Springs: spring stretches 0.5 ft when a 2-lb wt is attached

shivers20

Junior Member
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Mar 3, 2006
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I would like to know if I am on the right track with this problem.

A spring stretches 1/2 foot when a 2 pound weight is attached. If the weight is released from the equilibrium position with an initial upward velocity of 4 ft./sec and if the damping force is numerically equal to the velocity, find the equation of motion describing the position of the weight at time t.

a.) Set up the initial value problem which describes this system.

b.) Solve this IVP to find the desired equation of the motion.

c.) Discuss in terms of the displacement function you have found in part (b) the behavior of this system on the time interval from t = 0 to t = 5 sec. For what times, if any, in this interval, is the weight crossing the equilibrium position?

I started with a.)

m = W/g = 2/32 = 1/16

Find k: k = mg/L = 1/16*32 = 33/.5 = 66 <-this number seems kind of large?

Set up the IVP:
1/16u" + 66u = 0
u(0)= -1/16
u'(0)= 4

....cont'd How am I doing so far?
 
Don't forget that is says the damping force is numeerically equal to the velocity. So you would also have b=4, (4y').
 
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