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Forced Oscillator with unfamiliar forcing function and constants

  1. Oct 27, 2013 #1
    1. The problem statement, all variables and given/known data

    A force Fext(t)=F0[1-e(-a*t)] acts, for time t>0, on an oscillator which is at rest at x=0 at time 0. The mass is m; the spring constant is k; and the damping force id -b dx/dt. The parameters satisfy these relations:

    b=mq and k=4mq2 where q is a constant with units of inverse time.

    Find the motion. Determine x(t); and hand in a qualitatively correct graph of x(t).

    (B) Determine the final position.


    2. Relevant equations

    Typical form of a damped oscillator:

    [itex]\ddot{x}[/itex]+2β[itex]\dot{x}[/itex]+w2x=0

    3. The attempt at a solution
    This question is from an online, self-study course and we have no lecture for forced oscillators. I am left to figure this out with my book and the internet. I would greatly appreciate a few pokes in the right direction. All the examples I am finding on forced oscillators are of the more common form of: [itex]\ddot{x}[/itex]+2β[itex]\dot{x}[/itex]+w2x=Acos(wt)

    Here is what I am working with:

    [itex]\ddot{x}[/itex]+2β[itex]\dot{x}[/itex]+w2x=F[1-e(-at)]

    I understand that I need to find a solution to the homogeneous equation and also a particular solution and their sum will be my x(t). My solution to the homogeneous equation is simple if I can determine the nature of the damping (which I have problems with below). The tricky part then becomes the particular solution.

    I also need to determine the nature of the damping. Is this overdamped, underdamped, or critically damped? How do I relate b=mq to b2=4mk which is the form I am more accustomed to if it were critically damped, for example? I believe that if someone helps me figure out the nature of the damping I will be able to piece together a particular solution.

    (B) I see that the driving force quickly approaches F0 and I could take the limit as t->∞ to find the final position. I need x(t) first, though.

    Thank you for your time and help!
     
  2. jcsd
  3. Oct 27, 2013 #2

    vanhees71

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    First you should write down the equation of motion with the quantities given in the question and then compare it to your standard form in order to sort out the constants of your equation properly.

    For the particular solution, do you know the Green's function of the harmonic oscillator? Then you simply have to fold the external force with it.
     
  4. Oct 27, 2013 #3

    vela

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    You've been given the forces, so start by applying F=ma to get a differential equation for x(t) in terms of b, k, and m. When you solve the homogeneous equation, you should be able to deduce which case you have for the given values of b and k.
     
  5. Oct 27, 2013 #4

    rude man

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    That equation is dimensionally inconsistent. The equation is ƩF = mx''. Fix that first.

    Then rewrite 2β = 2ζw. What is the relation between ζ and the conditions of underdamping, critical amping and overdamping?

    BTW you should use w0, not w in your starting equation. w is the actual frequency whereas w0 is the natural (undamped) frequency.

    You should realize that it is unnecessary to know a priori if the system is underdamped, overdamped etc. But you need to be comfortable with imaginary numbers if the system is underdamped.
    Remember that exp(ix) = cos(x) + i sin(x).
     
  6. Oct 27, 2013 #5
    Thank you both for your help.

    m[itex]\ddot{x}[/itex]+b[itex]\dot{x}[/itex]+kx=0

    Dividing by m in the homogeneous form and replacing b/m=2B and k/m=w02

    Then plugging in the known relations b=mq and k=4mq2 I get:

    2B=q and w02=4q2

    B=q/2

    w0=2q

    If that is the case then W0 is clearly larger than B telling me this is underdamped.

    Is that what you mean vanheese71?

    -----

    Alternatively, I can look for a general solution to the homogeneous equation:

    m[itex]\ddot{x}[/itex]+b[itex]\dot{x}[/itex]+kx=0

    x(t)=C1e[itex]\frac{t(sqrt(b^2-4km)+b)}{2m}[/itex]+C2e[itex]\frac{t(sqrt(b^2-4km)-b)}{2m}[/itex]

    I can plug in b=mq and k=4mq2 but it is messy and I don't quite see where to go from there.
     
    Last edited: Oct 27, 2013
  7. Oct 27, 2013 #6

    vela

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    The square root will turn out to be imaginary, which gives rise to oscillating solutions, i.e., it's underdamped.

    For the particular solution, try using the method of undetermined coefficients.
     
