I've seen multiple questions about a pendulum on a train and most say to use $T = 2 \pi (L/F)^{1/2}$ and I have done this to compare the pendulum's periods before being on a train and then once its on the train I am aware the period on the train should be shorter however I am trying to prove this. My problem is resolving the Forces acting on the pendulum on the train the two forces being centripetal force and acceleration due to gravity. I've had a couple of ideas of how to do this being changing the view of the pendulum so that the equilibrium point has shifted such that the centripetal force is now acting vertically and the gravity is causing the oscillations but I can't get my head round resolving these forces

Can someone show me how to set up these forces ?

  • $\begingroup$ by $F$ you apparently mean the net force divided by the mass, right? (at least then the formula is true :) it is the new "effective $g$") $\endgroup$ – Ilja Apr 5 '16 at 12:55

I infer from the formula and your question, that you are talking about a train in a turn, that experiences centripetal force.

In this case, the "effective $g$" is greater than just the earth attraction.

In the simplest case you assume, that the new force is perpendicular to gravitation and that you know the parameters of the train motion: then the centripetal acceleration is $a = v^2/R$, and the net effective acceleration on the pendulum is $\sqrt{g^2+a^2}$. Put this into your formula as $F$ (it would be clearer to call this not $F$ but for example $g'$ ... the normal formula for a pendulum has an $\sqrt{l/g}$ in it), and there you have the solution.


Your question is incomplete. Are we talking about a train that is in a steady state with a constant speed and no turns, a train that is undergoing constant acceleration, or a train that is undergoing changing acceleration?

If the train is in a steady state then the pendulum is adequately modelled by the equation you posted.

If the train is undergoing constant acceleration, then relativity kicks in and you can use the equation if you adjust F to be not just the force of gravity but the acceleration of the train as well. If the train is in a turn, then this would be one of the rare occasions where you may use centrifugal force as opposed to centripetal force. :) Simply vector-add the centrifugal force with gravity to get your new F.

If the train is undergoing changing acceleration then that equation is no longer valid since it assumes that the pendulum is only undergoing constant acceleration. Then you will have to solve differential equations that deals with changes in accelerations.

  • $\begingroup$ where do you see relativity in the equation he posted? :) $\endgroup$ – Ilja Apr 5 '16 at 13:04
  • $\begingroup$ Heh, you are correct there. I was unintentionally confusing the comparison between an accelerating reference frame in low/no gravity vs a non-accelerating reference frame in gravity... with — in this case — a reference frame in constant motion vs one that is standing still. However... if his question involves a train in constant acceleration, then(!) relativity would indeed kick in and the equation can be used, where F in his equation need only be adjusted to the combined force vector of acceleration and gravity together. :) Answer adjusted according to your comment. $\endgroup$ – MichaelK Apr 5 '16 at 13:10
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    $\begingroup$ Your original reference to relativity kicking in when the train is moving with a constant velocity seemed fine to me. I assumed you were referring to Galilean relativity - the moving train frame of reference being equivalent to a stationary frame. $\endgroup$ – M. Enns Apr 5 '16 at 13:20
  • $\begingroup$ @M.Enns Well... yes, you are correct in that what I intended was Gallilean Relativity, even if I had the image of General Relativity in my mind. So I was right and wrong at the same time. :) $\endgroup$ – MichaelK Apr 5 '16 at 13:23
  • $\begingroup$ well, normaly you call relativity things that are connected to the theory of relativity, dont you? $\endgroup$ – Ilja Apr 5 '16 at 13:23

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