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Springs differential eqn

  1. Oct 20, 2015 #1
    1. The problem statement, all variables and given/known data

    After a mass weighing 8 pounds is attached to a 5-foot spring, the spring measures 6.6 feet. The entire system is placed in a medium that offers a damping constant of one. Find the equation of motion if the mass is initially released from a point 6 inches below the equilibrium point with a upward velocity of 1 ft/sec.

    2. Relevant equations
    F=kx
    my''+cy'+ky=0

    3. The attempt at a solution
    I got k=8/(6.6-5) = 8/1.2
    c= 1 as given ?

    setting up the eqn I got 1/4 y'' + 1y' + 8/1.2 y = 0
    or y'' +4y' + 80/3 y = 0

    is this actually right? I tried 3 times and I keep getting this, i dont know i feel suspicious ?!
     
  2. jcsd
  3. Oct 20, 2015 #2

    HallsofIvy

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    "8 pounds" is weight, or force, not mass. What does "damping constant" mean? What are its units and how does it give force?
     
  4. Oct 20, 2015 #3
    of course 8 lbs is weight, dividing by 32 gives me .25 which is my mass! damping i believe is given in the problem "damping constant of one" .. otherwise if not given i would find it by sqrt of 4*m*k
     
  5. Oct 20, 2015 #4

    HallsofIvy

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    "Damping force" is always opposite to the motion and is, approximately, proportional to the speed or the speed squared. I presume here you are told that it is proportional to speed. But because it is opposite to speed, the force must be -kv.
     
  6. Oct 20, 2015 #5
    and of course my iniital conditions would be y(0)=-6 y'(0)=1
     
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