Non-harmonic oscillation of pendulum

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The discussion centers on the formula for non-harmonic oscillation of a pendulum, contrasting it with the harmonic oscillation formula. The solution for non-harmonic oscillation is expressed using a Jacobian elliptic function, with specific parameters defined for initial angles. The small-angle approximation is confirmed to be accurate for small initial angles, while deviations increase with larger angles. The conversation also touches on the importance of proper thread management and posting etiquette within the forum. The thread concludes after the original question is satisfactorily answered.
nneutrino
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Hi,
I would like to ask what is the formula for non-harmonic oscillation of pendulum? I know that formula for harmonic oscillation of pendulum is: (d^2 φ)/(dt^2 )+g/r sinφ=0 where φ is angle, t is time, g is gravitational acceleration, r is length of a rope. I know that harmonic oscillation means that sinφ=φ.
 
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For an arbitrary initial angle, it can be shown that the solution to such a differential equation is given by
$$\theta(t) = 2 \sin^{-1}\bigg\{k ~\text{sn} \bigg[\sqrt{\frac{g}{L}}(t-t_0);k\bigg]\bigg\}$$
where ##k = \sin(\theta_0/2)## and ##t_0## is the time when the pendulum is vertical (##\theta = 0##). The function ##\text{sn}(x;k)## is a Jacobian elliptic function, which is defined as follows:

Given the function
$$u(y;k) = \int_0^y \frac{\mathrm{d}t}{\sqrt{(1-t^2)(1-k^2t^2)}}$$
the Jacobian elliptic function in question is defined as the inverse of this function:
$$y = \text{sn}(u;k)$$
Values for such functions are often found in tables.
The derivation of this result is non-trivial but certainly possible, if you remember the chain rule and integrate twice.
 
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Likes sergiokapone and nneutrino
Thanks for answer :smile:. Basically is it correct when I use α=-g/L*φ; T=2π*sqrt(L/g) for small amplitude where sinφ=φ and α=-g/L*sinφ; T=2π*sqrt(L/g)*(1+(1/16)*φ*φ+(11/3072*φ*φ+...) for large amplitudes? α-angular acceleration; g-gravitational acceleration; L- length of rope; φ- angle; T- period
 
Yes, that is correct. The harmonically oscillating solution and associated initial angle-independent period (##T = 2 \pi \sqrt{\ell/g})## are always approximations. The point is, the small-angle approximation solution deviates very little from the actual solution when the initial angle is small. Figure three here illustrates this nicely; as the initial angle is increased, the full equation for the period and the approximation deviate more and more.
 
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Great! Thanks a lot for explanation :smile:.

Edit (fresh_42): The rest of the post has been deleted, because it belongs to a separate thread.
 
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Multiple posting is not allowed at the PF, but maybe the other post is not a duplicate. It would be better if you would notify the Mentors before creating what looks like a duplicate post. We are dealing with post reports about this -- please give us a few hours to work this out.

Edit (fresh_42): The new subject which this warning belongs to is now in a separate thread.
Since the original question has been answered, this thread remains closed.
 
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