Deriving the Time Period of a Pendulum with Small Angle Approximation

In summary, the conversation is about a problem involving a pendulum and finding the time period without the small angle approximation. The question is then asked to calculate the time period for small angles, up to second order in the small angle approximation. The person is not sure how to interpret the question and wonders if they should derive the simplified expression or expand the integral using a Legendre polynomial. The expert confirms that the question is asking for the second order contribution in the small angle approximation and suggests making a normal series expansion to find the answer. The person then confirms their result as ##T(\varphi_0) = 2 \pi \sqrt{\frac{l}{g}} \big(1 + \frac{\varphi_0^2
  • #1
JulienB
408
12
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
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
@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.
 

1. What factors affect the time period of a pendulum?

The time period of a pendulum is affected by the length of the pendulum, the mass of the pendulum bob, and the acceleration due to gravity.

2. How does the length of a pendulum affect its time period?

The time period of a pendulum is directly proportional to the square root of its length. This means that as the length of the pendulum increases, the time period also increases.

3. Does the mass of the pendulum bob affect its time period?

Yes, the mass of the pendulum bob does affect its time period. A pendulum with a heavier bob will have a longer time period compared to a pendulum with a lighter bob.

4. How does gravity affect the time period of a pendulum?

The acceleration due to gravity affects the time period of a pendulum by increasing it. This is because the gravitational force acts as a restoring force on the pendulum, allowing it to swing back and forth.

5. Is the time period of a pendulum affected by the amplitude of its swing?

No, the time period of a pendulum is not affected by its amplitude, which is the distance the pendulum swings from its starting position. The time period only depends on the factors mentioned above, such as length and mass.

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