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Simple Harmonic Motion of a plank of mass

  1. Apr 10, 2006 #1
    I don't understand the solution to the following problem:

    A horizontal plank of mass m and length L is pivoted at one end. The plank's other end is supported by a spring of force constant k. The moment of inertia of the plank about the pivot is (1/3)m(L^2). The plank is displaced by a small angle theta from its horizontal equilibrium position. Show that it moves with simple harmonic motion with angular frequency omega = square root (3k/m).

    Here is my work:
    net torque = I[I]*alpha [/I]
    net torque = (1/3)m(L^2)*alpha
    kxd = (1/3)m(L^2)*(second derivative of theta)
    x is the vertical distance from the horizontal equilibrium position.
    kxLsin(theta) = (1/3)m(L^2)*(second derivative of theta)
    kxL(theta) / (mL^2) = second derivative of theta
    because theta is small, sin(theta) = theta
    kx(theta)/ mL = second derivative of theta
    using this, I would get omega as the square root of (3k/mL)
    How do I get rid of that L in my formula for omega? Where was I supposed to drop it?

  2. jcsd
  3. Apr 10, 2006 #2

    Physics Monkey

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    Your formula for angular velocity can't be right because it doesn't have the right units to be an angular velocity. So what did you do wrong?

    You have this equation [tex] \frac{d^2\theta}{dt^2} = \frac{3 k }{m L} x [/tex], but you can't just conclude that everything in front of the x is equal to [tex] \omega^2 [/tex]. To see why not, think about the units. What you need to do is relate x to [tex] \theta [/tex], then try to extract the angular frequency.

    Hope this helps.
    Last edited: Apr 10, 2006
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