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


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    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?
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