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How do I use a taylor series expansion to find effective spring constant

  1. Nov 1, 2014 #1

    grandpa2390

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    1. The problem statement, all variables and given/known data
    Let's pretend I am given a potential energy function and nothing else. I need to find the effective spring constant for oscillation about the equilibrium point using a taylor series expansion. I can't find an example or explanation anywhere on how to do this. the potential energy function I have has a stable equilibrium point.

    forgive me for not posting the exact problem. I am hoping someone can walk me through an example or point me to one, that way I can learn how to do it, then apply that to my homework rather than just doing the homework problem first.

    thankyou in advance for the trouble.

    2. Relevant equations
    taylor series expansion is
    http://www.wolframalpha.com/input/?i=f(r)+++f'(r)*(x-r)+++f"(r)/2!+*(x-r)^2

    and it goes on continually to the desired precision

    as for the potential energy function for this system, I am hoping that one with a stable equilibrium can be invented just for the purpose of explaining to me how to do it. Then I can attempt to apply that to my homework problem. if not I guess I can change the function I am given by setting all the constants equal to 1, but I'd really prefer to have to think about the problem rather than just plugging and chugging it into an algorithm created by an explanation. If you understand what i mean. but if I must, I will.

    3. The attempt at a solution

    1. I took the first derivative of my potential energy function.
    2. I set it equal to 0 and solved for my independent variable. This gave me the equilibrium point (and the answer to part a)

    3. I plugged it into a taylor series expansion, but I am uncomfortable with this because it makes no sense as to why that would give me the effective spring constant. Spring constant is Force / mass. I feel like there is probably an equation that I am supposed to plug the equilibrium value into that will give me my spring constant. and then I plug that into the taylor series.

    Please advise me, and I appreciate the advice.
     
  2. jcsd
  3. Nov 1, 2014 #2

    rude man

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    Hmm. For a spring-mass oscillating system there is no need for a Taylor series expansion, assuming inter alia Hooke's law holds for the spring over its entire swing.

    In other words the potential function is V = -1/2 kx2 so F = -grad V = + kx.

    It would be different for a swinging pendulum where the approximation sin(θ) ~ θ is typically made.
     
  4. Nov 1, 2014 #3

    grandpa2390

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    so what do I do. the question requires a taylor series. Does it help if I tell you that the constants and coefficients are all variables? I don't know if it is necessarily a spring. I think it is just referring to k= N/M
     
  5. Nov 1, 2014 #4

    ehild

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    The constant term in the Taylor series expansion is the potential energy at the equilibrium position. It can be the gravitational potential energy, for example, when a mass hangs on a spring.
    The second term is zero as you expand about the point where f'=0.
    The third term corresponds to the elastic potential energy 1/2 f"(r)(x-r)2=1/2 k (Δx)2. So the equivalent spring constant is k=f"(r).
     
  6. Nov 1, 2014 #5

    grandpa2390

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    That's interesting. I never seen that before. so I was heading in the right direction, even if i had no clue what I was doing.
    the y value at the equilibrium point at the bottom of this... parabola-ish shaped graph is the potential energy?

    the third term is the elastic potential energy.
    is this always true?

    or only for stable equilibriums. or what.
     
  7. Nov 1, 2014 #6

    rude man

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    That WOULD make life rougher, to be sure. You'd need to define the coefficients and constants.
     
  8. Nov 1, 2014 #7

    ehild

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    It is the equilibrium potential energy. If the position of the particle moving in the given potential "well" deviates from the equilibrium position, it has also elastic potential energy.
    Yes, the equilibrium must be stable otherwise oscillations about it can not occur. So the potential need to have a minimum at the equilibrium position. At a minimum, the second derivative of the function is positive, The elastic potential energy has to be also positive.
     
  9. Nov 1, 2014 #8

    grandpa2390

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    Ok I think that makes sense. at the bottom of the well, it has the equilibrium potential energy of f(r).
    does that mean that if it is not oscillating, then all the other terms in the series equal zero. but if it is oscillating, then the third term has some value greater than zero and that is the elastic potential energy (which contains the spring constant) is being added to the equilibrium potential energy.

    right because if it is unstable, like a ball at the top of a hill, and I give it a kick, it won't oscillate it will just fall.

    so what I need to do in my solution is take my third term and set it equal to 1/2 f"(r)(x-r)2 = 1/2 k (dx)^2 and show that k is equal to f"(r)?
    I can't help but admire the beauty. the simplicity of this and how it works out.
     
