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I am working on a particular type of pendulum called the Flying Pendulum. I did not find anything on the internet on the physics behind its working and so I am trying to prove that

Here is a flying pendulum video that shows the working of the pendulum : https://www.youtube.com/watch?v=xWCmXvWM0ZY

1) In order for a clock to function, the movement has to be isochronous! The pendulum should follow SHM. BUT DOES BEING ISOCHRONOUS MEANS THAT A PENDULUM IS FOLLOWING SHM?

2) I did an experiment and I have data with different lengths, time period, velocity, acceleration and the angle is 20 degrees. I used small angle approximaion for it and I know about the second differential equation.

However, I want to do a general conjecture kind of explanation by proving LHS = RHS showing that a flying pendulum also follows SHM. But I am really confused how to do that! Could someone please help me with this?

Thanks for your help

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    $\begingroup$ Hi Rosalin. There must be thousands of articles on the Internet about analysing the motion of a pendulum. Can you give us some idea of what you've already read on the subject? As it stands your question is likely to be closed on the grounds of insufficient prior research. $\endgroup$ – John Rennie Aug 25 '16 at 8:25
  • $\begingroup$ Actually I am working on a particular type of pendulum called the Flying Pendulum. I did not find anything on the internet on the physics behind its working and so I am trying to prove that in order for a clock to function, the movement has to be isochronous! So the pendulum should follow SHM. I know about the second differential equation with an angle of 20 degrees and I want to do a general conjecture kind of explanation by proving LHS = RHS that it follows SHM. But I am really confused how to do that! :@ Thanks for your help $\endgroup$ – Rosalin Aug 25 '16 at 8:42
  • $\begingroup$ Then you need to edit your question accordingly. Also bear in mind that most of us have no idea what a Flying Pendulum is so your question needs to explain this. $\endgroup$ – John Rennie Aug 25 '16 at 8:46
  • $\begingroup$ If you're asking about the definition of simple harmonic motion, i.e. what do the equations of motion look like for SHM, then this too should be extensively covered on the Internet. If you write down what you believe the EOM to be then we would be in a better position to comment. $\endgroup$ – John Rennie Aug 25 '16 at 8:48
  • $\begingroup$ See this question for a related discussion. This references Susskind's book The Theoretical Minimum, and you'll find the discussion of the Langrangian/Hamiltonian treatment in the chapter Lecture 8: Hamiltonian Mechanics and Time-Translation Invariance. $\endgroup$ – John Rennie Aug 25 '16 at 8:52
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BUT DOES BEING ISOCHRONOUS MEANS THAT A PENDULUM IS FOLLOWING SHM?

No. Isochronous motion is a synonym for periodic motion, i.e. $$ f(t + T) = f(t)\ \forall t$$ given the period $T$. SHM is a particular type of periodic motion where $f(t) = A \cos\left(\frac{2\pi}{T} t + \phi\right),$ and other forms related by trig identities and definitions of quantities.

What you want to look into are anharmonic oscillators (eg $F = -k x -\lambda x^3$).

As for your second question, I don't see how you can do a small angle expansion for a flying pendulum like the one depicted in the video. The angles the pendulum swings through are anything but small, and things like the rod width are non-negligible.

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    $\begingroup$ Thanks so so so much. I haven't really learnt anything in that depth about the anharmonic oscillators. But I will try to use some basic knowledge by studying it!! :) ALSO, IF THE OSCILLATION ANHARMONIC, I CANT'T USE THE T = 2pi root(l/g) ??? Thanks. $\endgroup$ – Rosalin Aug 25 '16 at 19:28
  • $\begingroup$ You may be interested to know that we cannot solve anharmonic oscillators exactly. In fact, the type of perturbation theory used in quantum mechanics was developed for solving classical anharmonic oscillators approximately. If I recall correctly, the first application for perturbation theory was calculating orbits of planets. Sorry if this is rambly - I'm very tired. $\endgroup$ – Sean E. Lake Aug 25 '16 at 19:33
  • $\begingroup$ $T= 2\pi \sqrt{\frac{l}{g}}$ is based on the small angle approximation of an ordinary pendulum, so if it gets the right answer it's a coincidence. Worse, for anharmonic oscillators the period depends on the amplitude. Consider the simple, real, rigid pendulum (eg a rigid bar with a pivot at one end). You can solve the motion using torque: $$\begin{align}\tau &= I \alpha\\ &\propto -\sin \theta,\end{align}$$ with $\theta$ the angle the bar makes from rest. Notice how $\tau=0$ when $\theta=0,180^\circ$? That means the bigger the swing, the longer $T$, up to $\infty$, if you can balance it. $\endgroup$ – Sean E. Lake Aug 26 '16 at 7:13

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