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 ?
Answer
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.
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