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How to find damping ratio, revision questions.

  1. Jan 20, 2013 #1
    Hi guys,
    Revision for an exam tomorrow, I hope someone can help?

    I've been given the answers but I can't work out how to find the damping ratio without being given a velocity or distance for the mass to travel?


    A mass of 6kg is suspended on a spring and set oscillating. it is observed that amplitude reduces to 5% of its initial value after 2 oscillations, which takes 0.57 seconds.

    Determine the following.

    The damping ratio (0.43)
    The natural frequency (3.51)
    The actual frequency (3.17)
    The spring stiffness (2915N/m)
    The critical damping coefficient (264.5Ns/m)
    The actual damping coefficient (113.7Ns/m)

    I can use the damping ratio to help solve some of the other values once I know how to work it out.

    Any help appreciated.
     
    Last edited: Jan 20, 2013
  2. jcsd
  3. Jan 21, 2013 #2
    OK this is what I have,
    Natural frequency = 2/0.57 = 3.51Hz
    Spring stiffness 3.51Hz = 1/2pi x square root of K/m (k works out to be about 2915)
    Critical damping square root of 4mk = 264.5

    I need the damping ratio to find actual damping and actual frequency, I don't think I can do it without a velocity?
    The 5% of amplitude must come into it somehow?
     
  4. Jan 21, 2013 #3

    rude man

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    We start with the diff. eq. F = ma:
    mx'' = -kx - cx'
    where k = spring constant
    Assume an initial x = x0 and solve the equation for x(t).
    Then, specify that at t = 0.57s two full oscillations have occurred, and that x(0.57) = 0.05x(0). This will enable you to solve for c, k, and the actual ω. You should already know the formula for the natural frequency as a function of m and k.

    The damping coefficient ζ = c/cm where cm is the damping coefficient for critical damping. You can look up that formula on Wikipedia (it's a function of k and m also), or derive it as being the value of c for which the roots of your auxiliary equation are both just barely real (any lower c would give you complex-conjugate roots). OK, I'll give it to you: cm = 2√(km).

    (I haven't actually solved the problem but I can if necessary).
     
  5. Jan 22, 2013 #4
    Thanks!

    Could you solve please so I can see how it's done?
     
  6. Jan 22, 2013 #5

    rude man

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    We are not allowed to do that. But if you post your work we can tell you where, if at all, you went wrong. The numerical answers are given to you already, apparently.

    Start with the expression for x(t) as I suggested.
     
  7. Jan 22, 2013 #6
    I mentioned it in the second post.
    I can work out Natural frequency,Spring stiffness and Critical damping.
    And when given the one of the other 3, (damping ratio, actual frequency, actual damping coefficient), I can solve for the other 2.

    I now need to differentiate to find x, once I have x I can find velocity and solve the rest from there.

    I've just been on Khan Academy to study differentiation but there is a lot to learn and I have to be up early.

    Cheers for the push in the right direction.
     
  8. Jan 22, 2013 #7

    rude man

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    OK, you probaly solved this equation in class, and it's a chore to do it again so I will give you the sol'n:
    x = x0exp(-ct/2m)cos(ωt)

    where
    ω = √(ω02 - c2/4m2)
    and the natural frequency ω0 = √(k/m).

    Now you should be able to use the givens in your problem to arrive at the missing parameters ω, ω0, c, cm, k, and ζ.

    P.S. you don't need to solve for velocity.
     
  9. Jan 24, 2013 #8
    Thanks, but is there a simpler way to explain what is going on?

    It's all getting very complicated.

    I'm going to have to dedicate my weekend to it I think.
     
  10. Jan 24, 2013 #9

    rude man

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    It's really just algebra at this point.

    For example: what's the simple relation connecting ω and 0.57 sec.?
    What about the 5% and exp(-ct/2m)?
    Then ω0 pops right out, as does k.
    And so do c and ζ.
     
  11. Jan 26, 2013 #10
    what's the simple relation connecting ω and 0.57 sec.?
    Natural Frequency is square root of k/m = 22.04 (radians) divide by 2pi to get 3.51Hz
    0.57 for 2 revs = 0.285 secs per revolution. 1/0.285 = 3.51Hz

    What about the 5% and exp(-ct/2m)?
    I cant for the life of me find a value for c (unless you mean critical?)

    x = x0exp(-ct/2m)cos(ωt)
    I dont understand where cos(ωt) comes into it.

    This is the best I can do so far
    2915 x e1 = 7923.79
    2915 x e0.95 = 7537.34
    7923.79 - 7537.34 = 386.45
    386.45/3.51 = 110.1 (nearest I can get to 113)

    Any chance you can solve it or solve a similar problem so I can see what you did?
    Then I have a chance of working out what's going on.

    I'm running out of time, the exam is on Monday and I've already dodged it twice.
     
  12. Jan 26, 2013 #11

    rude man

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    Where did you get k?
    Anyway, ω is not the natural frequency. ω is the actual frequency = 2πf. So period T = 1/f and 0.57s = 2T according to what you're given.

    The natural radian frequency is √(k/m).

    Solve exp(-ct/2m) = 5% = 0.05 for c. No, c is the actual damping coefficient, not the critical one.
    As I said, it's the solution to the differential equation based on F = ma. You can understand that only if you've followed its derivation, which is lengthy & you should have covered it in class or in your text.

    You need to understand this stuff rather than trying to memorize solutions to "similar" problems. Is the test going to have this very problem? Anyway, I'm not allowed to go any further than I have. Try to understand what I've written so far. I've given you every single formula you need to get all six parameters asked for.
     
  13. Jan 26, 2013 #12
    i got k from the equation
    Natural freq = 1/2pi x √k/m
    (3.51 x2pi)2 x 6= k
    =2915

    "You need to understand this stuff rather than trying to memorize solutions"
    I agree totally, but unfortunately we are given little more than equations to memorise.
    I find myself having to spend hours online trying to teach myself something that would have taken just a few minutes to explain.

    "You can understand that only if you've followed its derivation, which is lengthy & you should have covered it in class or in your text."
    No the hand outs we were given only cover questions that have force and velocity to find the damping ratio.

    Could you help me out with the calculus part?
     
  14. Jan 26, 2013 #13
    I think I've stumbled onto something.

    Critical damping = 264.5Ns/m

    But thats per second, so 264.5 x 0.57 = 150.765
    leaving 113.735 actual damping?

    I'm not sure if it makes sense but its the right number that I need?
     
  15. Jan 26, 2013 #14

    rude man

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    And where did you get the natural frequency from?
    I have. You don't need any calculus now. Just use the formulas I gave you and try to understand the equation for x(t): it's a damped sinusoid. Sorry I can't draw it for you. Lots of people on this forum could.

    If you want a derivation of x(t) look up damped harmonic oscillations in Wikipedia.
     
  16. Jan 26, 2013 #15
    I have natural frequency given on the revision sheet as 3.51Hz
    Are you hinting that this is wrong and 3.51 is the actual frequency?

    Makes more sense to me, and it wouldnt surprise me if they were giving us wrong answers.
     
  17. Jan 26, 2013 #16

    rude man

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    I haven't computed any actual numeical answers.
     
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