Mass/Spring System Damping Constant

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A mass weighing 32 pounds stretches a spring 6 inches. The spring constant is equal to 64 lb/ft.The mass moves through a medium offering a damping force that is numerically equal to \beta times the instantaneous velocity. Determine the values of \beta>0 for which the spring/mass system will exhibit oscillatory motion.



2*\lambda=\frac{\beta}{m}

\omega^{2}=\frac{k}{m}

\lambda^{2} - \omega^{2}>0 is overdamped

\lambda^{2} - \omega^{2}=0 is critically damped

\lambda^{2} - \omega^{2}<0 is underdamped

1 slug = 32 pounds

I've solved that \beta is equal/less than/greater than 2*\sqrt{k*m}=32, but I don't understand when it will or will not have oscillatory motion.
 
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Underdamped --> There is an oscillatory motion
Overdamped --> No oscillatory motion (gradually goes to final position)

So the limit that separates the two is the critically damped situation.
 
Thank you!
 
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