Non-harmonic oscillation of pendulum

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SUMMARY

The formula for non-harmonic oscillation of a pendulum is given by the equation $$\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. The discussion highlights that the small-angle approximation, where sinφ=φ, leads to the period formula ##T = 2\pi\sqrt{L/g}##, while larger amplitudes require more complex calculations. The derivation involves the chain rule and integration, emphasizing that the small-angle approximation is valid for small initial angles but diverges for larger angles.

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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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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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