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 Author Topic: Critical damping of spring  (Read 18246 times) 0 Members and 1 Guest are viewing this topic. Click to toggle author information(expand message area).
Fu-Kwun Hwang
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 « Embed this message on: April 12, 2009, 04:41:25 pm » posted from:Taipei,T\'ai-pei,Taiwan

For a spring with spring constant k, attached mass m, displacement x.
The equation of motion is F=m d2x/dt2= -k*x;
The nature frequence w0=sqrt(k/m);
If damping is introduced with a form of -b*v;
The equation become m d2x/dt2+ 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.

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