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Based on my own research, I found a general solution that can model a pendulum found on an accelerating train. The following solutions are based on small angle approximations.

F=-mgsin θ

F≈-mgθ, applying small angle approximation

F=-(mg/L) s, by applying arc length formula

This is in the form, F = -kx(hooke's law)

Therefore k = mg/L

Applying newton's laws

F = ma = -kx

Solving this eqn, ma = -(mg/L)x

we get,

enter image description here

This general equation can help plot the sway angle of the pendulum. Another known fact is that the length of a pendulum affects its period.

I am wondering if the equation will hold if the length of the pendulum is changing. i.e Becoming shorter progressively.

I expect the period to get shorter as the length shortens. How will its amplitude be affected?

Hari
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  • Do let me know if the question is unclear. I will perform the necessary edits if required. – Hari Mar 01 '21 at 06:56
  • How did you did you arrive at your1st formula? What's the meaning of the firrdt symbols? – Gert Mar 01 '21 at 07:26
  • Hi @Gert, the first formula can be derived by making use of newton's second law as well as making use of Hooke's law for a mass attached to a spring. Given by -kx = ma. By solving this differential equation and substituting k= mg/L which is the restoring force for a pendulum, the first equation can be derived. – Hari Mar 01 '21 at 07:43
  • can you please write your EOM i don’t think that your solution is correct? – Eli Mar 01 '21 at 07:59
  • Hi @Eli, I have added the method to derive the equations, do let me know if you need further elaboration – Hari Mar 01 '21 at 08:42
  • @Hari i got this equation https://physics.stackexchange.com/questions/616539/how-can-we-find-such-an-angle-possible-for-this-q – Eli Mar 01 '21 at 11:16
  • @Eli I believe there are multiple ways to end up with the solution. The method that you have presented, involves coming up with the differential equations which represents the motion of the system. If we were to apply small angle approximation and solve it, we can arrive at the same equation as the equation mentioned in my post. – Hari Mar 02 '21 at 05:14
  • you didn’t mention small angle in your post – Eli Mar 02 '21 at 07:25
  • @Eli sorry about that I will make the necessary edit – Hari Mar 02 '21 at 07:35

1 Answers1

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Your initial equation of motion is of the form: $$m \mathbf{a} = \mathbf{F}(L, g, a). $$ However, when $L$ varies in time, $F$ also becomes time dependent. Thus, your initial differential equation differs from your new one in dynamics. Therefore, you should solve the problem again.

By the way, your solution seems not quiet working. Namely, using equivalence principle, a pendulum in accelerating train plus gravity is equivalent to a pendulum in a gravitational field that is the vector sum of $-\mathbf{a}$ and $\mathbf{g}$. Hence, the frequency of the pendulum should be dependent at least to $g$ and $a$. Moreover, the axis of oscillation should be change by something like $\tan^{-1}(\frac{-g}{a})$.

Appo
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  • Hi @Appo, You are right, However, i believe the new axis of oscillation should be atan(ma/mg) = atan(a/g). I am dealing with small angles, hence I applied the small angle approximation. Furthermore, since the accelerations I am working with are 0.5m/s^2 . I once again approximated it to just taking g instead of the vector sum of a and g since the difference between the two are <1% – Hari Mar 01 '21 at 07:54
  • Based on your explanation above, what if instead of solving the differential each time, I make use of the general equation. Wouldn't that provide me with the same equation except that that the length variable has changed? – Hari Mar 01 '21 at 08:03
  • @Hari The only thing I can say in general, if the variation of $L$ is adiabatic (its energy variation in one period is negligeable toward the energy of the system), then you can keep the same form of solution. That was based on adiabaticity theorem of quantum mechanics which should remain true in such classical periodic problems. – Appo Mar 03 '21 at 11:19
  • That makes sense thank for your help – Hari Mar 03 '21 at 11:36