1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Vertically oscillating spring-mass system

  1. Jun 6, 2009 #1
    Consider a mass hanging freely from a spring, oriented vertically downwards. We know that because the restoring force acting on the mass is directly proportional to the displacement of the mass from the equilibrium position, i.e. the extension of the spring (F=kx), the system exhibits simple harmonic motion.

    Suppose that we pull it down by a distance of a from its equilibrium position, and then release it. Using energy considerations, and letting b be the maximum displacement of the mass from the equilibrium position during the upward portion of its motion,

    (1/2)ka2 - mga = (1/2)kb2 + mgb, and solving,
    a - b = 2mg/k

    So, a must be greater than b.

    Is this analysis correct, and if so, does this mean that systems executing SHM need not have equal amplitudes in the different oscillation directions? (e.g. during the upward motion, its amplitude is different from that of its downward motion)
  2. jcsd
  3. Jun 6, 2009 #2
    No. Try solving the equation again, and don't cancel out factors.

    The amplitude is the same on either side of the equilibrium because this is how it is defined. It is the position with minimum potential energy so greatest kinetic energy.
  4. Jun 13, 2009 #3
    (k/2) (a2 - b 2) = mg(a+b)

    (k/2)(a+b)(a-b) = mg(a+b)

    Cancelling a+b and rearranging, a-b = 2mg/k

    I don't see what's wrong with cancelling the a+b factor. It's not zero, so unless my initial equation was wrong, what's wrong with the analysis? In any case, at the amplitude position the system has greatest potential energy, not kinetic energy as you said.
  5. Jun 13, 2009 #4


    User Avatar
    Homework Helper

    Here's the catch: the equilibrium position of the mass is not at [itex]x = 0[/itex]. Of course, you could define your coordinate system such that [itex]x = 0[/itex] corresponds to the equilibrium position of the mass, but then the potential energy is not
    [tex]U = \frac{1}{2}kx^2 + mgx[/tex]
    but rather
    [tex]U = \frac{1}{2}kx^2 + mgx + Cx + D[/tex]
    where [itex]C[/itex] and [itex]D[/itex] are constants. I leave it to you to figure out what their values would be.

    EDIT: actually, I thought about it some more, and maybe some more detail is in order. There are really two equilibrium positions involved in this problem: (1) the equilibrium position of the spring itself, with no mass on it, and (2) the equilibrium position of the spring with the mass on it. When you define the potential energy of the system as
    [tex]U = \frac{1}{2}kx^2 + mgx[/tex]
    you're implicitly deciding to measure distances from equilibrium position #1. But when the mass is on the spring, that position is no longer an equilibrium. That's why you get a difference between the maximum height and the minimum height, because you're measuring from a point which is not an equilibrium point of the weighted spring.

    If you are careful about defining your distances and reference points, you should find that equilibrium position #2 is in fact equidistant from the highest and lowest points of the motion.
    Last edited: Jun 13, 2009
  6. Jun 13, 2009 #5
    Ah. That explains everything perfectly. So that's what I was missing.

    If the spring extension at equilibrium is A, then if we define x to be the displacement from the equilibrium position, we have
    U = (k/2)(A+x)2 - mgx, assuming x to be positive in the downward direction.

    Expanding this expression gives the constants C and D.

    I followed through my earlier analysis using a and b as the amplitudes in the upward and downward directions, and did manage to obtain a result showing a and b are equal. Thanks for your help, diazona!
  7. Jun 13, 2009 #6


    User Avatar
    Homework Helper

    You're welcome :wink: That's a surprisingly tricky situation when you think about it in detail - it took me a while to come up with my last post.
  8. Nov 21, 2009 #7
    I was trapped in this same tricky oscillation and I'd like to show you my http://www.scribd.com/doc/22857822/Oscilacion-vertical" to the problem, which is grounded solely on Newton's Second Law and Hook's Law, with no calculus at all.
    I found this discussion really helpful.
    Last edited by a moderator: Apr 24, 2017
  9. Nov 2, 2011 #8
    my apologizes for resurecting this old thread, but this concept has confused me for much of my high school and undergraduate years!

    i've browsed dozens of pf threads and this one has the best answer!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Vertically oscillating spring-mass system