Barton's pendulum, phase relationship

Glenn G
Hi community,
The phase relationship is 0 for the shorter pendulae, 1/4 cycle for the pendulum in resonance and in anti-phase for the longer pendulae; relative to the driver pendulum.
I have observed this but I can see it conceptually to an extent but wondered if anyone knows of a resource for the mathematical basis of this. I've tried to search for it but to no avail.
Would really appreciate any help.
regards,
G

Gold Member
Hyperphysics is often a good source of information like this. This link shows the result without too much of the derivation for a mass spring oscillation. It starts with the equation of motion and shows the solution in terms of the transient solution and the steady state solution which is how things settle down. The transient part (with e-γt) dies away and leaves you with a fairly simple expression for Amplitude and Phase.
If you look at the expression for phase, you see when k=mω2, the phase difference is 90° and it is on one side or the other, depending which of the two is greater.
Is that sufficient?

Glenn G
Glenn G
Thanks, quite involved but I get it.

sophiecentaur
Gold Member
There's a reasonably intuitive way of thinking about the phase of a driven oscillator (mass on spring). When the natural frequency is equal to the driving frequency, there is a 90° phase difference between the two oscillations. If the mass is lighter, its motion will 'follow' the driving frequency easier so you can look at the phase shift as being less. If the mass is greater then it will lag behind because it never quite makes it to the 90° phase point.
Of course, Barton's Pendulum is one stage harder than this because the pendulums all have different frequencies (there's not a particular 'driving' frequency. The shorter ones will be racing ahead of the rest and the longer ones will be lagging behind. It's a case of Coupled Pendulums, rather than Driven Pendulums.

Gold Member
Hi community,
The phase relationship is 0 for the shorter pendulae, 1/4 cycle for the pendulum in resonance and in anti-phase for the longer pendulae; relative to the driver pendulum.
I have observed this but I can see it conceptually to an extent but wondered if anyone knows of a resource for the mathematical basis of this. I've tried to search for it but to no avail.
Would really appreciate any help.
regards,
G
I have tried to illustrate the similarities between mechanical and electrical resonators. It might make it easier. The electrical circuit is an LCR series circuit and for the mechanical system I have assumed a spring and a mass. In both cases the drive power is kept the same as frequency is varied. All the drive power is absorbed in friction (mechanical) or resistance (electrical) once steady state conditions are reached. These are only analogies and other comparisons are possible.