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Dimensional analysis problem

  1. Dec 26, 2009 #1
    1. The problem statement, all variables and given/known data

    hi im trying out a question on dimensional analysis , im having trouble finding the solution when the force is non linear , the situation is a mass resting on a smooth surface attatched to a spring , the spring is attached to a wall on its other end ,

    2. Relevant equations
    the expression for the non linear force is
    [tex] f= -kx + bx^2[/tex]

    3. The attempt at a solution
    , what will be the expression for the frequency? [tex]\omega=m^d *k^e* (bx)^f*x^g[/tex]? or maybe not since its not giving me the expected answer. i found the expression for the case when the force is linear [itex]f=-kx[/itex]: [itex]\omega= c \sqrt{k/m}[/itex] , the epected answer for the non linear case is of the form [tex] \omega = f(x_o \frac{b}{k}) \sqrt{k/m} [/tex]
  2. jcsd
  3. Dec 26, 2009 #2
    If I understand this correctly, then you're looking for the frequency of the motion up to multiplication by a constant, right?

    Well, your approach here should be purely one of dimensional analysis, without referring to the analog of SHM.

    The one thing you can say about the frequency is that it is independent of the displacement, as such, you expect it to be a function of the parameters of your system.

    [tex]\omega = f(m, k, b)[/tex] with dimensions of [tex][time]^{-1}[/tex] (This approach doesn't give a result since you're stuck with the length dimension of [tex]b[/tex])

    Another way to approach it would be to assume that it is some multiple of the frequency of the linear system, [tex]\omega _0 = \sqrt{\frac{k}{m}}[/tex] which could prove fruitful here. (The same problem as above persists)

    I think the insight required here is to notice that the force is a restoring force only for [tex]-x_c < x <x_c[/tex] where [tex]x_c=\frac{k}{b}[/tex]

    And then using that as a clue to incorporate [tex]x_c[/tex] in your solution.

    Is the expected answer you posted the textbook's answer? I'm asking since it makes absolutely no sense.. If we are to understand that [tex]x_0=x_c[/tex], then [tex]f(x_0 \frac{b}{k})=f(1)[/tex] and that doesn't make sense dimensionally. But let's assume that the meaning was just 1, then all you get is just the linear frequency.

    Where did you get this question..?
    Last edited: Dec 26, 2009
  4. Dec 26, 2009 #3
    yes your right,
    i hope i did not refer to SHM,

    im trying to figure out what you mean here, ill keep on it.

    Introductory Classical Mechanics,
    with Problems and Solutions
    David Morin

    in the appendix pg XIV-6
  5. Dec 26, 2009 #4
    I was hoping it was a book I had access to. ^^; Could you please scan in the question, or write it in full? Since as it stands, it doesn't make sense. Especially the answer from the answer sheet which would be dimensionally inconsistent.

    The force is a function of distance, you can't plug in something which has the dimensions of a pure number ([tex]x_0\frac{b}{k}[/tex] would be a pure number) and get something that makes sense.
  6. Dec 26, 2009 #5
    here is a screenshot of the book , hope its clearer than my explanation

    Attached Files:

  7. Dec 26, 2009 #6
    Oh, I think I understand now. The lowercase [tex]f[/tex] just stands for some function whose argument is a pure number.

    This stems from two assumptions. The first is that there a dependence of the frequency on the parameter [tex]b[/tex].

    This is problematic since standard dimensional analysis (Assuming [tex]\omega = c\cdot m^{\alpha}\cdot k^{\beta} \cdot b^{\gamma}[/tex]) shows that there cannot be a dependence on [tex]b[/tex] because of the length dimension it introduces.

    We solve this problem by assuming a dependence on the amplitude of the motion as well. This way we can negate the dimension of length by saying that we have a dependence on [tex]x_0 b[/tex]

    We do this by saying that the new frequency is the same as the old, times some pure number, [tex]f(x_0\frac{b}{k})[/tex] which is a function of a pure number just as well.
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