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Title: Critical damping of springPost by: Fu-Kwun Hwang on April 12, 2009, 04:41:25 pm
For a spring with spring constant k, attached mass m, displacement x.
The equation of motion is F=m d ^{2}x/dt^{2}= -k*x;The nature frequence w0=sqrt(k/m); If damping is introduced with a form of -b*v; The equation become m d ^{2}x/dt^{2}+ c dx/dt + k x =0;The behavior of the system depends on the relative values of the two fundamental parameters, the natural frequency ω0 and the damping ratio ζ=c/ (2*sqrt(m*k)); When ζ = 1, the system is said to be critically damped. When ζ > 1, the system is said to be over-damped.when 0 ≤ ζ < 1,the system is under-damped. The following simulation let you play with different parameters to view the differece between those 3 modes: Initially, the system is set up at under-damped condition. Drag the blue ball to the spring, you will find how under-damped look like. Click b=b_critical to set it to critically damped condition, then click play to view the behavior. When it is paused again, drag b to larger value to find out how over-damped look likes. |