Proving the period of a pendulum

In summary, you are trying to solve an ODE that has initial conditions that are undefined. You have an equation for the position (##\theta(t)=-\theta(4t_2-t))##, but you don't know what the initial condition is. You need to show that the initial condition going forward from 4t2 are the same as those going forward from t=0. You need to solve the equation for ##\theta## and ##\dot\theta##, and then use a similar argument from before to show that the values of ##\theta## and ##\dot\theta## define the state of the system.
  • #1
GetSchwifty
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0

Homework Statement


I'm writing a small project as part of my degree, and I am stuck on proving one (probably simple!) thing towards the end. I've shown that the pendulum is periodic and has a total period of 4t2.

Homework Equations


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The Attempt at a Solution



I think I need to rewrite the same ODE with starting point t=0 but I'm having a complete mindblank on how to proceed from here. Any help would be greatly appreciated!
 
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  • #2
Not entirely sure about your notation, nor what your starting points are.
You seem to be drawing on some earlier equation ##\ddot\theta=csin(\theta)##, but what is csin there? Do you just mean a constant c multiplied by sine? "csin" is a standard abbreviation for complex sine in some computer languages.
You also seem to know that ##\theta(t_2)=\dot\theta(t_2)=\dot\theta(0)=0##.
Anything else?
 
  • #3
haruspex said:
n
Hi,

Sorry I should've given more information. The equation I'm drawing on as is as you said is ##\ddot\theta = -csin\theta## and this is just the regular sine function multiplied by a constant, c. No complex sine - sorry for the confusion.[PLAIN][PLAIN]http://blob:http://imgur.com/d719fbec-3e87-4f2c-a28f-955630eb054a [PLAIN][PLAIN]http://blob:http://imgur.com/d719fbec-3e87-4f2c-a28f-955630eb054a [PLAIN][PLAIN]http://blob:http://imgur.com/d719fbec-3e87-4f2c-a28f-955630eb054a [PLAIN][PLAIN]http://blob:http://imgur.com/d719fbec-3e87-4f2c-a28f-955630eb054a
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Here is everything I know that's relevant to periodicity in my paper. Thanks for your reply!
 
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  • #4
The latex is clearer if you use \sin rather than just sin.

As far as I can see, nowhere did you define the initial conditions (t=0). But in the second stage (post #1) you use ##\dot\theta(0)=0##.

You have an equation ##\theta(t)=-\theta(4t_2-t)##.
What would that imply at t=2t2? Is the minus sign a mistake?
 
  • #5
haruspex said:
The latex is clearer if you use \sin rather than just sin.

As far as I can see, nowhere did you define the initial conditions (t=0). But in the second stage (post #1) you use ##\dot\theta(0)=0##.

You have an equation ##\theta(t)=-\theta(4t_2-t)##.
What would that imply at t=2t2? Is the minus sign a mistake?
Noted for the future.

The minus sign is a typo, thanks for pointing that out!

At time t=2t2, we have that it's equal to ##\theta(-t)## by symmetry, since it's the furthest "left" position of the mass (taking moving from left to right as being negative).
 
  • #6
I think I need to prove that ##\theta(4t_2) = \theta(0)## but I'm not entirely sure... I think to do that I would need to basically run through a very similar argument again and then use cauchy-lipschitz to show uniqueness
 
  • #7
GetSchwifty said:
I think I need to prove that ##\theta(4t_2) = \theta(0)## but I'm not entirely sure... I think to do that I would need to basically run through a very similar argument again and then use cauchy-lipschitz to show uniqueness
You need to show that the initial conditions going forward from 4t2 are the same as those going forward from t=0. So that's the values of ##\theta## and ##\dot\theta##.
 
  • #8
Thanks. It's very late here in the UK so I'll actually write this up tomorrow - the flu really isn't helping my thought process!

So in short I need to show that ##\theta(0) = \theta(4t_2)## and that ##\dot\theta(0) = \dot\theta(4t_2)##? And then use a similar argument from before, from my second post with:

$$\begin{cases}

\ddot y& = -csiny(t)\\

y(t_2)&=-\theta(2t_2-t_2)=-\theta(t_2)=0 \\

\dot y(t_2)&=\dot\theta(2t_2-t_2)=\dot\theta(t_2)

\end{cases}$$

Or am I way off?

And then I would need to change this to reflect 0 and ##4t_2##?

$$-\theta(2t_2-t)=\theta(t), t\in[t_2, 2t_2]$$
 
  • #9
GetSchwifty said:
I need to show that ##\theta(0) = \theta(4t_2)## and that ##\dot\theta(0) = \dot\theta(4t_2)##?
Yes.
GetSchwifty said:
then use a similar argument
As you wish, but to me it would be suffice to say that the values of ##\theta## and ##\dot\theta## define the state of the system, so if the value pair recurs then the behaviour is cyclic. Likewise, for your antisymmetric result in post #3, I would just observe the effect of time reversal on the differential equation.
These are mechanistic arguments rather than purely mathematical ones.
 
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FAQ: Proving the period of a pendulum

1. What is the period of a pendulum?

The period of a pendulum is the time it takes for the pendulum to complete one full swing, from its starting position, back to that same position.

2. How do you prove the period of a pendulum?

The period of a pendulum can be proven using a simple formula: T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity. This formula was first discovered by Galileo Galilei in the 17th century.

3. What factors affect the period of a pendulum?

The period of a pendulum is affected by three main factors: the length of the pendulum, the mass of the pendulum, and the acceleration due to gravity. The longer the pendulum, the longer the period. The heavier the pendulum, the shorter the period. And the higher the acceleration due to gravity, the shorter the period.

4. How does the period of a pendulum relate to its amplitude?

The period of a pendulum is independent of its amplitude, which is the distance the pendulum swings from its starting position. This means that no matter how far a pendulum swings, its period will remain the same as long as the other factors (length, mass, and gravity) stay constant.

5. What is the importance of proving the period of a pendulum?

Proving the period of a pendulum is important because it allows us to accurately measure time. Pendulums have been used in timekeeping devices, such as clocks, for centuries. Understanding the factors that affect the period of a pendulum can also help us design more accurate timekeeping devices and study the laws of motion.

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