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Simple Harmonic Motion Pendulum, can we use PE=1/2kAmplitude^2?

  1. Aug 31, 2014 #1
    I was wondering if we can somehow use the formula Potential Energy = 1/2K(x(x=Amplitude))^2 for a pendulum if we are only given the angle of displacement?

    Would the problems normally just say the PE at the top of the pendulum is such and such, please find max Velocity,
    Or also the max velocity is such and such please find max potential energy, or max amplitude.
    Last edited: Aug 31, 2014
  2. jcsd
  3. Aug 31, 2014 #2
    Your formula for potential energy is not the one usually given in descriptions of a harmonic oscillator, and it is generally unclear what you are asking about. Being more specific could help.
  4. Aug 31, 2014 #3


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    No, you can't. ##k## is the spring constant. There is no spring involved with a pendulum.
  5. Aug 31, 2014 #4

    Philip Wood

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    The k in Hong's formula is an equivalent spring constant. The gravitational PE of the pendulum bob relative to its lowest position is, mgh, in which h is its height relative to its lowest point. It's easy to show by Pythagoras that for small displacements with horizontal component x, for a pendulum of length L, [itex]x^2 = 2hL[/itex]. So the PE of the bob is [itex]mgh = \frac{1}{2} \frac{mg}{L} x^2[/itex], so the effective spring constant, k, is [itex]\frac{mg}{L}[/itex].
  6. Aug 31, 2014 #5
    Thanks Everyone!
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