Longrightarrow \frac{d^2x}{dt^2} = -\omega^2 \cos{\omega t} = -\omega^2 x(t) = -\frac{k}{m}x(t)[/itex]

    as you wanted.

    This makes sense from a physics perspective, since I don't think you'd really expect periodic things in nature to have periods of [itex]2\pi[/itex] (this would make things way too easy!!! :smile:)
    Last edited: Mar 27, 2005
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