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Longitudinal Wave Dynamics

  1. Sep 1, 2011 #1
    I am doing some study into longitudinal wave dynamics. I am using theoretical models of wave motion in continuous bar and comparing this to numerical analysis using a lumped mass model.

    So far I have discovered that the continuous bar vibrations, caused by base support motion (i.e. vibration caused by the entire bar moving) excites and infinite number of natural modes (with the lower ones having a higher contribution).

    When modelling this with mass-spring-mass-spring...etc. of the same length and wave speed (i.e. a=sqrt(E/ro) ) only the first natural mode seems to be excited. Am I modelling my numerical system wrong, or can it only capture the first mode?
     
  2. jcsd
  3. Sep 2, 2011 #2
    To visualize this I have included the following plots from a video of the velocity of the bar. Why is the discretised bar not the same?

    [PLAIN]http://www.test1.ausalive.com/images/640_bar2.jpg [Broken]
    [PLAIN]http://www.test1.ausalive.com/images/640_bar3.jpg [Broken]
    [PLAIN]http://www.test1.ausalive.com/images/640_bar1.jpg [Broken]
     
    Last edited by a moderator: May 5, 2017
  4. Sep 2, 2011 #3

    Philip Wood

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    There's quite a lot I'm not clear about here.
    What's being plotted on the horizontal graph axes?
    Is the bar or its lumped-mass quasi-equivalent infinite or of finite length?
    Presumably finite because you talk of modes, suggesting standing waves. But the graphs suggest progressive waves.
    The method of exciting waves strikes me as unusual - more usual to force one end of the system to oscillate sinusoidally, but I'm happy to be told I'm wrong.
     
  5. Sep 3, 2011 #4
    The horizontal axis is the length of the bar (the left side has the base support motion).

    The three images are taken at different time intervals.

    The discretised model was created by applying Newton 2 to the mass-spring-mass...etc system (i.e. sum(F)=ma ). These equations were then converted to state-space format and solved numerically with the same system configuration as the continuous bar.
     
  6. Sep 3, 2011 #5

    Philip Wood

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    Right, so we have a progressive wave travelling down the bar (or lumped system). [I was confused by your talk of modes, which suggested standing (or stationary) waves.] Now we need to know exactly how the waves were excited. Is one end of the bar subjected to a force F(t), and what is the form of the function F(t)?

    Since the continuous mass distribution obeys a liner wave equation with no dispersion (all harmonic components travel at the same speed) I'd expect F(t) to be the same shape as the wave profiles you've plotted for the continuous bar. The form of this seems to be that of a sum 1 + 2 cos(wt) + 2 cos(2wt) + 2cos(3wt) .... + 2cos(nwt).
    Can you confirm?
     
    Last edited: Sep 3, 2011
  7. Sep 3, 2011 #6
    They are both excited in the same manner.

    The excitation is from a base support motion (i.e. a bar with one end fixed and one end free) with a displacement profile u_g = F_g(t) applied to the 'fixed' end (which moves according to u_g).
     
  8. Sep 3, 2011 #7

    Philip Wood

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    Sorry, I think I was modifying my last post while you were replying. Please give it another read.

    The key point about lumped mass systems is the phenomenon of cut-off frequency: a frequency above which harmonics won't propagate. So I'm not surprised to see higher harmonics not appearing on your right hand graphs, though I don't quite see why only a half cycle seems to be present. I may well be missing something.
     
  9. Sep 4, 2011 #8
    The bar was subject to a base support motion in the form v = -(t-sqrt(a))^2 + a. That motion is what has excited the bar. I can give you more details on this method of excitation if you require.
     
    Last edited: Sep 5, 2011
  10. Sep 5, 2011 #9

    Philip Wood

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    Sorry, but I'm still very confused.

    (1) Can you confirm that the bar is subjected to this 'base support motion' at just one end?
    (2) I can't make any sense of your expression for v. First, it can't be correct as it is dimensionally inhomogeneous. [I'll explain further if you want.] Second, v is, presumably a function of time, but your formula, as it stands, doesn't tell us how time is involved.
     
  11. Sep 5, 2011 #10
    I have updated the expression to include time.

    I have found a solution for the problem: the analogue model was a summation of terms; when processing this infinite summation was truncated to 5 terms. This caused the phenomenon in the bar which dissappeared when the amount of terms was increased to 20.

    Thank you for your interest in finding the answer.
     
  12. Sep 5, 2011 #11

    Philip Wood

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    I'm really pleased that you've found your solution.
     
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