Wilberforce Pendulum - electrical equivalent - Page 1
Forum section: electronics / projects

author: moffy
author_rank: Super Contributor
timestamp: August 07, 2024, 04:32:20 am
content: I ran across something new for me called the Wilberforce Pendulum:https://www.youtube.com/watch?v=c0YShREnsWs&pp=ygUUd2lsYmVyZm9yY2UgcGVuZHVsdW0%3Dwhich is a coupled pendulum, where you start one part oscillating and all the energy gets transferred to the second part then backwards and forwards. So I wondered if I could do the same electrically, so I coupled two LC tanks with a capacitor and started one of them oscillating by stopping a DC current going through one of the coils. Not sure if it has any uses but it is interesting and replicates the mechanical system well.

author: moffy
author_rank: Super Contributor
timestamp: August 08, 2024, 12:24:07 am
content: Sorry, I just realised that I left out an explanation for the equivalence of the electrical model with the Wilberforce pedulum.You can think of it this way:1. L1 is the mass hanging at the end of the spring, no rotational component.2. C1 is the main spring in extension, linear motion.3. L2 is the rotational moment of inertia of the mass at the end of the spring.4. C2 is the torsional component of the spring.5. C3 is the spring coupling between the linear and rotational motions.With these equivalences in place then:a. I(L1) is linear velocity. e.g. k1*L1*I*I/2 = m*v*v/2b. V(C1) is the linear displacement of the spring e.g. k2*C1*V*V/2 = k*x*x/2*k1 and k2 are scale factors.Since the voltage across L1 and C1 is the same, that means their displacement is always the same.For the torsional component mass becomes moment of inertia and displacement becomes rotation.The only slight anomaly is how the oscillation is started by a current suddenly switched off which would be equivalent to imparting a fixed velocity to the mass, I could have used a voltage source and some switches, equivalent to stretching the spring and releasing, but electrically the current source is more convenient. I hope this fills in some gaps from the first post.

author: RoGeorge
author_rank: Super Contributor
timestamp: August 11, 2024, 06:36:37 am
content: Nice demo!For the docs only, a link summarizing the equivalence between electric circuits and mechanical systems:https://lpsa.swarthmore.edu/Analogs/ElectricalMechanicalAnalogs.html

author: moffy
author_rank: Super Contributor
timestamp: August 11, 2024, 07:46:39 am
content: Conservation of EnergyIt is possible to make the mass equivalent to the capacitance and the inductance equivalent to the spring constant because their energy equations, C*V*V/2 and L*I*I/2 have the same form. By swapping them, voltage becomes equivalent to velocity and current becomes equivalent to displacement. The current source is now equivalent to stretching the spring and releasing it.e.g.

author: iMo
author_rank: Super Contributor
timestamp: August 11, 2024, 08:36:56 am
content: Your sim oscillation should stop after a while as the default LTspice series resistance in L is 1m, afaik..

author: moffy
author_rank: Super Contributor
timestamp: August 11, 2024, 08:45:02 am
content: Quote from: iMo on August 11, 2024, 08:36:56 amThanks very much for that, I did notice the oscillation decaying and was less than half by 4ms but I couldn't find any resistive components. I put it down to convergence losses because when I changed the solver to 'Gear' the decay became much faster, but I prefer your explanation.P.S. Found the reference in their help file under 'L:Inductor':"By default, LTspice will supply losses to inductors to aid SMPS transient analysis. For SMPS, these losses are of usually of no consequence, but may be turned off if desired. On the "Tools=> Control Panel=>Hacks!" page, uncheck "Supply a min. inductor damping if no Rpar is given." "P.P.S. There is a second hack along the lines of what you stated: "Always default inductors to Rser=0".

author: MrAl
author_rank: Super Contributor
timestamp: August 11, 2024, 11:29:34 am
content: The force current analogy is common. That makes the current equivalent to force and the capacitance equivalent to mass.

author: moffy
author_rank: Super Contributor
timestamp: August 11, 2024, 11:45:51 am
content: Quote from: MrAl on August 11, 2024, 11:29:34 amDepends upon what is being modeled mechanically. If modelling a DC motor where the capacitance represents the moment of rotational inertia and rotational velocity is represented by voltage, just like a real DC motor, then current equals force/torque, but that is not modelling, that is just the way a DC motor works. In a DC motor current directly relates to torque or force because the magnetic field which produces the torque is related to current.In this case not so, a different mechanical system.P.S. You are correct since the force exerted by a spring is F = -k*x, and since x is represented by 'I' then so is force, but different scalings and different directions, but that only applies to the second model.

author: MrAl
author_rank: Super Contributor
timestamp: August 12, 2024, 12:37:14 pm
content: Quote from: moffy on August 11, 2024, 11:45:51 amOh I am not saying it's the only analogy out there, there are lots of them. There's one instance where the force/current analogy is harder to get to work too but most of the time it makes sense. It's really the math that is the same that's all.