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Spring and gravitational energies (Simple harmonic motion)

  1. Apr 14, 2014 #1

    Show that the combined spring energy and gravitational energy for a mass m hanging from a light spring of force constant k can be expressed as 1/2 ky2, where y is the distance above or below the equilibrium position.

    Figure shows a block connected to spring, where equilibrium is, and where the spring's unstretched position is. The distance from unstretched position to equilibrium is defined as h. From equilibirum to mass is y.

    The answer starts off as

    U = 1/2 k(h+y)2 - mg((h/2) +y)

    From there it says mg=kh and substitutes for mg then solves.

    I'm confused how U equals all that. I get how potential energy of a spring is 1/2 kx^2 for a spring or mgh in other cases, but this is quite confusing.

    Then where did the 1/2 from h/2 come from?
  2. jcsd
  3. Apr 15, 2014 #2
    The answer I can think of is that all energy is relative. So if the datum is conveniently chosen so that some variables drop out then that is still valid.

    As an alternate approach, why dont you assume a datum distance of "d" and assume the two terms can be set to ky^2. Then solve for d.
  4. Apr 15, 2014 #3

    Simon Bridge

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    The potential energies are relative - but the position the values are relative to is provided as part of the problem setup.
    (Well - sort of...)

    The idea is that the equilibrium position is defined to be where the force of gravity is the same as the force of the spring ... since gravity is a constant, and the spring is extended by length h, this means: kh=mg

    To understand the original equation, first sketch out what they are describing and then work out the energies, and add them up.
  5. Apr 15, 2014 #4
    Nonsense. The gravitational potential energy datum is not given at all in the set up.
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