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Modeling a Spring

  1. Oct 1, 2012 #1
    1. The problem statement, all variables and given/known data
    A brick of mass 6 kg hangs from the end of a spring. When the brick is at rest, the spring is stretched by 3 cm. The spring is then stretched an additional 4 cm and released. Assume there is no air resistance. Note that the acceleration due to gravity, g, is g=980 cm/s2.

    Set up a differential equation with initial conditions describing the motion and solve it for the displacement s(t) of the mass from its equilibrium position (with the spring stretched 3 cm).


    2. Relevant equations
    my'' + by' + ky = Fexternal

    y(0) = y0 + initial displacement
    y'(0) = V0

    m = mass
    b = damping constant
    k = spring constant

    F = -kx

    3. The attempt at a solution
    I get:

    k = 19.6N/cm
    y(0) = -4cm
    y'(0) = 0

    I'm assuming b = 1 since it wasn't given.

    6y'' + y' + 19.6y = 0, solving for the initial conditions
    .
    .
    .
    y(x) = e^(-.08333x)(-0.1846sin(1.8055x)-4cos(1.0855x))
    But that comes back as incorrect. I had a similar problem and didn't have any trouble so I'm having a hard time seeing where I went wrong with this one.
     
  2. jcsd
  3. Oct 1, 2012 #2

    BruceW

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    That is correct.

    Ah, but they do give b, and it is not 1. You've said b is the damping constant. And it says in the question that there is no air resistance, so they are implying something about the value of b.
     
  4. Oct 1, 2012 #3
    I spoke with the TA and b=0 unless stated otherwise. Reworking from 6y''+19.6y=0, I get y(t)=-4cos(1.8074t) and it's still incorrect. This is driving me nuts. I have two similar problems and am running into the same issue with the general solution. I know the amplitude and frequency are correct but something is wrong with my general solution.
     
  5. Oct 2, 2012 #4

    BruceW

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    Your equation was my'' + ky = fext since there was no damping, which is correct, but then you assume that the external force is zero. But this is wrong, because there is an external force.
     
  6. Oct 2, 2012 #5
    Would gravity on the mass be an external force? I say Fext = 0 because we haven't covered springs with external forces on them yet and I'm not sure how I would work that into the equation considering I'm solving for displacement as a function of time.
     
  7. Oct 2, 2012 #6

    BruceW

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    Gravity on the mass definitely would be an external force. Perhaps the reason why your answer is marked as 'incorrect' is because you haven't explained why the presence of gravity doesn't affect the amplitude and frequency. Have you had any classes where they explained the effect of gravity on this system?
     
  8. Oct 2, 2012 #7
    Not yet, no. This is actually for my differential equations class but it's applicable to this subforum. The goal of this homework set is modeling springs without any external forces.
     
  9. Oct 5, 2012 #8

    BruceW

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    I guess this problem is an exception to the homework set. As you have said, you already have the correct solution. So the only reason I can think it was marked as 'incorrect' is because the marker didn't like your working. And I would guess that it is because they didn't like your explanation of how gravity affects the solution. Did the marker give any hint to what was wrong with your work? And how did you explain gravity's effect?
     
  10. Apr 1, 2013 #9
    I had this same hw problem, use g=980 cm/s2. instead of g=9.8m/s2. This changes your answer from y(t)=-4cos(1.8074t) to y(t)=-4cos(18.074t). They want the answer in terms of cm and seconds not meters and seconds.

    Also in the class im taking, we referred to negative y as positive direction , i don't know how you do it but that would make the -4 coefficient become +4.

    TL;DR y(t)=4cos(18.074t)
     
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