Deriving the Time Period of a Pendulum with Small Angle Approximation

[JulienB]

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.

 

[Orodruin]

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 $\varphi_0^2$.

 

[JulienB]

@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.