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Damped spring problem with laplace transform

  1. Nov 9, 2014 #1
    1. The problem statement, all variables and given/known data
    A 4-pound weight stretches a spring 2 feet. The weight is released from rest 15 inches above the equilibrium position, and the resulting motion takes place in a medium offering a damping force numerically equal to 7/8 times the instantaneous velocity. Use the Laplace transform to find the equation of motion x(t). (Use g = 32 ft/s2 for the acceleration due to gravity.)

    2. Relevant equations
    my''+(beta)y'+ky=0

    m=4/32=1/8
    beta=7/8
    k=4/2=2
    3. The attempt at a solution
    1/8y''+7/8y'+2y=0, y(0)=-18, y'(0)=0

    I know the answer which is:
    -1/10e^(-7t/2)[7√(15)sin(√(15)t/2)+15cos(√(15)t/2)]
    https://www.webassign.net/latexImages/5/f/f1c0fb013b3d1d7c4fddd1209a1d60.gif

    but my awnser came out to be:
    http://www4c.wolframalpha.com/Calculate/MSP/MSP104701e9a70abfi51f43a000045edi0bi5e950fi2?MSPStoreType=image/gif&s=14&w=373.&h=45 [Broken].

    which makes me fell like my y'(0) is incorrect. I would think that y'(0) would be a valid initial condition but the problem seem to not state anymore about velocity.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Nov 9, 2014 #2

    vela

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    I take it the weight is released from 18 inches above the equilibrium point, not 15 inches like you typed above. The link to your answer doesn't work, but I suspect the problem is simply that you didn't specify y(0) in the correct units. The statement does say that the weight was released from rest, so your condition of y'(0) = 0 is correct.
     
    Last edited by a moderator: May 7, 2017
  4. Nov 9, 2014 #3
    my answer came out to be

    y(x) = -6/5 e^(-7 x/2) (7 sqrt(15) sin((sqrt(15) x)/2)+15 cos((sqrt(15) x)/2))
    which is different and yes the problem is suppose to be released at 18 not 15.
     
  5. Nov 9, 2014 #4
    my answer came out to be
    y(x) = -6/5 e^(-7 x/2) (7 sqrt(15) sin((sqrt(15) x)/2)+15 cos((sqrt(15) x)/2))
    which is different. and yes the problem is suppose to be released at 18 not 15.
     
  6. Nov 9, 2014 #5

    vela

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    Unless you show your work, we can't figure out what you're doing wrong.
     
  7. Dec 7, 2014 #6
    sorry for the late reply but I found that the released spring distance was in inches so I just had to convert the 15 in to ft.
     
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