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Damped oscilations

  1. May 20, 2003 #1
    Hello people!!!

    I have quite a complex problem... I need to graph (on my PC, using excel or mathcad) the following motion:
    a cube on a spring with given k and given initial Energy oscilates on a horizontal surface with friction coefficient u.
    Now, i know this are damped oscillations which die out with time, but in order to graph them, i need to present the stopping force as b dx/dt , or as a fraction of the velocity...
    what do i do? any ideas how to present friction through velocity, or am i completly on the wrong track??

    thanx
     
  2. jcsd
  3. May 20, 2003 #2
    I don't think you should model the friction as a function of velocity. That would be the approach to use for air resistance or fluid resistance, which logically are related to velocity.

    If the friction depends only on μ, m and g, why not just treat it as a constant term in your equation?

    (Unless there is some aerodynamic lift going on that would cause the friction to vary in some way with velocity. That would be an interesting problem, but I don't think you are given enough information to deal with it.)
     
  4. May 21, 2003 #3
    Yeah, I think that the kinetic friction coefficient is the proportionality constant between the normal force and the frictional force

    So you'd take the weight of the object and multiply it by the coefficient to get the friction, I believe (you oughtta check that in your book/with someone who's more on the ball, though, if this is really important)
     
  5. May 21, 2003 #4
    what you have is a damped, undriven harmonic oscillator. The differential equation which describes that is:
    from F = ma = -kx
    md2x/dt2 + bdx/dt + kx = 0
    this is a differential equation with characteristic equation
    mr2 + br + k =0
    r = -b/2m +- ((b/2m)2-(k/m))1/2
    there can be overdamped, underdamped, and critically damped cases dependant on the difference of (b/2m)2-(k/m).
    you graph:
    x = Ae-γtcos(ωt+θ)
    γ=b/2m
    ω=((k/m)-(γ)2)1/2
    A is initial amplitude
    you should get a decaying cos wave.
     
  6. May 28, 2003 #5
    Thanks a lot, but i have this solution and am trying to apply it to a specific problem.... my real problem is presenting friction in the form b dx/dt, because i do not have the quoeficient b, but u from Friciton=u.normal reaction
     
  7. May 28, 2003 #6
    Hello again, tudur. It's an interesting problem. It seemed suspicious that when it comes to harmonic motion, my physics textbook and two differential equations textbooks and everything I could find on the internet talk about air resistance and fluid resistance, and never any details about the ordinary kinetic friction between a sliding object and the surface it slides on.

    Not so strange, though, when you try to set up the equations, right? Everybody likes to treat this subject with damping forces that vary with speed, I guess because that makes the math relatively easy to deal with. Multiplying the damping coefficient by dx/dt accomplishes something that is obviously essential: it ensures that the direction of the damping force is always opposite to the direction of motion, and you get a result like the one that schwarzchildradius posted.

    But it doesn't really seem to be a good approximation for the situation you are trying to model. We would really like to have a force with constant magnitude μN, but whose sign changes twice each period so that at any given time it is opposite to the sign of the velocity. I don't know what function you can use to accomplish that.

    Maybe you should start with the equation schwarzchildradius posted
    x = Ae-(b/2m)tcos(ωt+φ)

    and experiment with different values for b, and maybe that will give you something that approximates what you need. It won't be easy but you can use energy considerations to check the results. If the frictional force is constant, the loss of energy will be a constant multiple of the cumulative displacement. So, (with spring constant=k and coeff. of friction=μ) assuming for example that you start at t0 with v0=0 velocity and x0=A (maximum displacement), you can compute the starting potential energy (1/2)kx02. Then you can compute what the displacement is at time t1/2=π/ω and at time t1=2π/ω so you know how much distance was traveled in the first period (D = x0 - x1/2 + x1). From that you can compute the work done by friction μND, subtract that from the initial potential energy, and the result is what the potential energy should be at time t1 (end of the first period). Hopefully, this will be very close to (1/2)kx12. If not, try changing the value you assigned to b in the equation and do it all again.

