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Simple Harmonic Motion and pendulum clock

  1. May 5, 2009 #1
    1. The problem statement, all variables and given/known data

    What will be better to take to the moon, a spring clock or a pendulum clock. Why?

    2. Relevant equations

    F= -k*(Delta)x

    k = g m/Dx

    T = 2*π*[square-root of:(m/k)]

    3. The attempt at a solution

    I know that there is less gravitational force while on the moon... Just dont see what the difference would be between block clocks. they both depend on gravity in the same manner...
  2. jcsd
  3. May 5, 2009 #2
    I'm not sure... but doesn't the equation read F=-kx, but k is NOT dependent on g? couldn't the spring be sideways?
  4. May 5, 2009 #3
    well Usually weve been doing mass spring problems vertically, else, he would have told us... but what would be the factor that would affect a change in k while on the moon???
  5. May 5, 2009 #4
    the only things that the equation involves are: mass, gravity, k constant
  6. May 6, 2009 #5
    A pendulum would be better. For a pendulum, period is about 2pi*sqrt(l/g) This is dependent on g, which is different in the moon. It is dependent on g, because g is the restoring force which brings the mass at the end of the string back to equilibrium position. However, for a spring block oscillator, the mass of the object (ideally) doesn't contribute to the restoring force, which is just -k(Dx), and the period is independent of g.

    Hope this helped.
  7. May 6, 2009 #6


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    The spring clock would be better because it will run at the same rate on the moon as it does on earth.
  8. May 6, 2009 #7
    thank you guys very much.
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