Large oscillations of pendulum

In summary, a equation of a particle moving under a force that depends only on position in one dimension can be reduced to t = an integral of a function of x by multiplying the equation by 2x' and then the first term is (x2)' . This equation can be solved for t using elementary integration when the one-dimensional force is inverse square.
  • #1
Hamal_Arietis
156
15
When I solve this problem i have the equation:
$$x''+Asinx=0$$
How solve this equation if x large?
I think we use some approximations
 
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  • #2
It's horrible. You cannot solve it exactly with familiar functions, and it involves things called elliptic integrals. Specialised stuff :eek:I think.
If you like that sort of thing :oldwink: I can tell you there is a treatment in a book so old it is probably available as free e-book somewhere "The Dynamics Of Particles" by A G Webster, pp 45-48. He does give a formula for the period, a series in squared signs of the amplitude.
Maybe you could follow the argument up to the point where Webster gets
t = something multiplied by ∫ between limits one of which is essentially the pendulum angle of something of form
dx/√(1 - k2 sin2(x/2)) . That is the elliptic integral.
I guess you could plot some solutions if you can understand what he is talking about and use Wolfram app or Mathematica to give you the elliptic integrals numerically.

Noting your age I would say this is surely not in your syllabus for a long time, if ever.
 
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  • #3
My problems said that:
$$\int_0^1 \frac{dt}{\sqrt{(1-t^2)(1-k^2t^2)}}=\frac{\pi}{2} \sum_{n=0}^{∞}(\frac{(2n)!}{2^{2n}(n!)^2})^2$$
I think it is the same form with elliptic integral that you said.
I solved it but i can't make they exactly alike.
https://www.physicsforums.com/threads/large-oscillations-of-pendulum.885701/
 
  • #5
If I am not mistaken this form has different subject on the left hand side than the form I gave which was t = ...
If you want to go further I am not the person to ask - but it seems he has these last minutes arrived!. :oldsmile:

What I think it is useful for you to be familiar with is that an equation x'' + f(x) = 0 , which is the equation of a particle moving under a force that depends only on position in one dimension, can be reduced to t = an integral of a function of x by multiplying the equation by 2x' and then the first term is (x2)' . So you can get an equation with x' as LHS, and on the RHS you'll get through just an integral with respect to t of something involving a square root. By the way before square rooting this is just the energy equation. Just if you're lucky this will be an integral that corresponds to functions you know the integral of.

If you're interested you could work this out for simple harmonic motion - second term kx. You should find you get the same results as what is usually done in a more direct way. You can also solve it with 'elementary' integration when the one-dimensional force is inverse square. But in other cases you may also find that the integrals are not any of the 'elementary' ones, as here.
 
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  • #6
thanks for all helping ^^
 

1. What is a pendulum?

A pendulum is a weight suspended from a fixed point that can freely swing back and forth under the influence of gravity.

2. What are large oscillations of a pendulum?

Large oscillations of a pendulum refer to the swinging motion of the pendulum when it swings back and forth with a large amplitude, or distance from its resting point.

3. What factors affect the frequency of a pendulum's oscillations?

The length of the pendulum, the weight of the suspended object, and the strength of gravity are the main factors that affect the frequency of a pendulum's oscillations.

4. How does the length of a pendulum affect its oscillations?

The longer the pendulum, the slower the frequency of its oscillations. This is because a longer pendulum has a greater distance to travel in each swing, so it takes longer to complete one full swing.

5. What is the relationship between the mass of a pendulum and its oscillation frequency?

The mass of a pendulum does not affect its oscillation frequency. This is because the force of gravity is directly proportional to the mass of the object, so any changes in mass will also be reflected in the gravitational force, resulting in the same frequency of oscillation.

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