# How do you know if an equation of spring motion is overdamped?

Looking at an equation, how can you know if if it overdamped, critically damped, or under-damped?

For example:

How can you tell that the equation $c_1e^{2x} + c_2e^{-2x}$ is overdamped?

How can you tell that the equation $e^{-x}(c_1+c_2x)$ is critically damped?

How can you tell that the equation $e^{-t}(c_1cos(3t) +c_2sin(3t))$ is underdamped?

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These are not "equations", let alone ODEs, but function terms. You can find out about their behavior by looking at their graphs. –  Christian Blatter Nov 19 '12 at 9:41

The shape of these is the key. An overdamped system will be pure exponentials (though they are usually all decreasing). Critically damped has a term in $xe^x$. And underdamped have oscillatory solutions, like yours with cosine and sine waves.

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Can you explain why? i.e. What does having a term in $xe^x$ have to do with returning to equilibrium quickly, etc? –  Imray Nov 19 '12 at 2:20
@Imray: it means you have a double root in the characteristic equation, which puts it on the boundary between two real roots (overdamped) and two complex roots (underdamped) –  Ross Millikan Nov 19 '12 at 2:46
Can you give me a real world example of overdamped and critically damped? I understand that under-damped is the motion of an ordinary spring system or pendulum that dies down over time, but I can't picture the other two. –  Imray Nov 19 '12 at 2:54
@Imray: they are still springs or pendulums (pendula?) but with so much friction they don't overshoot. that is why critically damped approaches equilibrium fastest. Overdamped is like moving through molasses-you just can't get there very fast, so reducing the damping is a good thing. Underdamped gets you to equilibrium faster, but then you overshoot. –  Ross Millikan Nov 19 '12 at 3:02
Just did a lot of studying on the topic yesterday, I understand it now. Thank you Ross –  Imray Dec 12 '12 at 20:55

Hint:

Just look at how your equations are set up.

• $e^{-x}(c_1+c_2x)$ means you have repeated roots.
• $c_1e^{2x} + c_2e^{-2x}$ means you have distinct roots.
• $e^{-t}(c_1cos(3t) +c_2sin(3t))$ means you have complex conjugates roots.

The roots will tell you whether it is critically damped, overdamped, or underdamped.

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But my question was deeper. Why do repeated roots mean it's critically damped? Why do distinct roots mean its overdamped? –  Imray Nov 19 '12 at 15:36
hyperphysics.phy-astr.gsu.edu/hbase/oscda2.html take a look at that site. –  diimension Nov 20 '12 at 1:38