# spring system-Damped

• Oct 9th 2009, 06:04 AM
latavee
spring system-Damped
A 4-foot spring measures 8 feet long after a mass weighing 8 pounds is attached to it. The medium through which the mass moves offers damping force numerically equal to the square root of two times the instantaneous velocity. Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 5 ft/s. After finding the time at which the mass attains its extreme displacement from the equilibrium position, what is the position of the mass at that instant?

mass=8lbs
x(0)=5 ft/s

8lbs=mg=ks
8=4k
2=K

x"+2(lambda)x+(omega)^2x=0

beta=damping force

2(lambda)=(beta/m)
omega^2=k/m

How do i find beta?
• Oct 10th 2009, 12:05 AM
CaptainBlack
Quote:

Originally Posted by latavee
A 4-foot spring measures 8 feet long after a mass weighing 8 pounds is attached to it. The medium through which the mass moves offers damping force numerically equal to the square root of two times the instantaneous velocity. Find the equation of motion if the mass is initially released from the equilibrium position with a downward velocity of 5 ft/s. After finding the time at which the mass attains its extreme displacement from the equilibrium position, what is the position of the mass at that instant?

mass=8lbs
x(0)=5 ft/s

8lbs=mg=ks
8=4k
2=K

x"+2(lambda)x+(omega)^2x=0

beta=damping force

2(lambda)=(beta/m)
omega^2=k/m

How do i find beta?

You do realise that this problem is imposible to do because of the way units are confused? Do we use pounds-feet-seconds (in which case the poundal is the unit of force) or slugs-feet-seconds (in which case the pound is the unit of force)?

From the context of the first paragraph it is the latter (we have a mass which weighs 8 pounds, but a weight is a force so our unit of mass is the slug), so then we are using slugs as our unit of farce.

So for our spring we get:

$\displaystyle 4=k8$

That is the extension in feet is the spring constant $\displaystyle k$ times the load in pounds force and so $\displaystyle k=1/2$ ft per pound force.

Now from the statement of the problem if $\displaystyle y$ is the displacement from the equilibrium extension the force on the load is:

$\displaystyle f(y)=-\frac{y}{k}-\sqrt{2}y'$

This is equal to the mass (that is $\displaystyle 32/8=1/4$ slugs) times the acceleration so:

$\displaystyle (1/4) y''=-\frac{y}{2}-\sqrt{2}y'$

or rearranging:

$\displaystyle y''+2y'+\sqrt{32}y=0$

(pheww.. and I hope I have the units right! Please suggest to you teacher that they update their text/s and teaching to use SI (mks) units)

CB