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Critically Damped Oscillator Spring Constant and Damping Parameter

  1. Jan 28, 2012 #1
    1. The problem statement, all variables and given/known data

    A mass of 1000 kg drops from a height of 10.0 m onto a platform of negligible mass.
    It is desired to design a spring and damper on which to mount the platform so that it
    will settle to a new equilibrium position 2.00 m below its original position as quickly
    as possible without overshooting.
    Find the spring constant k and the damping parameter
    if the system is critically
    damped.

    2. Relevant equations

    ω^2(frequency squared)=γ^2(damping parameter squared)
    E=U=mgh at equilibrium
    E=1/2kA^2
    x(t)=(A1+A2t)e^(-γt)

    3. The attempt at a solution

    First, I solved for energy:
    E=U=mgh=19400

    Then for the spring constant:
    k=2E/A^2

    But now I need amplitude, so this is where I taking a shot in the dark:

    x(t)=(A1+A2t)e^(-γt)
    Now I was thinking to say that if t goes to infinity, x is 2, but it gave me no information... I need help! please and thank you
     
  2. jcsd
  3. Jan 29, 2012 #2

    rude man

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    1. Realize that there is no actual "equlibrium" position. That position is realized only after infinite time has passed.

    2. The moment the mass hits the platform, you have a classical mass-spring-damper situation with an initial velocity x'(0+) easily determined from energy conservation. Just solve the 2nd order diff. eq. mx'' + kx + cx' = 0. When you do, you'll look for the solution for which c/m, where c is the damping coefficient, yields barely two real roots for the attendant algebraic equation. k, the spring constant, is easily derived form the 2m criterion.

    By 'barely' I mean just avoiding a complex-conjugate solution of your algebraic equation, which implies oscillatory behavior. So your time constants will be as short as possible without incurring oscillations.
     
  4. Jan 29, 2012 #3
    Ok then, well I am having problems with the DE since we have two unknowns, the spring constant k and the damping parameter...
     
  5. Jan 29, 2012 #4

    rude man

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    Think about it for a second. When the system is at rest, the spring supports 1000 kg at a depth of 2m. So what's k?
     
  6. Jan 29, 2012 #5
    I want to use F=-mg=-kx, yes?
     
    Last edited: Jan 29, 2012
  7. Jan 29, 2012 #6

    rude man

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    Right! So what is your number for k?
     
  8. Jan 29, 2012 #7
    ok so k=4900, thus damping parameter is 2.21, and the final equation is:
    x(t)=(10+22.1t)exp(-2.21t)
    Right?
     
  9. Jan 29, 2012 #8

    rude man

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    Right! But you didn't even need to solve the whole equation.

    If you write the equation in the standard form mx'' + nx' + p2x = 0 then
    critical damping → n = p

    But p = √(k/m) = 2.21 = n so that's all you were asked to do. (Assuming the instructor meant "n" as the "damping parameter", which he apparently did.)

    (I plead guilty to not solving the whole equation like you did. I never do what's not necessary!).
     
  10. Jan 29, 2012 #9
    Haha Thanks so much, I really appreciate it!
     
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