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Accelerated pendulum

  1. Dec 26, 2008 #1


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    Assume a pendulum, a mass M attached to a massless rod, which is oscillating in small angle. According to the oscillating equation

    [tex]\ddot{\theta} + \frac{g}{l}\theta = 0[/tex]

    We find that the frequency is proportional to [tex]\sqrt{g/l}[/tex]. Now if the pendulum is accelerated upward at acceleration [tex]a[/tex], since [tex]a[/tex] is along the opposite direction as the gravity [tex]g[/tex], the frequency of the accelerated pendulum should be

    [tex]\omega = \sqrt{\frac{g-a}{l}}[/tex]

    why in textbook, in said the frequency becomes [tex]\omega = \sqrt{\frac{g+a}{l}}[/tex]?
  2. jcsd
  3. Dec 26, 2008 #2
    Aha! Actually, it is not the pendulum that is accelerated upwards! It is the peg that is holding the pendulum that is accelerated. So the pendulum itself experiences a pseudo force downwards.

    If you stand in an elevator and it accelerates at "a" then your weight
    as you would experience will be m(g+a), because ma downwards is the pseudo force.

    Pendulum is no different: the effective downward acceleration of the pendulum of mass m is (mg + pseudo force)/m = (mg + ma)/m = (g+a).

    Makes sense?
  4. Dec 26, 2008 #3


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    Yes, it makes sense.
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