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Homework Help: Vertical spring energy transformation

  1. Apr 28, 2006 #1
    a mass is attached to a spring and released. it then oscillates in simple harmonic motion. what is the transformation of energy?

    i understand how it works horizontally (max Ee at the 2 ends, max Ek at the equilibrium position), but how does it work vertically now that Eg is also present?

    this is how i see it: setting the maximum stretch as h = 0, upon release of the mass, Eg = max (let's set max as 1 unit), Ee = 0, Ek = 0. at the bottom, Ee = 1, Eg = 0, Ek = 0. but at the middle, which is the equilibrium point, Ek should be max, and Eg and Ee should be 0, but it is half way of the unstretched position, so Ee = 1/2, and Eg is also 1/2 since it is half way between max height and min height (which i set to 0), so how can there be Ek? this is what's confusing me.
    Last edited: Apr 28, 2006
  2. jcsd
  3. Apr 29, 2006 #2

    Chi Meson

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    Ee=1/2kx^2. At the halfway point, the Ee is not half of that at the max strecth.
  4. May 5, 2006 #3
    OKay okay,let's try and clear up things once and for all. So in the case of the vertical mass and spring, there are three distinct kinds of energy transformations that are occuring. They are Kinetic energy, gravitational potential energy and elastic potential energy. In the simple case, thier formuals are (in order): 1/2mv^2, mgh and 1/2kx^2,where m is the mass of the BLOCK, v is the velocityof the block, h is the change in height of the block, k is the spring constant, and x is the stretching distance of the spring.

    OKay, that was a mouthful, now for the big picture. So when the mass is released. it falls...simple. This is due to the gravitaional potenial energy! As it falls, that energy is being converted to both kinetic AND elastic potential enegry, which should make sense; the block is picking up speed while the spring is stretching. Throughout the process, its a sort of balance scale of energy, where at the VERY top of the oscillation you have GRAVITATIONAL potential energy, and ELASTIC potential energy at the bottom. Kinetic is just "there" so to speak in the midst of transition.

    Mkay, I know that was a lot, but hopefully it put the whole picture together.
  5. May 5, 2006 #4


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    There is actually something interesting and nontrivial in the case of a vertical spring. In calculating [itex] {1 \over 2 } k x^2 [/itex], where is the x measured from? One can measure it from the equilibrium position corresponding to the length when the spring is horizontal (let's call this the "unstretched" equilibrium position) OR one can measure it from the new equilibrium position when the spring is vertical (the "stretched eq. pos.). This affects the way one applies conservation of energy.

    Usually, most people start working with respect to the new, "stretched" equilibrium position (so that x goes from +A to -A, where A si the amplitude of the motion). In that case, it turns out that when applying conservation of energy, one does not need to include mgh! One simply uses the potential energy stored in the spring and the kinetic energy.
    It's easy to rpove this explicitly and is ultimately due to the fact that the force of the spring is a linear force. Shifting the origin in [itex] {1 \over 2 } k x^2 [/itex] adds a linear term in the shift and this term is exactly cancelled by mgh. Quite neat to see at work, actually.

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