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Perturbation approximation of the period of a pendulum

  • Thread starter limofunder
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  • #1
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Homework Statement


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


Homework Equations




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

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?
 

Answers and Replies

  • #2
gabbagabbahey
Homework Helper
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Homework Statement


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


Homework Equations




so the relevant equations are the taylor expansion of θ0, which would be
θ000+εθ012 θ02/2
Huh?! :confused:

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

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?
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.
 
  • #3
15
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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?
 
  • #4
gabbagabbahey
Homework Helper
Gold Member
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6
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?
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.
 
  • #5
15
0
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)?
 
  • #6
gabbagabbahey
Homework Helper
Gold Member
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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
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].
 
  • #7
15
0
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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