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Spring damper system equation of motion

  1. Nov 16, 2013 #1
    1. The problem statement, all variables and given/known data
    A 10 kg block is displaced 20 mm and released. If damping coefficient is 100 N.s/m,
    how many cycles will be executed before amplitude is reduced to 1 mm or below? The stiffness
    of the spring is k=20000 N/m.


    2. Relevant equations

    3. The attempt at a solution
    I first moved the mass to the inner radius and equated the Kinetic Energy of the system.
    Ke1 = Ke2
    Where i found m2 is 4m1

    Next i equated the kinetic energy of the system and equated that to :
    [itex]\frac{1}{2}[/itex]*m[itex]_{2}[/itex]v[itex]^{2}[/itex] + [itex]\frac{1}{2}[/itex]I[itex]\frac{V^{2}}{r^{2}}[/itex]= [itex]\frac{1}{2}[/itex]* m[itex]_{eq}[/itex]*v[itex]^{2}[/itex]
    meq = m2 + [itex]\frac{I}{r^{2}}[/itex] where m2is 4*m1

    I then substituted the numbers in and found Meq= 190

    Next to find the amplitude i found the damping ratio of the system
    [itex]\zeta[/itex] =[itex]\frac{c}{Cc}[/itex]

    Cc = 2*m*Wn
    Wn = [itex]\sqrt{\frac{K}{M}}[/itex]
    Wn= [itex]\sqrt{\frac{20000}{190}}[/itex] Wn = 10.26
    Cc = 2(190) * (10.26) = 3899 ∴ [itex]\zeta[/itex] = [itex]\frac{100}{3899}[/itex]
    [itex]\zeta[/itex] = 0.0256
    Under damped system E.O.M =
    X(t) = e[itex]^{-\zeta*W_{n}*t}[/itex] { x[itex]_{o}[/itex]Cos(w[itex]_{d}[/itex]t) + [itex]\frac{x^{.}+W_{n}*X_{0}}{w_{d}}[/itex]*Sin(w[itex]_{d}[/itex]t) }
    I'm trying to find the t value that would make X(t) be less than 1mm, i'm not sure how i would do that without just picking random values of t, as the equation doesn't seem solvable just for t.
    Last edited: Nov 16, 2013
  2. jcsd
  3. Nov 16, 2013 #2

    Simon Bridge

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    Why did you move the mass to the inner radius and what is m2?
    Could the Q factor help you there?
  4. Nov 16, 2013 #3
    I moved the mass to the inner radius i guess to simplify the system, honestly not sure, its the way I've been taught this semester to do it.

    So if i had the original system the mass would be M1
    But now I've moved it to the inner radius The mass is now M2
    For it to still be the same [itex]\frac{1}{2}[/itex]M1*V1[itex]^{2}[/itex] = [itex]\frac{1}{2}[/itex]M2*V2[itex]^{2}[/itex]

    Where V1 = [itex]\dot{\theta}[/itex]2r
    Where V2 = [itex]\dot{\theta}[/itex]r
    [itex]\frac{1}{2}[/itex]M1*([itex]\dot{\theta}[/itex]2r[itex])^{2}[/itex] = [itex]\frac{1}{2}[/itex]M2*([itex]\dot{\theta}[/itex]r)[itex]^{2}[/itex]
    Through Cancelling 4M1 = M2

    I'm not sure what the Q factor is
  5. Nov 16, 2013 #4

    Simon Bridge

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  6. Nov 17, 2013 #5
    Still not 100% sure on this Q factor, hard to see how it applies without an example.

    Q = [itex]\frac{1}{2\zeta}[/itex]
    Q = [itex]\frac{1}{2*0.0256}[/itex] Q = 19.53

    Q = 2[itex]\pi[/itex]*[itex]\frac{Energy Stored}{Energy Lost Per Cycle}[/itex]

    [itex]\frac{Q}{2\pi}[/itex] = [itex]\frac{Energy Stored}{Energy Lost Per Cycle}[/itex]

    3.11 = [itex]\frac{Energy Stored}{Energy Lost Per Cycle}[/itex]

    Can you give a hint at what the next step would be, would i find the energy at the start? By using
    T(t) + V(t) = E, Where T is the kinetic energy and V is the potential, i guess at t = 0 there is no kinetic energy.

    Therefore [itex]\frac{1}{2}[/itex]kx[itex]_{0}[/itex][itex]^{2}[/itex] = E(0)
  7. Nov 17, 2013 #6

    Simon Bridge

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    How does the energy relate to the amplitude of the oscillations?
  8. Nov 18, 2013 #7
    Well as the energy is lost the amplitude decreases starting off at a maximum I guess, so there must be a time where the amount of energy lost means the amplitude falls below 2mm, i mean if I multiply both sides by 3.18 I get that 3.18 x energy lost per cycle = energy stored, at first I thought that meant it oscillates for 3.18 cycles but thst seems too little when the damping is that small. I guess directly the potential energy at the start is converted into kinetic energy so energy is somewhat proportional to velocity squared so if I could find velocity I could find the time through equation of motion.
  9. Nov 18, 2013 #8

    Simon Bridge

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    Which means that you need to know how the amplitude related to the energy ... i.e. is it an inverse-square law?

    You can do it more directly using potential energy ... all the kinetic energy ends up stored in the spring, and the energy stored in a spring is related to how far it is compressed ... which is what you want to know about.

    When the amplitude is 2mm, then the system has lost a certain percentage of amplitude ... which relates to a certain percentage of energy, and you have an equation for the rate of energy loss with time.

    You should check the extent of the damping though - is this underdamped, critically damped, what? And what does that mean for the general motion?
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