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Question about vertical oscillations and energy

  1. Aug 3, 2011 #1
    I have a few questions regarding vertical oscillations (not damped). Let's say you have a spring hanging from a ceiling with a mass attached to it. So, for example, if I had a mass attached to a spring and I pulled it down by 10 meters, if mg/k = 5 meters then the amplitude would be 5 meters.

    Let's say I know what A (amplitude) and k (spring constant) are. I know that energy is conserved and that for horizontal oscillations:

    E = (1/2)k(A^2)

    Is that true for vertical oscillations? Does gravity basically "disappear" during a vertical oscillation? In other words, for a vertical oscillation do we still have E = (1/2)k(A^2) or is there gravitational potential energy involved too?

    My book says that other than determining the new "equilibrium" position for the oscillations, "everything we have learned about horizontal oscillations is equally valid for vertical oscillations". I have never seen energy used in a vertical oscillation problem though and would like to double check.

    Also, Vmax is still omega times A, right? A = amplitude, omega = angular frequency
    Last edited: Aug 3, 2011
  2. jcsd
  3. Aug 3, 2011 #2


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    Maybe you can think of it in terms of how Hooke's law balances gravity with some additional stretch to the spring:

    The differential equation for the 1-d spring-mass system in the absence of an outside force is usually derived in some way like:
    [tex]F = ma = \frac{d^2 x}{dt^2} \text{ (Newton's 2nd law)},[/tex]
    [tex]F = -kx \text{ (Hooke's law)}, [/tex]
    [tex]\frac{d^2 x}{dt^2} = -kx.[/tex]

    Now turn the system so it's vertical (with the mass at position z above rest position):
    [tex]F = -kz - mg = \frac{d^2 z}{dt^2}.[/tex]
    You can make this look exactly as before by making a substitution:
    [tex]-kz - mg = -kx,[/tex]
    [tex]z = x - mg/k.[/tex]
    Nothing changes on the right hand side since
    [tex]\frac{d^2 x}{dt^2} = \frac{d^2 z}{dt^2}.[/tex]
    At x = 0, F = 0 as well, so all you're doing here is rewriting the vertical system in terms of x, the position around the new equilibrium point with gravity. This is exactly the place z such that the spring force due to the additional downward stretch of the spring cancels the gravitational force:
    [tex]x = 0: z = -mg/k[/tex]

    Of course, once you have the equation in terms of x, you see that everything else (e.g., the frequency of oscillation, potential energy, etc.) must be the same as before.
    Last edited: Aug 3, 2011
  4. Aug 3, 2011 #3

    Doc Al

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    That's a very interesting question. Yes, gravitational PE is involved but the reference point for measuring it is arbitrary. If you cleverly choose to measure gravitational PE from the midpoint between the unstretched position and the new equilibrium position, then you can show that the total PE is (1/2)k(x^2), where x is the displacement from equilibrium and thus the total energy is E = (1/2)k(A^2).

    As far as everything else goes, all is the same.
  5. Aug 3, 2011 #4
    Does that also mean that E = (1/2)m(Vmax)^2?
  6. Aug 3, 2011 #5

    Doc Al

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    Staff: Mentor

    Sure. (As long as you've chosen your zero point for gravitational PE correctly. It's not at x = 0.)
  7. Aug 3, 2011 #6
    OK. What bothered me is that my professor did a review question where you have a block hanging from a vertical spring and a bullet is fired into the block from the bottom up. The question asked "What is the maximum displacement of the spring?"

    Of course, you use conservation of momentum just before and just after the collision to find the speed of the two masses after the collision. I initially thought of this as a SHM problem and just thought to myself,

    Vmax = speed just after collision = Vtotal
    (1/2)(M+m)Vmax^2 = (1/2)k(A)^2
    and just solve for A.

    Here's what he wrote:

    (1/2)(M+m)Vtotal^2 + 0 + (1/2)k(y)^2 = 0 + (M+m)g(d) + (1/2)k(d-y)^2

    Where d is the max compression of the spring, d = 0 at the equilibrium point, and y is mg/k. Is that right, and am I wrong?

    Maybe this will be easier to answer: If after the collision I wanted to know how much the spring will be compressed/stretched by by relative to its unstretched length, can I do:

    (1/2)(M+m)Vmax^2 = (1/2)(k)(A^2)

    Solve for A, and say my answer is (mg/k)-A?
    Last edited: Aug 3, 2011
  8. Aug 3, 2011 #7

    Doc Al

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    Not a bad thought, but the problem is that the initial position of the 'block+bullet' after the collision is not the equilibrium position.

    That's correct. It's just conservation of energy.

    See my comment above.
  9. Aug 3, 2011 #8
    Ah, I see what you mean. If with the block and bullet the spring is allowed to rest, the new equilibrium position will be different. Correct?

    Boy, that's a tricky problem. I see why he did what he did now; for a second I was going crazy and I thought the problem contradicted everything I knew about vertical SHM. I guess these problems do help you understand the material even more though.
  10. Aug 3, 2011 #9

    Doc Al

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    Most definitely. Solve as many problems as you can.
  11. Aug 3, 2011 #10
    That's good advice. Thanks for all the help!
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