Simple Pendulum motion - Numerical Integration

In summary, the differential equation of motion for the simple pendulum can be shown to be##\ddot {θ} = -(g/L)sinθ##. Given that L=9.81 m and that the pendulum is releasedfrom rest at θ=60deg, determine the time required for the pendulum to reach the positionθ=0deg. Use Δt=0.10s, and compare your answer with the analytical solution of 1.686s.
  • #1
Alexanddros81
177
4

Homework Statement


The differential equation of motion for the simple pendulum can be shown to be
##\ddot {θ} = -(g/L)sinθ##. Given that L=9.81 m and that the pendulum is released
from rest at θ=60deg, determine the time required for the pendulum to reach the position
θ=0deg. Use Δt=0.10s, and compare your answer with the analytical solution of 1.686s

Fig P13_81.png

Homework Equations

The Attempt at a Solution



Pytels_Dynamics095.jpg


I guess I am missing something...
I need to compute θ value? How do I do that? Do I find θ by using x, L, s ?
 

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  • #2
What method are you using? Eulers? What is your question for us?
 
Last edited:
  • #4
Since ##u_{\theta} = \dot \theta##, ##\dot u_{\theta} = \ddot \theta##.
So, you are solving the equation ##\ddot \theta = -\sin{\theta}##. No need to introduce x or s.
 
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Likes scottdave
  • #5
Hi.
Can you check the following if it is correct?
By reading on a Dover book regarding differential equations and browsing on the web
i give the following as a linear solution for the undamped pendulum problem for small angle θ.
Pytels_Dynamics096.jpg


Do I proceed with the linear equation of the pendulum equation or with non-linear for this
particular problem? (Initial condition for θ = 60deg)

Also I want to use the 4th order Runge-Kutta method.
 

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  • #6
With an initial condition of θ=60 degrees, θ in radians is 1.05, and sin(θ) = 0.866, so you can see that the small angle approximation fails badly. You need to use the full non-linear equation in your Runge-Kutta code. That is the whole point of the exercise. With the small angle approximation, the equation can be solved analytically and there is no need for numeric integration.
 
  • #7
So the equivalent first-order diff.equations are

##u_θ=\dot {θ}## and ##\dot {u_θ} = -(g/L)sinθ##

Correct?
 
  • #8
Alexanddros81 said:
So the equivalent first-order diff.equations are

##u_θ=\dot {θ}## and ##\dot {u_θ} = -(g/L)sinθ##

Correct?

Correct.
 
  • #9
Pytels_Dynamics097.jpg


How do I implement the above runge-kutta 4th order to this specific problem?
I don't know how...

Is ##k1=f(θ_i)## ? and ## k2=f(θ_i + 1/2h) ## etc. ?
Is ##\dot {u} = f(θ)## ? (not according to the above)

Is ##k1=f(θ_ι, ω_i)## etc ?
 

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  • #10
Since this is a second order problem, your yi's and k's are two component vectors, with the components being [itex]y_i = (\theta, \dot \theta)[/itex]. Here your function [itex]f(t, y_i) = f(t, (\theta, \dot \theta)) = (\dot \theta, -\frac{g}{L} \sin(\theta))[/itex]. Note that f is independent of t, which simplifies things. So you start with [itex]y_0 = (\theta, \dot \theta) = (\theta_0, 0)[/itex]. So then [itex]k_1 = f(0,(\theta_0,0)) = (0, -\frac{g}{L} \sin(\theta_0)) [/itex] , [itex]k_2 = f(h/2,y_0 + h/2*k_1) = f(h/2,(\theta_0,-\frac{g}{L} \sin(\theta_0)*h/2))=(-\frac{g}{L} \sin(\theta_0)*h/2, -\frac{g}{L} \sin(\theta_0)) [/itex] and so on. Does this help?
 
  • #11
to phyzguy:
thanks this is the information I needed. Is there some more info related to this?
 
  • #12
A few questions:
1) since ##y_0 = (\theta, \dot \theta) = (\theta_0, 0)## we say that ##k_1 = f(0,(\theta_0,0)) = (0, -\frac{g}{L} \sin(\theta_0))##

the last expression for k1 is ##f(0,(\theta_0,0)) = (0, -\frac{g}{L} \sin(\theta_0))##
The above means that t=0 (first parameter) and ##y_0 = (\theta_0,0) = -\frac{g}{L} \sin(\theta_0)## (second parameter). Correct?

2) Is your expression for k2 correct? If yes how did you derived it?
 
  • #13
I think you're still not getting it. The function f returns the derivative of y, so since [itex] y = (\theta, \dot \theta)[/itex], [itex] f(y) = (\dot \theta,-\frac{g}{L} \sin(\theta))[/itex]. From there you just plug in. To find k2 you just plug in [itex] y_0 + \frac{h}{2} k1 [/itex] into f. You wrote:

Alexanddros81 said:
##y_0 = (\theta_0,0) = -\frac{g}{L} \sin(\theta_0)## (second parameter).

The first equality is correct, but the second is not.

Here is a link which explains in more detail. Are you writing some computer code to do this? If so in what language? You might try defining the code, maybe that will help. Define a function f that takes y as its input, and define another function that does one RK4 step.
 

1. What is a simple pendulum?

A simple pendulum is a weight suspended from a string or rod that is free to swing back and forth. It is a classic example of harmonic motion, where the motion is repeated at a constant frequency.

2. How is the motion of a simple pendulum described?

The motion of a simple pendulum can be described by its period, which is the time it takes for one complete swing, and its amplitude, which is the maximum angle the pendulum swings from its resting position.

3. What is numerical integration in the context of simple pendulum motion?

Numerical integration is a method used to approximate the motion of a simple pendulum by breaking it down into small time intervals and calculating the position and velocity at each interval. This allows for a more accurate prediction of the pendulum's motion compared to analytical methods.

4. What factors affect the motion of a simple pendulum?

The motion of a simple pendulum is affected by its length, mass, and the force of gravity. A longer pendulum will have a longer period, while a heavier pendulum will have a shorter period. The force of gravity also plays a role in determining the period of a pendulum.

5. How can numerical integration be used to analyze a simple pendulum?

Numerical integration can be used to create a graph or plot of the position and velocity of a pendulum over time. This can help to visualize the motion of the pendulum and make predictions about its behavior. It can also be used to investigate the effects of changing parameters, such as length or mass, on the pendulum's motion.

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