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Vertical Spring PE and KE energy

  1. Jan 24, 2013 #1
    Let's say we have a vertical spring. As it oscillates, the energy bounces between gravitational potential energy, spring potential energy, and kinetic energy, right? Why does my lab say that spring potential energy = kinetic energy at any moment in time? Is that a mistake on their part?
  2. jcsd
  3. Jan 24, 2013 #2
    Gravitational potential energy is probably negligible. More accurately, the difference in gravitational potential energy between different positions is negligible. You know the mass on your spring, calculate its gravitational potential energy at different positions and see how it compares to the spring's energy (presuming you know the spring constant).
  4. Jan 24, 2013 #3
    I question the assertion based on the idea that when the mass bounces across the spring's neutral point, it will have zero potential energy while the kinetic energy can be non-zero.
  5. Jan 24, 2013 #4


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    What they should be saying is that the total energy, PE + KE, is conserved and constant in time.
  6. Jan 27, 2013 #5
    Can someone just confirm that this is correct? I'm about to submit a lab tomorrow morning and I'm too tired at the moment to do what he suggested.

    Edit: Never mind. I understand why they ignored gravitational potential energy now... If you measure the distance from the hanging equilibrium position (without the mass) opposed to the equilibrium position with mass then gravitational potential energy is incorporated into spring potential energy.

    Edit2: Never mind. I think my lab is wrong. I'm so confused.
    Last edited: Jan 27, 2013
  7. Jan 28, 2013 #6
    If your lab script say PE = KE at all times then it is wrong. Otherwise there would be no transfer of energy and hence no bouncing of the spring!

    (With a contrived definition of where PE = 0 you could set up a situation where |PE| = KE, but I'm sure that wasn't their intention).

    You're also right about the GPE. If your spring is nice and linear, then you can simplify things by just considering PE (a combination of GPE and spring PE).
    Last edited: Jan 28, 2013
  8. Jan 28, 2013 #7
    What would the point of that be? :| They would still be two separate terms, right?
  9. Jan 29, 2013 #8
    Yes, but only one will be time dependent.

    If you consider first a horizontal spring with a mass on the end (i.e. ignore gravity). The total energy, at any time, is:

    [itex]E_{Total}= E_{K} + E_{SP}[/itex]

    where [itex]E_{K}[/itex] is kinetic energy and [itex]E_{SP}[/itex] is spring potential energy. More explicitly we have:

    [itex]E_{Total} = 1/2 mv^{2} + 1/2 kx^{2}[/itex]

    where [itex]v[/itex] is the velocity of the mass, and [itex]x[/itex] is the extension of the spring. Assuming we're not losing energy through friction etc., then the total energy is constant from the Conservation of Energy Principle. We can confirm that by substituting in expressions for [itex]x[/itex] and [itex]v[/itex] if we want.

    So we have quite a simple situation. In terms of problem solving, we know that:

    [itex]1/2 mv^{2} + 1/2 kx^{2} = [/itex] constant

    at all times, which is a pretty useful identity.

    What I want to show is that we can recover this simple situation even when gravity is involved. I'll do this by working relative to the initial, equilibrium extension of the spring due to gravity.

    When the spring is in equilibrium with gravity, forces are balanced - the downward force of gravity matches the upwards force of the spring. So:

    [itex]kl = mg[/itex]

    where [itex]l[/itex] is the equilibrium extension of the spring and [itex]g[/itex] is the acceleration due to gravity. So the initial extension is:

    [itex]l = mg/k[/itex]

    which we'll need shortly.

    Now, we set the spring bouncing. At any time we have the total energy:

    [itex]E_{Total} = E_{K} + E_{GP} + E_{SP}[/itex]

    where [itex]E_{GP}[/itex] is gravitational potential energy.

    To simplify things, I'm going to choose [itex]E_{GP} = 0[/itex] at the equilibrium position - I'm free to choose this anywhere I like.

    Now we can say:

    [itex]E_{Total} = 1/2 mv^{2} - mgx + 1/2 k(x + l)^{2}[/itex]

    where [itex]x[/itex] is the extension beyond the equilibrium extension at any give time. So the total extension is [itex]x + l[/itex], which is why that expression appears in the term for the spring potential energy.

    Now, expand out the final term:

    [itex]E_{Total} = 1/2 mv^{2} - mgx + 1/2 kx^{2} + kxl + 1/2 kl^{2}[/itex]

    substitute in the expression we calculated for the equilibrium extension, [itex]l[/itex] earlier:

    [itex]E_{Total} = 1/2 mv^{2} - mgx + 1/2 kx^{2} + mgx + 1/2 kl^{2}[/itex]

    cancel the [itex]mgx[/itex] terms, and we have:

    [itex]E_{Total} = 1/2 mv^{2} + 1/2 kx^{2} + 1/2 kl^{2}[/itex]

    So now, as you say, we still have three terms instead of two. The first describes the kinetic energy as a function of velocity, and hence of time. The second describes the potential energy as a function of extension from equilibrium, and so also as a function of time. And the final term describes the energy as a function of the equilibrium extension which is a constant in time.

    So, in terms of the time evolution of the system, we have got back to a nice, simple, identity:

    [itex]1/2 mv^{2} + 1/2 kx^{2}=[/itex] constant

    where [itex]x[/itex] is now the extension from equilibrium.

    This makes some problems easier to solve.
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