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Mechanical Motion of Springs Differential Equations

  1. Apr 23, 2012 #1
    1. The problem statement, all variables and given/known data

    Skjermbilde_2012_04_23_kl_6_50_05_PM.png

    3. The attempt at a solution
    So I've been interpreting the information in the problem as follows: [itex]F_{damping} = 4u' = μ(u')[/itex], [itex]k = \frac{4N}{m}[/itex]. If the system is critically damped then [itex]μ = 2\sqrt{km} = 2\sqrt{\frac{4N}{m}m} = 2\sqrt{4N}[/itex]. Now it seems as though the spring constant is cancelling out my mass, so if [itex]μ =4[/itex] then simply force [itex]N =1[/itex] and the system is critically damped for any mass. But then, this doesn't seem quite right. I must be misunderstanding the problem.
     
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  3. Apr 23, 2012 #2

    HallsofIvy

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    The "m" in N/m is NOT the mass. Nor is N a parameter. That is just saying that the spring constant is 4 Newtons per meter.
     
  4. Apr 23, 2012 #3
    Let me see if I can do better with this now. I write,

    Critically damped implies [itex]μ = 2\sqrt{km}[/itex]. Given that [itex]F_{damping} = μ(u') = 4u'[/itex], then [itex]μ = 4[/itex] and we then say [itex]4 = 2\sqrt{4m}[/itex] which implies that m = 1.
     
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