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2nd Order Linear - Modeling Spring Oscillation

  1. Jul 20, 2012 #1
    To my knowledge I assume:

    Newton's second law of motion :

    F = ma = mx''

    Hooke's Law

    F = ks where s is the distance displaced by the mass.

    When a mass is attached to the spring, the new spring force is:

    F = k(s+x)

    While the downward force is still:


    If the two forces are equal and opposite, then:

    mg = k(s+x) after simplifying



    My question is: From kx=0, how do I get:


    I know how this leads to the final equation mx''-kx=0

    But do we just assume that kx is a force and set it equal to F. I just don't understand the reasoning.
  2. jcsd
  3. Jul 20, 2012 #2


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    What do "s" and "x" mean here? In the first two equations, they are essentially the same. Do you mean that s is the "natural length" of the spring?

    What? mg= ks+ kx so kx= mg- ks. How did that become 0? At the equilibrium position, if s is the length of the spring when hanging with mass m attached (NOT quite the same as the "natural length"), then that will be correct and the conclusion would be that x= 0- that we are at the equilibrium position.

    You don't. F= k(s+ x)- mg.

    Perhaps because that is NOT the correct result. There is a gravitational force of -mg, which is always negatve and so always downward, and a spring force of -k(s+x), which is negative if x< -s, if x is above the "natural length", and so downward, but positive if x> -s, s is below the "natural length", and so upward.
    "mass times acceleration equals force" gives mx''= -k(s+x)- mg which is the same as mx''+ kx= -ks- mg.

    Now, with that given, the standard method of solving such an equation is to find the general solution to the "associated homogeneous equation", x''+ kx= 0, and add to that any specific solution to the entire equation. That may be where "x''+ kx= 0" is coming from.
  4. Jul 20, 2012 #3
    Start with any spring attached to the ceiling of room. Attach a mass to this spring. The spring will stretch downward due to the mass attached to it. Eventually the mass, attached to the spring, will reach an equilibrium point. At this point, there are two forces acting on the mass: The force of gravity puling the mas down, and the force of the spring puling the mass up.

    For this problem, we assume that the downward direction is positive, while the upward direction is negative.

    Downward force: F=mg

    Upward force: Using Hooke's Law: the spring itself exerts a restoring force, F, opposite to the direction of the elongation, s. S is the distance the spring stretched (the length of the spring after attaching the mass - the length of the spring before attaching the mass).

    *See figures 1 and 2 of my attachment.

    Simply stated: F=ks, where k is a constant of proportionality.

    At equilibrium, the two opposing forces are equal.

    mg=ks --> mg-ks=0

    Then, we use our hands to pull the mass down, past it's equilibrium point.

    The two forces acting on the object are now:

    Downward: F=mg

    Upward: F=k(s+x) = ks+kx

    If the two forces are equal:

    mg = ks + kx
    mg - ks - kx = 0 In this system, we know that mg-ks=0, so
    0 - kx = 0

    Attached Files:

  5. Jul 20, 2012 #4
    Oops, That's wrong but I figured it out.

    The total force acting on the mass, when it's stretched down past equilibrium, is:

    F = mg - k(s+x)

    The upward force "-k(s+x)" is negative and the downward force "mg" is positive because we assumed upward direction was negative and downward direction was positive. Simplifying:

    F = mg - ks - kx Again, in this system we know that mg-ks=0, so
    F= -kx The total force acting on the mass at a distance x from the

    Once we let go of the mass, the spring will bounce up and down, or oscillate. Force in motion is characterized by F=ma

    ma = -kx Acceleration can be expressed as the 2nd derivative of x with respect
    to time
    mx'' = -kx

    mx'' + kx = 0

    Thanks for trying to help. I'm sorry I'm not good at explaining this stuff. I had assumed that this problem was a standard 2nd order linear differential equation application and that most people would know what I was trying to say.
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