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Perturbation approximation of the period of a pendulum

  1. May 18, 2009 #1
    1. The problem statement, all variables and given/known data
    Find the perturbation approximation of the following in terms of powers of θ0.
    [tex]T=\sqrt{\frac{8L}{g}}\int^{\theta_0}_{0} \frac{d\theta}{\sqrt{cos\theta - cos\theta_0}}[/tex]
    It is helpful to first perform the change of variable u = θ/θ0 in the integral


    2. Relevant equations


    so the relevant equations are the taylor expansion of θ0, which would be
    θ000+εθ012 θ02/2

    3. The attempt at a solution
    so then making the change of variable suggested, we have
    [tex]T=\sqrt{\frac{8L}{g}}\int^{\theta_0}_{0} \frac{\theta_0 du}{\sqrt{cos(u\theta_0) - cos\theta_0}}[/tex]
    is it correct to say that the perturbation approximation is then
    [tex]\frac{\theta_0_0 + \epsilon \theta_0_1+
    \frac{\epsilon^2 \theta_0_2}{2}...}{\sqrt{cos(u\theta_0_0)-\epsilon sin(u\theta_0_1)-\epsilon^2cos(u\theta_0_2)-cost\theta_0_0)-\epsilon sin(theta_0_1)-\epsilon^2cos(\theta_0_2)...-cos\theta_0}}=0[/tex]

    or am i doing something wrong?
     
  2. jcsd
  3. May 18, 2009 #2

    gabbagabbahey

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    Huh?! :confused:

    What are [itex]\theta_{0_0}[/itex], [itex]\theta_{0_1}[/itex] and [itex]\theta_{0_2}[/itex] supposed to represent?

    What you have written seems nonsensical to me.

    The way I would interpret the question is that you are supposed to Taylor expand the integrand about the point [itex]\theta=\theta_0[/itex], then integrate your Taylor Series term by term in order to end up with a series of the form:

    [tex]\sum_{n=0}^\infty a_n \theta_0^n[/tex]

    that represents your integral.
     
  4. May 18, 2009 #3
    i erroneously thought that the
    θ00+εθ01+ε2 θ02/2
    terms represented the expansion about θ0.

    so then following the help you gave, the expansion of the integrand would be
    following Taylor series:
    f(0)+f'(0)+f''(0)... where we only go the second order since the original problem asks us to go to O(θ2) (I forgot to list this in the original question).
    we get
    [tex]\frac{\theta_0}{\sqrt{cosu\theta_0-cos\theta_0}}+\frac{2 (cosu\theta_0-cos\theta_0)-\theta_0(sin\theta_0-usinu\theta_0)}{2 (cosu\theta_0-cos\theta_0)^{3/2}}*\theta_0 [/tex]
    [tex]+ \frac{2 (cosu\theta_0-cos\theta_0)^{3/2}[(sin\theta_0-usinu\theta_0)-\theta_0(cos\theta_0-u^2cosu\theta_0)]-6[(cosu\theta_0-cos\theta_0)-\theta_0(sin\theta_0-usinu\theta_0)]((sin\theta_0-usinu\theta_0)(cosu\theta_0-cos\theta_0)^{1/2}}{4(cosu\theta_0-cos\theta_0)^3}[/tex]
    of course, last term would also be multiplied by θ02/2.when substituting in 0 for fn, however, everything becomes zero, so what did i do wrong?
     
  5. May 18, 2009 #4

    gabbagabbahey

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    Since you know [itex]0\leq u\leq 1[/itex] for your entire integration interval, try expanding [itex]f(u)=\frac{1}{\sqrt{\cos(u\theta_0)-\cos(\theta_0)}}[/itex] about [itex]u=0[/itex] instead.
     
  6. May 18, 2009 #5
    So the expansion of
    [tex]
    f(u)=\frac{1}{\sqrt{\cos(u\theta_0)-\cos(\theta_0)}}[/tex]
    about u=0 gives:
    f(u)+θ0f'(u)+θ02/2
    yielding
    [tex]\frac{1}{\sqrt{cosu\theta_0-cos\theta_0}}+\frac{u\sin(u\theta_0)}{2 (\cos(u\theta_0)-\cos(\theta_0)^{3/2}}+\frac{\sin(u\theta_0)+u^2\cos(u\theta_0))(\cos(u\theta_0)-\cos(\theta_0))+\frac{3}{2}u^2\sin^2(u\theta_0)}{4(\cos(u\theta_0)-\cos(\theta_0)^{3/2}}[/tex]
    thus about u=0, everything but the first term becomes 0, giving:
    [tex]\frac{1}{\sqrt{1-\cos(\theta_0)}}}[/tex]
    how can we solve about bigOH(θ02)?
     
  7. May 18, 2009 #6

    gabbagabbahey

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    No, the expansion about [itex]u=0[/itex] is:

    [tex]f(u)=f(0)+f'(0)u+\frac{f''(0)}{2!}u^2+\frac{f'''(0)}{3!}u^3+\ldots[/tex]

    And since [itex]f(u)[/itex] is a function of [itex]u[/itex], the differentiation is with respect to [itex]u[/itex], not [itex]\theta_0[/itex].
     
  8. May 18, 2009 #7
    I think I finally understand it, then we use the fact that u=θ/θ0, and rewrite the expansion in terms of θ0?
     
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