  8. Oct 27, 2013 #7

    rude man

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    It's messy but it can be done.

    However, the best way is for you to find the solution to the homogeneous equation in the underdamped case. Many websites can give you this.

    As for the forcing function, you probably haven't had Green's function, and I assume also not the Laplace transform, so you need to guess the form of x(t) for the forcing function case. To help you with this, note that F contains a constant plus an exponential, and that you're likely to have sin and cos in there also since it's underdamped.

    HOWEVER: don't even try to come up with the exact x(t)! It's VERY messy! (It would take me 1/2 hr to write it down for you!) All you're asked is to produce a qualitatively correct curve.

    So, realizing that F = 0 for t = 0 you know that initially it's just the homogeneous response. And for the steady-state response, which means many time constants of the homogeneous solution, F = constant = F0. So you can assume the steady-state equation to be just mx'' + bx' + kx = F0.

    For part B: if you use 'common sense' and think of the physics of the situation as in a mass-spring situation so that damping is zero, x(infinity) should be obvious.(hint: x'' = x' = 0 at t = infinity.)

    Put all the pieces together and you should be able to come up with a plausibly accurate graph of x(t) vs. t.
     
    Last edited: Oct 27, 2013
  9. Oct 27, 2013 #8
    mx''=-bx'-kx+F[1-e(-at)]

    mx''+bx'+kx=F[1-e(-at)]

    x''+2Bx'+w02=[1-e(-at)]/m, 2B=b/m, w02=k/m

    Is that what you meant by fixing the equation?

    I'm not sure where zeta comes in. I have yet to come across that symbol in my study of oscillators that I can recall.

    You're absolutely right. Sorry I missed it!

    You're right. I was subconsciously trying to skip right to the typical form for a particular type of damping. I did work out a general solution to the homogeneous equation in my last post (I missed your post when replying) which I think allowed me to relate b, m, k, and q with and find out that it is underdamped. If my conclusion is correct I hope that I arrived at the correct conclusion in a sensible way because I definitely want/need to understand it instead of happening across the correct solution.

    EDIT: Just noticed two new responses. Checking them out now!
     
    Last edited: Oct 27, 2013
  10. Oct 27, 2013 #9
    Ahh, fantastic. Both methods have brought me to the same conclusion. I feel that the first route was more natural for me. Does what I did seem to hold water because that's the work I will show if so?

    Yes, that is sort of why I was trying to jump ahead earlier and guess the nature of the damping. I'm pretty comfortable with finding the solutions to the homogeneous portion.

    I did happen across a solution that used Green's theorem which is, you are correct, unfamiliar to me. I should learn that method because I will definitely use it eventually but as noted below it may not be immediately necessary. Thank you for pointing that out.

    I'm not entirely sure what you mean by many time constants of the homogeneous solution but I can see that the 1-e(-at) portion quickly approaches 1 as t increases making the forcing function F0 after short t. So, I agree, the steady state equation is what you have shown.

    How, though, do I produce a graph that considers that short period of dynamic force? Sure, the initial response is simply the homogeneous solution and after short t it is what you have shown but between t=0 and t=short?

    In addition to producing the graph I have to show x(t). Can I represent x(t) as a homogeneous solution plus the steady state function solution? I thought I had to use a particular solution that included the 1-e(-at)
     
  11. Oct 27, 2013 #10

    vela

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    For the particular solution, assume it has the form A+Be-at. Plug it into the differential equation and then solve for A and B.
     
  12. Oct 27, 2013 #11
    Thank you.

    I was told my original differential equation was dimensionally inconsistent. Is the equation below correct?

    x''+2*Bx'+w02x=[itex]\frac{F_0(1-e^{-at})}{m}[/itex]

    Of course, at t=0 there is no driving force so it would be homogeneous.
     
    Last edited: Oct 27, 2013
  13. Oct 27, 2013 #12

    rude man

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    Yes.
    It's called the damping coefficient. There is a very simple relationship between it and the condition of damping.
    OK, will keep tabs.
     
  14. Oct 27, 2013 #13
    I'm trying to figure out which way to go. You have suggested, if I am interpreting your response correctly, that finding a particular solution to the transient function is unnecessary. Whereas Vela has suggested an educated guess on what the solution would look like, plugging it in, and solving for the particular.

