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Homework Help: Grandfather clock

  1. Feb 10, 2005 #1


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    A grandfather clock contains a pendulum of lenghth 0.99336 m with a weight of 1.4kg. The driving mechanism comprises anohter mass of 2kg which goes down by 0.8m in 7 days in order to keep the amplitude of oscillation of the pendulum to give theta=0.08(<<1). What is the natural period of the pendulum? What is the value of Q for this system? Take the gravitational acceleration equal to 9.80665m/s^2?

    For the period part of the question, I simply assumed that this was a simple pendulum and used the equation T=2pi*sqrt(l/g) where l is the length of the pendulum (ie 0.99336). Can someone please check if this is right :shy:
  2. jcsd
  3. Feb 10, 2005 #2


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    Q: What is the pendulum's natural period T?
    Yes, period T is given (for the simple small oscillation pendulum) by:

    [tex] T = 2\pi\sqrt {(L/g)} [/tex]

    where L is pendulum length and g is gravitational acceleration. Would it surprise you to learn that almost every Grandfather Clock has period T=(2 sec)??

    Q: What is Q factor for this pendulum??
    The Q factor is a measure of pendulum efficiency and reflects the pendulum's energy loss per period as a fraction of its total energy (which will be a mix of potential and kinetic energies during the pendulum's oscillation). Q is defined by:

    [tex] Q = \frac {(Total Energy)} {(Energy Lost Per Period)} [/tex] <--- Make sure this Q def is the same used by your text

    where (Total Energy) is given by the following for pendulum mass "m" and displacement angle (theta):

    [tex] (Total Energy) = mgL(1 - cos(\theta)) [/tex] <--- Study the derivation in your text

    The (Energy Lost Per Period) can be inferred from the work required to keep it oscillating provided by the falling driving weight "M":

    [tex] (Energy Lost Per Period) = (Work Rate)*(Period \ T) [/tex]

    [tex] (Work Rate) = Mg(Fall Velocity Of Driving Weight) [/tex]

    Some values provided by the problem statement are:
    (Fall Velocity of Driving Weight) = (0.8 meters)/(7 days) = 1.323x10^(-6) m/sec
    M = 2 kg
    [tex] \theta = 0.08 \ radians [/tex]
    L = 0.993 meters
    m = 1.4 kg
    g = 9.8 m/sec^2
    T = Should be close to 2 sec
    Q = Should be near 840

    Last edited: Feb 10, 2005
  4. Feb 10, 2005 #3


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    ^^^Ooooo I see it now. The thing is that my textbook actually had an equation for Q but I really didn't understand the equation itself or how to use it because there were so many missing variables. But yah, I think I like your approach to finding it. It's a lot more logical than my text. Thanks a lot =)
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