[solved] Pendulum - How to calculate the time it takes to go from A to B?

Thread Starter

atferrari

Joined Jan 6, 2004
4,672
SOLVED

There is an ideal pendulum. Length L and period T are known.

Amplitude is 15°.

20190313_191942.jpg

How could I calculate the time it would take to reach B, starting at A?
 
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Papabravo

Joined Feb 24, 2006
19,572
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.
 

Thread Starter

atferrari

Joined Jan 6, 2004
4,672
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.
Two dimensions in a perfect plane. Yes, non-linear. That's why me asking for help.
 

Papabravo

Joined Feb 24, 2006
19,572
Two dimensions in a perfect plane. Yes, non-linear. That's why me asking for help.
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.
 

Thread Starter

atferrari

Joined Jan 6, 2004
4,672
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 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.
 

Papabravo

Joined Feb 24, 2006
19,572
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.
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:

\(t\;=\;\frac{1}{\omega^{}_0}\int\limits_{0}^{\theta}\frac{d\theta}{\sqrt{1\;-\;\left(\frac{2\omega^{}_n}{\omega^{}_0}\right)^{2}sin^2\left(\frac{\theta}{2}\right)}}\)

https://en.wikipedia.org/wiki/Elliptic_integral

I should add that \(\omega^{}_0\;\) is the velocity at \(\theta\;=\;0\;\)

and that \(\omega^{}_n\;=\; \sqrt{\frac{g}{l}} \)
 
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Janis59

Joined Aug 21, 2017
1,523
About pendulums there is only one thing I am more than sure. Every year when I teach 10-graders, we realize the experimental g measuring by means of pendulum. And in last 15 years none of my pupils never have got anything nearer than 7,8...8,3 m/s^2 instead of 9,81.
Just weight of rope is very essential there, and if to substitute it with fishing thread, then elasticity gives the same fault.
 

mvas

Joined Jun 19, 2017
539
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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Thread Starter

atferrari

Joined Jan 6, 2004
4,672
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.
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 it like that?
 

mvas

Joined Jun 19, 2017
539
No, not two pendulums ... one pendulum & the time elapsed travelling from Point A to Point B

STEP #1: Do you understand this formula ?

The time "t" for any small Partial_Angle as follows ...
Partial_Angle = Max_Angle x sin ( Sqrt ( t x g / L ) )

What is the Partial_Angle ?
What is the Max_Angle ?
 

Thread Starter

atferrari

Joined Jan 6, 2004
4,672
Hi @mvas

Gracias for your interest.
Back at home after 3 weeks working in Patagonia, really busy with no time nor energy to look at this.
Is this what your are proposing?

Pendulum mvas.jpg

Edit to add this question:

Is this formula (from your previous post) correct?

Partial_Angle = Max_Angle x sin ( Sqrt ( t x g / L ) )
 
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visionofast

Joined Oct 17, 2018
104
I think you'd better to solve the problem in Cylindrical axis instead of xy-Cartesian for a better understanding.
so, there's a constant F=mg force that should be divided to tangential force FΦ and orthogonal force Fr.
the Fr cancels by pendulum's string and just a FΦ makes the effort in a circular distance.
 
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