  10. Nov 1, 2014 #9

    grandpa2390

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    so in the taylor function. what is x? is it just distance the spring is stretched relative to the origin? so that x-r is the distance it is stretched from equilibrium. the further you stretch the mass, the amount of energy it contains increases exponentially.
    or with a pendulum, x would be the distance from ground. and r would be the lowest point of the pendulum swing (equilibrium point).

    and is there a difference between effective spring constant and spring constant?
     
  11. Nov 1, 2014 #10

    grandpa2390

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    I don't know if I am allowed to tack on another question. but I hate to keep starting threads.

    part c is to find the frequency of oscillation for small displacements form equilibrium of a mass m.
    my spring constant is a formula (I don't have a number)
    and they don't give me a value for mass m.

    Do I just say sqrt(k/m) and just write my formula for k over m and put it under a square root sign? Forgive me, it just seems to simple to be correct. So I just want to ask to be sure.
     
  12. Nov 2, 2014 #11

    ehild

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    It is not exponential, but it increases at least quadratically :)

    The effective spring constant refers to something that is not a spring. The pendulum is an example. The potential energy depends on the height of the bob above the deepest position, and it is PE= mgL(1-cosθ) if you choose the zero of the potential energy there.
    You have to note that to get the spring constant, you need a variable with dimension of length. θ=s/L where s is the length of the arc along the circle the bob moves on. So you need the Taylor expansion with respect to s of the function f(s)=mgL(1-cos(s/L)). f"(s) = mgLsin(cos(s/L)/L2 which is mg/L at s=0. So k = mg/L .

    An other example: consider a bead with mass m=1g and positive charge q =0.1 mC able to move along the x axis between two other positive charges, Q1=0.2 mC and Q2 = 0.8 mC, 1 m apart. What is the potential energy of the bead as function of x? Where is the equilibrium position ? Displaced slightly, how would it move? What is the period of small oscillations?
     
  13. Nov 2, 2014 #12

    ehild

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    sqrt(k/m) is equal to the angular frequency, ω.
    ##ω=\sqrt{k/m}##. The frequency is ##f=ω/(2π) ##.
     
  14. Nov 2, 2014 #13

    grandpa2390

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    So I am confused. k is the second derivative of this: f(s)=mgL(1-cos(s/L)) not my potential energy function? I am supposed to take my equilibrium value and plug it into the expansion of f(s)=mgL(1-cos(s/L))?
    I am confused because the 2nd term of expansion of the potential energy function did turn out to be zero like you said. so I thought that that was what you were saying to expand.

    also I am not told if this is a spring or a pendulum. so I don't know if it has an arc.

    or are you just showing me the proof for mg/l
     
    Last edited: Nov 2, 2014
  15. Nov 2, 2014 #14

    ehild

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    f(s)=mgL(1-cos(s/L)) is the potential energy function if the potential energy is chosen zero at the bottom of the trajectory. We use s= Lθ instead of the angle, as we want the Taylor expansion with respect a variable of length dimension. Only this way is the second derivative at the equilibrium position (s=0) equal to the spring constant.
    The fist term, of the expansion is the potential energy at the equilibrium: it is zero in this case.
    The second term is proportional with f', and it is zero at equilibrium, as the potential energy has minimum there.
    The third term gives the elastic potential energy.
     
  16. Nov 2, 2014 #15

    grandpa2390

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    ok so that is just an example if a potential energy function was in terms of theta, such as a pendulum with an equilibrium point at zero potential energy. my potential energy function is not written in terms of cosine or theta. just in terms of x. so I am good. I to the second derivative of my potential energy function and used f"(r) as k. I didn't have to put it in terms of cosine or anything.

    so if I am ever given a potential energy function that is in terms of theta, I'll have to convert that theta to a variable of length (such as s/l) in order to find k.

    then found the angular frequency w
    and converted it to frequency v
     
    Last edited: Nov 2, 2014
  17. Nov 2, 2014 #16

    ehild

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    Correct!
     
  18. Nov 2, 2014 #17

    grandpa2390

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    yay I learned something! Thanks. for all your help. now I just need to learn how to set up extremization problems
     
  19. Nov 2, 2014 #18

    ehild

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    It was showing how the "spring constant" of the pendulum is mg/l.

    And I am pleased that you understood something. :)
     
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