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Pendulum and SHM?

  1. Sep 22, 2004 #1
    Why is it that a simple pendulum only shows SHM at small angles close to the equilibrium position?
  2. jcsd
  3. Sep 22, 2004 #2


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    Because the restoring force is not linear and is only approximately linear at small amplitudes.
  4. Sep 22, 2004 #3


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    Consider that if the launch angle of the bob is 180 degrees (the "string" would have to rigid to maintain the length), the bob would balance there. Effectively this means the period is infinite for amplitude 180 degrees. It is easy to see then that as the amplitude becomes larger, the period is no longer given by the simple formula (i.e. constant independent of amplitude), but will increase.
  5. Sep 22, 2004 #4


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    If you resolve the gravitational force on a pendulum bob at angle &theta; to the vertical into a component perpendicular to the circular path (which just puts tension on the pendulum arm) and the component parallel to the path (which accelerates and decelerates the pendulum) you find that, from the right triangle set up, the parallel acceleration is proportional to sin(&theta;). For small values of &theta; that is very close to &theta; itself. It is that simplicity: acceleration= &theta; that gives "simple harmonic motion". For larger angles, we would have to use sin(&theta); instead of &theta; and that gives a much harder problem.
  6. Sep 22, 2004 #5
    Also, something I've noticed is when the angle is fairly large, the pendulum begins to slightly vibrate as it swings. This indicates it's no longer undergoing SHM alone.
    Last edited: Sep 22, 2004

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