Two dimensions in a perfect plane. Yes, non-linear. That's why me asking for help.Is the problem in 2 dimensions or 3? In 3 dimensions, the Foucault pendulum is subject to the Earth's rotation and the Coriolis acceleration. The equation of motion is non-linear and IIRC it follows an elliptical path when it's motion is projected on a surface perpendicular to the radius vector from the center of the Earth to the pivot.
For small displacements you can linearize the equation with the substitution sin θ = θ. That doesn't wok for θ ≥ 7° so not much help on your problem. Since there is no closed form solution we are left with a numerical approach.Two dimensions in a perfect plane. Yes, non-linear. That's why me asking for help.
Could you elaborate briefly? At lost here.Since there is no closed form solution we are left with a numerical approach.
You can linearize the equation of motion for small values of the angle. For larger values you can apply numerical methods to derive an approximate solution.Could you elaborate briefly? At lost here.
The equation of motion for a simple pendulum is:Could you elaborate briefly? At lost here.
I can see that all this is well over my head. No matter how much I try, I cannot go any further. Being honest I should give up.The equation of motion for a simple pendulum is:
\(\frac{d^2\theta}{dt^2}\;+\;\omega^{}_ 0^{2}sin\theta\;=\;0\)
The introduction of the sin function makes the differential equation non linear.
I have one reference that says you can derive an expression for time as a function of angle using elliptic integrals of the first kind. AFAIK these are evaluated with tables or numerical methods. The expression is:I can see that all this is well over my head. No matter how much I try, I cannot go any further. Being honest I should give up.
I expected to have someone showing the calculation I should do to actually obtain a value. Gracias anyway.
Tagging @WBahn so maybe he could help. Gracias.As it's a pendulum, I really expected bahn to show up in this thread.
Could not make sense of your explanation yet but I fear the temptation to calculate this as the difference between two pendula one going up to A and the other up to B.Is the time "t" for any small Partial_Angle as follows ?
Partial_Angle = Max_Angle x sin ( Sqrt ( t x g / L ) )
and the Period is defined as ...
T = 2 x Pi x sqrt ( L / g )
Next, combine the above two equations to eliminate: "L" and "g".
Given we know ... T, Partial_Angle & Max_Angle
Finally, solve for ... Elapsed Time "t" to Point A.
The Elapsed Time to Point B can be computed from "T"
So, the time elapsed from Point A to Point B, is then the difference between the two elapsed times.
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