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Question about simple harmonic motion

  1. Oct 9, 2005 #1
    A square block, with a mass of 3.40 kg and edge lengths d = 6.00 cm, is mounted on an axle through its center. A spring of spring constant k = 1190 N/m connects the block's upper corner with a rigid wall. Initially the spring is at its rest length. If the block is rotated by 3° and then released, what is the period of the resulting SHM?

    What type of problem should this be treated as?
  2. jcsd
  3. Oct 9, 2005 #2


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    It oscillates by rotation ... it's called a torsional oscillator.
    You look at "restoring torque" which returns the object
    (which responds slowly due to its rotational Inertia) to
    the equilibrium orientation angle.

    set torque = I alpha , get torque as function of theta.
    Now it should operationally look like an oscillator eq'n.

    Be careful to keep the omega_(orientation_change_rate)
    distinct from the omega_(forward trig function argument)
    omega_ocr has amplitude 3 degrees, while
    omega_tfa is multiplied by time.

    Enjoy it, this one is fun!
  4. Oct 9, 2005 #3
    there are two different omegas? I'm slightly confused. I know for a torsion oscillator, period is usually found using T = (2*pi)*(I/kappa)^(1/2)
    Inertia can be calculated...but how should I go about getting kappa, setting the net torque = -k*theta?
  5. Oct 10, 2005 #4


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    Torque is force multiplied by perpendicular distance from axis of rotation.

    Tau = -K(d/2)^2 @ sin@ =@ approx
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