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Length of a spring constantly accelerating through space?

  1. Dec 24, 2014 #1


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    Imagine if I were to apply a constant force, F, to a spring of length L, spring constant K, mass m, and uniform density. The force causes the spring to accelerate with acceleration a.
    For the sake of imagination, I will say I'm applying the force with my hand (which is pushing one end of the spring).
    Surely the spring will compress, (how else would the other end of the spring, away from my hand, accelerate?) but, we shouldn't expect the spring to compress according to Hooke's law, because Hooke's law applies to when the force is acting on both sides of the spring, (right?) and in this case the force acts only on a single side.

    So my question is, by how much would the spring be compressed? (What is ΔL?)

    My approach:
    Pretend the spring consists of infinitely many infinitesimal springs, each of identical length dx. I'm thinking that we can then treat each infinitesimal spring as having a spring constant of [itex]k\frac{L}{dx}[/itex] (right?).
    Let's now put an x-axis along the direction of motion, with x=0 being the end of the spring where my hand is.
    One of the infinitesimal springs at a distance x will have to apply a force of [itex]ma\frac{L-x}{L}[/itex] in order to accelerate the springs ahead of it. To then find the compression of that single spring we would now use Hooke's law and divide the force by it's spring constant, to get a change in length of [itex]-ma\frac{L-x}{kL^2}dx[/itex]
    Then, to get the change in length of the entire spring, we would sum up all the length changes of all the infinitesimal springs to get [itex]ΔL=∫_0^L\frac{-ma(L-x)}{kL^2}dx=\frac{ma}{2k}=\frac{F}{2k}[/itex]

    I might be making some simplifying assumptions. For example I am assuming the spring length is in equilibrium (the spring is not oscillating). I am not sure if this would be true or not (maybe someone can comment on if the spring would oscillate if you accelerated it from rest?) but I'm just trying to simply analyze a massive spring (because every spring problem I've come across has treated the spring as massless...).

    The answer seems reasonable to me: The spring would compress half as much as if the force was acting on both sides (i.e. half as much as if we used Hooke's law).
    The reason this seems reasonable is because the density is uniform, and so the force applied by the infinitesimal pieces of the springs varies linearly, so we should be able to simply average it out to the middle, effectively treating the spring as if all of the mass were at the midpoint. (I know that this logic is not very precise, but it's just a post-rationalization).

    I made up this problem, and so I have no way of checking my answer, so I am asking you helpful folks.
  2. jcsd
  3. Dec 24, 2014 #2


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    I just thought of an alternate (yet very similar) method of solving it, and it yields [itex]ΔL=0[/itex] :confused:

    Nevermind, I miscalculated. :redface: The new method yields [itex]ΔL=-L\frac{ma}{2kL+ma}[/itex]
    Last edited: Dec 24, 2014
  4. Dec 24, 2014 #3
  5. Dec 24, 2014 #4


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    Okay, thanks. According to that thread my first solution is correct. But now I'm a bit curious why my second method is incorrect.

    For the second method I considered the already compressed spring. So instead of having linear density [itex]\frac{m}{L}[/itex] it would have a linear density of [itex]\frac{m}{L-ΔL}[/itex], and instead of integrating from 0 to L, I would integrate from 0 to (L-ΔL). The infinitesimal spring elements would still have a spring constant of [itex]k\frac{L}{dx}[/itex] so you then get the integral [itex]\int_0^{L-ΔL}\frac{-ma(L-ΔL-x)}{kL(L-ΔL)}dx[/itex] which gave me the answer in my second post, [itex]ΔL=-L\frac{ma}{2kL+ma}[/itex]

    Does anyone perhaps have an explanation of why this method is wrong? I was actually thinking it makes more sense than my first method... :confused:
  6. Dec 24, 2014 #5
    Just thinking. About whether starting from the compressed equilibrium point, that you would have to use the spring constant of k(L-ΔL)/dx for this analysis.
  7. Dec 24, 2014 #6


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    The spring is not uniformly compressed - the compression is greatest at the end where the force is applied. Thus the density is not a uniform ##m/(L-\Delta{L})## across the entire compressed length, as you were assuming in the second method. Your first method automatically compensates for this because the integration variable ##l## isn't actually the length - it's the number of infinitesimal elements to the left of ##l##.
  8. Dec 24, 2014 #7


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    Oh, of course! Thank you Nugatory! Thanks especially for explaining why the first method is logical :)
  9. Dec 25, 2014 #8
    Thanks for this very interesting thread!

    The only way a spring can not oscillate is when it is already in its equilibrium position. Since the equilibrium position of the spring is different when it is resting and when it is accelerated it should oscillate. This is because of the sudden acceleration - there is no way the spring can get into equilibrium with respect to the new force in zero time.

    Lets check the mathematics to see if there is evidence for this wall of words. I will consider a mass attached to a spring obeying Hooke's Law to which suddenly a force is applied.

    Take the harmonic oscillator equation

    ## m \ddot x+ kx = 0 ##

    before applying the force and

    ## m \ddot x + kx = F ##

    after the sudden onset of the force. We know that the only equilibrium position of the first equation is ## \dot{x}(t) = 0, x(t) = 0 ## for all t. Lets say that the sudden onset of force occurs at time ##t_0##. Then ##\dot{x}(t_0) = 0, x(t_0)=0 ## are the initial values for the second equation. The second equation can be rewritten as

    ## m \ddot x' + kx' = 0 ## with ##x'=(x-\frac{F}{k})##.

    This yields the only equilibrium solution ## x'(t) = 0, \dot x'(t)=0 ## or ##x(t) = \frac{F}{k}, \dot{x}(t) = 0## for all ##t>t_0##. Therefore the given initial values derived from the first equation are not compatible with the equilibrium position of the second equation, and the spring should oscillate. This, of course, is for the harmonic oscillator equation without damping. You may 'jump' into an overdamped or critically damped solution in the damped case when applying the force in the damped case. Even a damped oscillation will eventually oscillate with an unnoticable amplitude after a long time. Maybe I will dive into that later after the holidays.
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