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 [tex]\beta[/tex] times the instantaneous velocity. Determine the values of [tex]\beta[/tex]>0 for which the spring/mass system will exhibit oscillatory motion.



2*[tex]\lambda[/tex]=[tex]\frac{\beta}{m}[/tex]

[tex]\omega[/tex][tex]^{2}[/tex]=[tex]\frac{k}{m}[/tex]

[tex]\lambda[/tex][tex]^{2}[/tex] - [tex]\omega[/tex][tex]^{2}[/tex]>0 is overdamped

[tex]\lambda[/tex][tex]^{2}[/tex] - [tex]\omega[/tex][tex]^{2}[/tex]=0 is critically damped

[tex]\lambda[/tex][tex]^{2}[/tex] - [tex]\omega[/tex][tex]^{2}[/tex]<0 is underdamped

1 slug = 32 pounds

I've solved that [tex]\beta[/tex] is equal/less than/greater than 2*[tex]\sqrt{k*m}[/tex]=32, but I don't understand when it will or will not have oscillatory motion.
 

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  • #2
jack action
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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.
 
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Thank you!
 

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