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Author Topic: Barton’s Pendulum  (Read 15404 times)
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Fu-Kwun Hwang
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on: April 24, 2010, 05:33:56 pm » posted from:Taipei,T\'ai-pei,Taiwan

Quoted from: http://www.fas.harvard.edu/~scdiroff/lds/OscillationsWaves/BartonsPendulum/BartonsPendulum.html

All objects have a natural frequency of vibration or resonant frequency. If you force a system - in this case a set of pendulums - to oscillate, you get a maximum transfer of energy, i.e. maximum amplitude imparted, when the driving frequency equals the resonant frequency of the driven system. The phase relationship between the driver and driven oscillator is also related by their relative frequencies of oscillation.

Barton’s Pendulum consists of eleven pendulums hanging from a single thread that is connected between the two ends of a wooden rod (figure 1). The thread sags in this asymmetric way because the driver pendulum is a wooden ball 5cm in diameter, and the other ten are inverted Belmont Springs® drinking cups. The lengths of the driven pendulums range from 1.0m to 0.1m in 10cm intervals; the driver is 0.5m in length to the center of the ball. When the driver is given a swing, it sets into motion the other ten pendulums, with the result that the 0.5m driven pendulum has the largest amplitude and the other amplitudes being smaller and smaller the further away from the 0.5m they are.

You also get a very clear illustration of the phase of oscillation relative to the driver. The pendulum at resonance is π/2 behind the driver, all the shorter pendulums are in phase with the driver and all the longer ones are π out of phase.

You can change the length of the driven pendulum (change ID from 1-10).

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