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

#### atferrari

Joined Jan 6, 2004
4,672
SOLVED

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

Amplitude is 15°.

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.

#### 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.

#### atferrari

Joined Jan 6, 2004
4,672
Since there is no closed form solution we are left with a numerical approach.
Could you elaborate briefly? At lost here.

#### shortbus

Joined Sep 30, 2009
9,757
The pendulum expert should be along soon.

xox

#### Papabravo

Joined Feb 24, 2006
19,572
Could you elaborate briefly? At lost here.
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.

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#### 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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#### shortbus

Joined Sep 30, 2009
9,757
As it's a pendulum, I really expected bahn to show up in this thread.

#### atferrari

Joined Jan 6, 2004
4,672
As it's a pendulum, I really expected bahn to show up in this thread.
Tagging @WBahn so maybe he could help. Gracias.

#### shortbus

Joined Sep 30, 2009
9,757
Learned my lesson well in another pendulum thread recently. He is the man when there is a question on them.

#### 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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#### 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 ?

#### atferrari

Joined Jan 6, 2004
4,672
Hi @mvas

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?

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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