    I have responded since your most recent post and I believe that more recent post better expresses my current understanding and weaknesses.
     
  15. Oct 27, 2013 #14

    rude man

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    The homogeneous solution is x(t) = 0 for all t since the initial conditions are both zero x(0) = x'(0) = 0.

    So divide the forcing function into two sections: F1 = F0 and F2 = -F0 exp(-at), t > 0. Then the complete x(t) is the sum of the effects of F1 and F2.

    There is no separate 'transient' solution. There is only one solution, caused by F1 + F2. The steady-state solution is x(infinity), when x = x' = 0. So x(infinity) = a constant ≠ 0.

    Choosing trial x(t) = A + B exp(-at) will not work. (I know 'cause I worked the problem!). There will be a simple ~ exp(-at) term but then also a second term ~ exp(-dt) sin(g)t where d and g are very involved expressions. It should not be surprising to find sin terms for an underdamped system.

    Anyway - pick whatever floats your boat!
     
  16. Oct 27, 2013 #15
    That should have been obvious from the get go. I'm sort of embarrassed to not have realized that earlier. Thank you. Looking back at the thread, though, I realize we may have all be missing that at first. :tongue:

    If that is not a typo (did you mean F0[1-e(-at)]? otherwise, where did that come from?) then at t=∞, F2=0 and the sum of F1 and F2 is simply F0. This is a bit confusing, really, because the way I'm interpreting your comment regarding the homogeneous solution is that, effectively, there is no oscillation. We're dealing with a driving force and a resisting force. I don't know how else to interpret this besides simple motion equations:

    F=ma => a=F/m => dv/dt=F/m => dv=F/m dt

    Plug in F and solve.

    Or do you mean proceed with x''+2Bx'+w02=Forces and just don't expect a contribution from the homogeneous solution. If that is the case then I still am not sure how to solve this differential equation. Sorry for the misunderstanding... I've never come across a problem like this before. Perhaps I am over thinking it. An entire day of crunching physics problems tends to make me a little loopy.

    (B) Given the above I imagine this problem as a spring attached to wall and someone grabbing the spring, giving it a pull, and holding it at its final stretched position. The response to the 'driving' force is the spring constant which is dependent on x. Once the force of kx=F0 the system will stop moving. I just need to solve for x at that point which I guess will have to be in terms of k and F0 seeing as how I'm not given any values. Does this sound right?
     
  17. Oct 27, 2013 #16
    Eh, just realized when I lied down for bed where those two pieces F1 and F2 came from. Multiply through... I'll have to re-think the problem tomorrow but my confusion still stands.
     
  18. Oct 27, 2013 #17

    vela

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    No one missed that point because it's not correct. The initial conditions need to be satisfied by the complete solution. You can't apply it to only the homogeneous solution to solve for the constants.

    As I said earlier, just set the particular solution to ##x_p(t) = A + Be^{-at}##, plug it into the differential equation, and solve for A and B. Once you have the complete solution, then apply the initial conditions x(0)=x'(0)=0 to solve for the constants c1 and c2.
     
  19. Oct 28, 2013 #18

    rude man

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    I have to say that this problem is hard to address without invoking the complete expression for x(t) which is very elaborate, involving an exponential and an exponential multiplied by a sine term. This is due to the fact that unfortunately the system is underdamped.

    Vela's approach will not work unless his "a" can be complex. This is obvious because no products of exponentials and sines can otherwise ensue. But maybe he did have a complex "a" in mind. I'm not saying that will work either, but it might.

    I solved this by the Laplace transform which makes it a piece of cake IF you have a good table of transforms or are willing to shell out $$ to Wolfram Alpha.
     
  20. Oct 28, 2013 #19

    vela

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    The particular solution doesn't contain any oscillating terms. I'm not sure why you keep insisting it does. Those terms are part of the homogeneous solution.
     
  21. Oct 28, 2013 #20

    vanhees71

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    It's not so difficult. Either you use the Green's function (which leads to a somewhat elaborate integration) or you use the already suggested "ansatz of the form of the right-hand side", i.e.,
    [tex]x(t)=A+B \exp(-a t)[/tex]
    and determine values for [itex]A[/itex] and [itex]B[/itex]. That's perhaps even simpler than the method with the Green's function, although you have to add the general solution of the homogeneous equation and determine the coefficients to fulfill the initial condition.
     
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