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Math Help - Simple Harmonic Motion problem

  1. #1
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    Question Simple Harmonic Motion problem

    Yet another unsolved homework problem, this one involving simple harmonic motion:

    "The speed of a point moving along the x-axis is given by

    v^2 = 108 + 72x - 36x^2

    Find the mean position, the amplitude and the frequency of the motion."

    What do they mean find the mean position? Shouldn't it be at the centre? I know that I can use the formula

    x = Acos(ωt + α)

    to find amplitude (A), and

    f = 1/T where T = 2π/ω

    to find frequency (f)

    But I can't even seem to get started. Thanks for any help
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  2. #2
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    at max displacement, v = 0

    108 + 72x - 36x^2 = 0

    36(3 + 2x - x^2) = 0

    36(3-x)(1+x) = 0

    the particle oscillates between x = 3 and x = -1

    because of symmetry, equilibrium will midway between these two x-values, at x = 1.

    at equilibrium, the particle is at v_{max} = A\omega

    v^2 = 108 + 72(1) - 36(1^2) = 144

    v = 12

    12 = A\omega = 2\omega

    \omega = 6

    f = \frac{\omega}{2\pi} = \frac{3}{\pi} hz
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  3. #3
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    Cool thanks skeeter
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