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Time period of a pendulum

  • Thread starter JulienB
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  • #1
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Hi everybody! I have a quick question about a pendulum. The first question of a problem asked me to find an integral expression for the time period of a pendulum without the small angle approximation, which I did and I got that:

##T(\varphi) = 4\sqrt{\frac{l}{g}} \int_{0}^{\pi/2} \frac{d\xi}{\sqrt{1 - \sin^2 (\varphi_0/2) \sin^2 \xi}}##

which seems correct. But then I am asked: "Calculate T for small angles ##\varphi_0 << 1## until the second order in ##\varphi_0##" (translated from german but quite accurate I believe).
I am not sure how to interpret the question: do they want me to derive ##T = 2\pi \sqrt{\frac{l}{g}}## (but then I don't do anything in second order) or do they want me to expand the integral until second order, with the Legendre polynomial for example (but then I don't do any small angle approximation)?

For info, this problem takes place in the context of a course about advanced mechanics. We're between Lagrangian and Hamiltonian at the moment.

Thanks a lot in advance for your answers.


Julien.
 

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  • #2
Orodruin
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They want you to do exactly what they say, find the second order contribution in ##\varphi_0##. In order to do this you will have to make a normal series expansion in ##\varphi_0## and keep terms up to order ##\varphi_0^2##.
 
  • #3
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@Orodruin Okay thanks that's what I thought. For info I get ##T(\varphi_0) = 2 \pi \sqrt{\frac{l}{g}} \big(1 + \frac{\varphi_0^2}{16}\big)##.

Thanks a lot for your answer.


Julien.
 

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