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Ok so I am an A2 physics student, and for one of my pieces of coursework I conducted a practical investigation, my topic being the factors affecting the period and swing of a bifilar pendulum.

The only useful information I was able to find on the topic was here: http://voyager.egglescliffe.org.uk/physics/gravitation/bifilar/bif.html

What I need to do is explain simply why increasing d in the diagram below, the distance between the threads on support A, while keeping d on bar B the same, will increase the period.

http://physics.dorpstraat21.nl/images/expts/bifilar%20pendulum1.png

In the link I posted above, Faysal Riaz seems to explain this in this section:

"Therefore the restoring Couple, CR, which acts towards the equilibrium position so negative, is given by:

CR = (-Mgθd2)/4y 


Applying Newton’s Second Law for the rotational motion of the rod, which is of constant mass:

(Id2θ)/dt2= (-Mgθd2)/4y

∴(d2θ)/dt2= (-Mgθd2)/4Iy"

The problem is, I have no clue what this means. Can anyone simplify this so I could understand it better, or explain why decreasing d in the way I described above, increases the period in a few lines of basic mechanics if that's possible?

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You know that swinging is oscillation of the total energy between kinetic and potential energy.

The kinetic energy in the bifilar pendulum is mainly generated by the horizontal elongation of the bar.

In the following we always consider oscillations with the same horizontal elongation angle amplitude. Furthermore, we approximate the oscillation of the horizontal elongation as sinusoidal. Under these assumptions the maximal kinetic energy is proportional to the squared oscillation frequency.

The potential energy is proportional to the lifing of the bar. For one and the same horizontal elongation angle the bar is lifted the higher the larger d is.

Therefore, for larger d you have higher maximal potential energy in the oscillation with constant amplitude therefore you have higher maximal kinetic energy which implies higher oscillation frequency.

This is the scheme, now you can try to get approximate formulae if you like.


Note: The pendulum would even work with d=0 where there is no lifting of the bar. But in this case the working principle is different. With d=0 you must consider torsion stiffness of the thread and the potential energy is stored in the torsion of the thread.

For our considerations above we have neglected this effect.

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  • $\begingroup$ @ρσݥzση "Kinetic energy proportional to oscillation frequency" was not quite right. I added "squared". On the other hand potential energy proportional to lifting is right. The lifting is approximately proportional to the squared horizontal elongation angle. This way you get something like a "spring constant" w.r.t. the elongation. $\endgroup$ – Tobias Mar 19 '14 at 12:35
  • $\begingroup$ @ρσݥzση The answer describes another case (d changes on the support and on the suspension). But, I hope you can adapt the ideas to your case. $\endgroup$ – Tobias Mar 19 '14 at 13:04

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