Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B Barton's pendulum, phase relationship

  1. Nov 8, 2018 #1
    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
     
  2. jcsd
  3. Nov 8, 2018 #2

    sophiecentaur

    User Avatar
    Science Advisor
    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?
     
  4. Nov 24, 2018 #3
    Thanks, quite involved but I get it.
     
  5. Nov 24, 2018 #4

    sophiecentaur

    User Avatar
    Science Advisor
    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.
     
  6. Nov 25, 2018 #5

    tech99

    User Avatar
    Gold Member

    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.
    upload_2018-11-25_19-53-34.png
     
  7. Nov 25, 2018 #6

    sophiecentaur

    User Avatar
    Science Advisor
    Gold Member

    Actually, that's not quite right. The middle pendulum has a heavy mass on it and the others are light. The middle pendulum can be regarded as the 'driver' and the others as 'driven'.
    I agree that the Maths may be easier (more familiar) perhaps for someone happy with EE but it is much easier to produce a row of coupled pendulums than a row of coupled electrical oscillators. The concepts of Electrical Impedance and Reactance are things that we EE's took in with our Mother's milk but I do wonder about how 'intuitive' those concepts are. The left hand side of your table would actually be enough, I feel. (It's nicely put, too)
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?