    You could also try different powers of t in the exponential term (not in the cosine term). In other words, maybe
    x = Ae-(b/2m)t2cos(ωt+φ) or x = Ae-(b/2m)t(3/2)cos(ωt+φ) or some other power of t.
    I can't give you any mathematical support for that last idea. It's just a guess that maybe it will give you a curve that fits the facts better.)

    I'm just groping in the dark here, so if anybody really knows how to deal with this, please clue us in.

    I guess we can see why your professor thought this was a worthy problem to give you. Let us know what you come up with. Good luck.
     
  8. May 29, 2003 #7
    Forget about my previous post on this thread.
    Not only is it too much work -- it's just wrong. It'll never come out with the right answer since your curve is not decreasing exponentially.
    If Ui is the given initial (potential) energy, with CONSTANT friction opposing the motion, your velocity (x') has to go to 0 (and STAY at 0) within some finite number of oscillations. It's just a matter of time until fk*(dist. traveled) = Ui. Right?

    Also, you CAN'T represent the friction force as b*dx/dt, because ordinary friction between a block and a surface is not a function of velocity. (Unless you're talking about air resistance, or fluid resistance, which usually are treated as a function of velocity.) In that case, this solution is no good and you should just solve the basic differential equation and graph the result. Or just use the formula that schwarzchildradius posted. But it was interesting to think about this anyway.

    If we're actually talking about ordinary friction, I think you have to define the function piecewise.
    Assuming you start at t=0, x=xmax, x' = 0
    and coeff. of kinetic friction is μ
    and coeff. of static friction is also μ (usually greater than kinetic friction, but you only mentioned one μ, so maybe you want to simplify it & treat them as the same. Otherwise, adjust the solution accordingly.
    You are given the initial energy, so using U = (1/2)kx2 solve for x and that gives you the initial displacement xmax

    Then, solve the diff. equation:
    mx" + kx = μm
    letting ω = √(k/m)
    and using the initial conditions, at t=0, x(0)=xmax, x'(0) = 0

    (You do the math based on your actual numbers.)

    Then solve your equation for x and t when x'=0 (which is the next time that the block comes to a stop, after 1/2 cycle); call them x1 and t1.
    This lets you calculate the spring force at that time using f = -kx1. The block will resume moving only if that force is greater in magnitude than μm. (If you want to be more precise, use μs here).

    If it's going to continue moving, now you need a new function to describe the next half-cycle. Now the direction of the friction is reversed so the differential equation to solve is:
    m" + kx = -μm
    with new initial condition: x(t1) = x1 using the values you found when the block stopped.

    And so on...

    For each half-cycle there will be a new equation describing that segment of the graph, and at the end of each segment, you have to compare the spring force to the frictional force to see whether the block starts moving again, or remains at rest. Eventually, the block just comes to a stop someplace (not necessarily at x=0).

    Ugh. Have fun.
     
  9. Jun 2, 2003 #8
    Hey, gnome, thanks for the long answers, i'm sorry i wasted so much of your time... actually i'm not yet a university student (that would be next year)and this problem is not given by my professor, it's just an analogy of a more complex situation i was trying to describe.... obviously i'll have to make another analogy, because this one won't work very well, even if i manage to calculate the result cycle by cycle....
    thanx a lot anyway, if i come up with something on that problem, i'll let you know
     
  10. Jun 2, 2003 #9
    No need to apologize. I don't feel that it was a was of time at all. Thank you for raising what turned out to be a surprisingly interesting question. Throughout my Physics I course, friction between a sliding object and a solid surface was fairly simple to deal with, while the resistance of air or liquids seemed much more complicated because the resistive force varied with speed. Surprisingly, in harmonic motion it seems that the opposite is true. Your question pointed out that while it is not too difficult to find an equation to describe harmonic motion of an object that is being slowed down by air resistance, it is actually extremely difficult to model harmonic motion involving "simple" friction mathematically.

    By the way, what is the "more complex situation" you were trying to describe?
     
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