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Homework Help: Simple harmonic motion on an incline

  1. Feb 4, 2013 #1
    1. The problem statement, all variables and given/known data

    2. Relevant equations
    F = -dU/dx

    3. The attempt at a solution
    [itex]U = \frac{1}{2}kx^2 + mgxsin\theta \\\\
    F = -(kx + mgsin\theta) \\\\
    F = -kx - mgsin\theta \\\\[/itex]

    We want to set the force = 0 because that's when the block is in equilibrium with no forces acting on it.

    [itex]0 = -kx - mgsin\theta \\\\
    x = -\frac{mgsin\theta}{k} \\\\
    x = -\frac{\frac{14.0}{g}gsin\theta}{k} \\\\
    x = -\frac{14.0sin(40 deg)}{120} \\\\
    x = -0.075[/itex]

    So since 0.075 m is the equilibrium position, it is the distance from the top of the incline to the equilibrium position, but the solution says that 0.075 m is the displacement from the position 0.450 m to equilibrium. I didn't even include 0.450 m in the equation, how can I assume that x is the distance from 0.450 to 0.075? Why is it not some arbitrary position to 0.075?
  2. jcsd
  3. Feb 4, 2013 #2


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    Staff Emeritus
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    The solution you refer to is correct. The 0.075 m is the amount the spring is stretched from its unstretched length.

    Check Hooke's Law again.

    It may say [itex]\ \ F_\text{Spring}=-k(x - x_0)\,, \ [/itex] where x is the length of the spring, and x0 is the unstretched length.

    Or it may say [itex]\ \ F_\text{Spring}=-k\,x\,, \ [/itex] where x is the amount the spring is stretched (from its unstretched length).
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