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Figuring out the kinetic energy of springs with mass

  1. Dec 8, 2004 #1
    Please feel free to ask me questions if anything doesn't make sense.
    I have this problem that's driving me crazy. The system I'm looking at is a 2 mass, 3 spring system configured like this,

    _____
    |
    \
    /
    [Mass]
    |
    \
    /
    [Mass]
    |
    \
    /
    ___|___
    (x1 = top mass' distance from springs equilibrium, ie x1=0 spring not stretched, x2 = top mass' distance from springs equilibrium. x1 and x2 illustrated below)
    |
    |---[Mass]-----[Mass]----|
    |

    |.......|---->.......|---->
    ...........x1.............x2
    (disregard the periods, I just put them in to space it out to look right)

    With the first spring attached to a support, the other end attached to a mass, a second spring attached to that mass, the other end is attached to a second mass, and a third spring is attached to the second mass, and the ground.
    The two masses are both the same, and the 3 springs all have the same masses and k's.
    I'm trying to come up with the differential equations for this system, and then find the eigenvectors and eigenvalues. I've solved this problem before assuming that the springs are massless, but now I'd like to solve it taking their mass into consideration. We can also assume that the spring velocity at a point on the spring linearly interpolates the velocity of the wall and mass (or mass and mass). I've been able to get fairly far, but I keep getting stuck. Here's what I've come up with so far.
    I figured that I should add up the total energy for the top mass and the two springs attached to it. We already know the kinetic energy for a mass (1/2m(x1')^2), and the potential energy for a spring (1/2k(x1)^2), so we need to figure out the kinetic energy for the springs. For the top spring I said that the velocity at a point a distance x from the top is v=x1' *(x/L), if we look at a small piece of the spring delta x we can say that it's mass is (ms/L)dx, were L is the length of the spring, and ms is the total mass for 1 spring (9 grams). So then I said that the kinetic energy is the integral from 0 to L of 1/2(ms/L*dx)(x1' *x/L)^2 by using 1/2mv^2. After taking that integral I got that the kinetic energy for the to spring was (ms*x1'^2)/6, which according to what I can see seemed to make sense. Then for the middle spring I followed a similar type of logic. For my velocity equation I used v=x1'((2L-x)/L)-x2'((L-x)/L) measuring the distance x from the top. My m was the same, and I integrated from L to 2L, I end up with ms/6 *(x1'^2+x1'*x2'+x2'^2). According to the few basic consistency checks I did on it, this one also seems to make sense. For instance if one of the velocities (x1' or x2') is zero then it simplifies to (ms * x1'^2)/6 which is what we got for the top spring. Or if both of the springs are moving with the same velocity (x1' = x2') we'd expect to get something like 1/2mv^2, and we do get 1/2 ms*x1'^2. And the kinetic energy from the bottom spring is similar to the top spring, (ms*x2'^2)/6.
    So then I added up these energys for the top mass and got 1/2mx1'^2 + 1/2kx1'^2 + 1/2k(x2-x1)^2 + (ms*x1'^2)/6 + ms/6 * (x1'^2 +x1'*x2'+x2'^2) = E. E is just a constant which is the total amount of energy. Then if we take the derivative we get mx1''x1' + kx1x1' - k(x2-x1)x1' +(ms*x1'x1'')/3 + (ms*x1'x1'')/3 +(ms*x1'')/6 = 0. Here is where I run into a problem. When I did the same problem using massless springs then I ended up with an x1' in each term that can be cancelled out, but this time I don't. So I'm not sure if one of the kinetic energy equations I came up with is invalid or what. My teacher told us that the eigenvalues should be -k/(m+5/6ms) and -3k/(m+1/2ms), and the eigenvectors should be [1,1] and [-1,1]. So I'm trying to come up with the correct differential equation to get to this point.
    Sorry this is pretty long. If you have any questions or are unsure about any of my stuff please ask. Your help is greatly appreciated, Thanks.
     
    Last edited: Dec 9, 2004
  2. jcsd
  3. Apr 27, 2010 #2
    Did you solve this problem? I am currently struggling with it.
     
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