For a spring with spring constant k, attached mass m, displacement x.
The equation of motion is F=m d[sup]2[/sup]x/dt[sup]2[/sup]= -k*x;
The nature frequence w0=sqrt(k/m);
If damping is introduced with a form of -b*v;
The equation become m d[sup]2[/sup]x/dt[sup]2[/sup]+ c dx/dt + k x =0;
The behavior of the system depends on the relative values of the two fundamental parameters, the natural frequency ?0 and the damping ratio ?=c/ (2*sqrt(m*k));

When ? = 1, the system is said to be [b]critically damped[/b].
When ? > 1, the system is said to be [b]over-damped[/b].
when 0 ? ? < 1,the system is [b]under-damped[/b].

The following simulation let you play with different parameters to view the differece between those 3 modes:
Initially, the system is set up at [u]under-damped[/u] condition.
Drag the blue ball to the spring, you will find how under-damped look like.
Click b=b_critical to set it to [u]critically damped[/u] condition, then click play to view the behavior.
When it is paused again, drag b to larger value to find out how [u]over-damped[/u] look likes.

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