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Base excitation of undamped Spring-mass system

  1. Oct 27, 2016 #1
    1. The problem statement, all variables and given/known data
    I am a little stuck/unsure in an homework question for my signals and systems course. See question sheet attached. I have been able to do question 1 for the first section using FBD, and I have solved the DE using Laplace Transforms (See below for my solution). I am both unsure if my solution is correct, and what to do for question 2 and 3 of Section 1A. It might be because I did not learn most of this is English, but I am just not sure exactly what those two questions are asking?Please advise, thank you! Just a place to start at the very least would suffice!
    2. Relevant equations


    3. The attempt at a solution
    y(t) = [(F0 * ω02)]/[(ω02 - ω2)] * sin(ωt) + C1*cos(ω0t) + C2*sin(ω0t)
     

    Attached Files:

  2. jcsd
  3. Oct 27, 2016 #2

    BvU

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    Hello again, Rian,

    Well, at least the problem statement is legible now :smile:
    Part 1 asks to show something, but you ask if your solution of the DE is correct (without showing how you got it, so you make it a little difficult). What is ##F_0, \ \omega, \ \omega_0 \ \ ## ? (I know, but you apparently assume some things you don't mention)

    For part 2: you know (or can google) what a characteristic equation for a differential equation is. Can you confirm ?

    For part 3: the frequency response is (amplitude of y) divided by (amplitude of h) as a function of ##\omega##. Clearly things become problematic at some value of ##\omega##. Something to fix that is attempted on page 2, so you have a long way to go after Section 1A.

    The subject at hand is very interesting and extremely useful in many fields of science !
     
  4. Oct 27, 2016 #3
    Thanks for the advice regarding posting. Well for the DE, I just proved it using a FBD, and I'm confident the actual DE is solid. I then asumed:

    h(t) = F0cos(ωt)

    ω ≡ Forcing frequency
    ω0 ≡ Natural Frequency of the system
    F0 ≡ Amplitude of the base excitation

    I took the laplace transform of of each term, rearranged for Y(s) and then did the inverse transform to get y(t).

    For part 2, correct me if I'm wrong, but I know normally the characteristic equation of a DE is when you arrange the DE in standard form and then use the coefficients ans polynomials to find the roots?
    I've just been unable to figure out how to do it with a 2nd order non-homogeneous function like this one?

    So with part 3, should I then be taking the fourier transform of each term for my solution for y(t) and h(t), treating h(t) as an input and dividing the two? If so, how do I find the amplitude of each? Yes, I get that much, at resonance the transient part goes to infinity (ω=ω0).
    I know is still have a bit to go, but I feel like if I can grasp this bit, it sort off repeats itself in the next parts? Thank you again.
     
  5. Oct 27, 2016 #4
    Also, for the characteristic equation, even though it's non-homogeneous, is that just referring to the characteristic equation of the homogeneous variant when your finding the solution to the complimentary solution (DC)?
     
  6. Oct 27, 2016 #5
    Also, in that case the homog soln is:

    assuming y(t) = eλt:
    y''(t) = λ2eλt

    thus

    λ2eλt + ω02eλt = 0

    λ = ±iω0?
     
  7. Oct 28, 2016 #6

    BvU

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    Part 1 is just fine. Part two too: yes, for the characteristic equation you look at the homogeneous equation (and indeed find ##\omega_0## as you did)

    For part 3 you make use of the linearity of the equation: if f(t) is a solution of the homogeneous equation and g(t) one particular solution to the inhomogeneous equation, then f(t) + g(t) is a general solution to the equation + initial conditions. The initial conditions (your C1 and C2) can be applied to f(t) .

    The g(t) that goes to infinity in the undamped case is not the transient part. That term will become clearer when you consider the damped case: it is the f(t) that is the transient response; the C1 and C2 are irrelevant for the long term. But in the undamped case the transient response isn't "transient" :smile:
     
  8. Oct 28, 2016 #7
    Gotchya! Yea sorry I misspoke, and I realized after posting that the undamped condition is unbounded and will go to infinity without the dampener in the improved case. I'll have a further crack at it and post back afterwards.
     
  9. Oct 28, 2016 #8
    So this is what I got. Did both part 1 and 2 (See attached file), I am just unsure regarding Part 3 still?
     

    Attached Files:

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