Anonymous
Anonymous asked in Science & MathematicsPhysics · 2 months ago

Simple Harmonic Motion Question?

At a certain harbor, the tides cause the ocean surface to rise and fall a distance d (from highest level to lowest level) in simple harmonic motion, with a period of 12.9 h. How long does it take for the water to fall a distance 0.330d from its highest level?

When I solved this, I got 2.15 but that is wrong.

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  • Anonymous
    2 months ago

    0.33*2 = 0.66

    Θ = arccos 0.66 ≅ 48.7°

    t = 12.9/4*48.7/90 = 1.745 h

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  • 2 months ago

    The period is12.9 h. half is rise and half is fall = 6.45*.33=2.13h.

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  • 2 months ago

    You are given a period T = 12.9 hr and the fact that this is simple harmonic motion.  So chose to represent the height of the water, y by

    y = d*cos(2*pi*t/T)  where t = time in hours.  Note that at t = 0, the tide is at its highest point

    Now the water falls a distance 0.33d so the height of the water at that time is y = d - 0.33d

    y = 0.67 d = d*cos(2*pi*t/T)   now take the arccos and solve for t

     arccos(0.67) = 2*pi*t/T -->  t = T*arccos(0.67)/(2*pi) = 1.72 hours

    Remember to get arccos back in radians, not degrees

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