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SHM with unusual spring

  1. Dec 21, 2011 #1
    1. The problem statement, all variables and given/known data
    This was a question on my 2nd midterm that I am still curious about.

    An unusual spring has a force law given by F = -cx^3 where x is in meters, F is in Newton's, and c has units of N/m^3.

    The spring is used to launch a 3.00 kg block across a horizontal frictionless surface. If c = 600 N/m^3 and the spring is initially compressed by 0.500 m, what will the velocity of the block be when it reaches the equilibrium position?


    2. Relevant equations
    F = ma
    x(t) = Acos(ωt)
    v(t) = -Aωsin(ωt)
    a(t) = -A(ω^2)cos(ωt)

    3. The attempt at a solution
    First, I find the acceleration with the equation ma = -cx^3 (with the direction positive to the left of the equilibrium position) during initial position because that is where maximum acceleration occurs. I get a value of a = -25m/s^2.

    I plug this into the a(t) = -A(ω^2)cos(ωt) with t = 0, and I get a value of ω = 7.071.

    Now since I have ω, I can use v(t) = -Aωsin(1) to find maximum speed which occurs at the equilibrium position. I get an answer of 3.54 m/s, which fits into one of the multiple choice questions.

    The problem is that the actual answer is 2.50 m/s, and I have no idea what else I can do to solve this.
     
  2. jcsd
  3. Dec 21, 2011 #2

    Doc Al

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    Instead of trying to use the equations for SHM (do you think they apply?), use conservation of energy. What's the energy stored in the compressed spring?
     
  4. Dec 21, 2011 #3
    By conservation of energy, all the spring potential turns to kinetic.

    0.5cx^2 = 0.5mv^2

    The c constant is given in N/m^3 though, and I don't how to convert that to N/m if it's possible.

    I try to do that anyways and get a wrong value of v = 7.071, which is by coincidence (?) what I got for ω in my earlier attempt.
     
  5. Dec 21, 2011 #4

    vela

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    You missed Doc Al's point. You can't use your normal spring formulas because you're working with a non-standard spring. You need to go back to the basics.
     
  6. Dec 21, 2011 #5
    This time, I'll use W = ∫Fdx

    I find the integral of F, then I get a value of -9.375 for the area under the curve from x = 0 to x = 0.5 therefore ...

    W = -9.375J

    I'm not sure how to interpret the negative value, but I assume I just use the magnitude of it, and this all gets turned to kinetic energy when the spring reaches equilibrium position.

    9.375J = 0.5(3.0kg)v^2
    v = 2.5 m/s

    Hmm I actually got it this time.. Sure wish I knew how to do integrals back then.
     
  7. Dec 21, 2011 #6

    vela

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    You calculated the work done by the spring as it was stretched, which is indeed negative. The potential energy of the spring is negative of the work done by the spring.
     
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