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Blocks and Compressed Spring

  1. Dec 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Two blocks of equal mass, connected by a masslless spring of constant ##k## are on an air track. her unstreched length of the spring is ##L##. at ##t=0## block ##b## is compressed to distance from block ##a## at ##l/2## (block ##a## is pressed against a wall). Find the motion of the center of mass of the system as a function of time.


    2. Relevant equations



    3. The attempt at a solution

    I think I've got most of this problem worked out, aside from the end.

    Our center of mass is ##R=\frac{1}{2}(r_a + r_b )##. We have set up new labels. ##r' _a = r_a - R ## and likewise for ##r' _b##, we have ##r' _a = -r' _b ##.

    The springs displacement from it's equilibrium length is given by ##r_a - r_b - L = r' _a - r_b - L##, ext we set up the equations of motion and subtract ##m\ddot{r}'_b## from ##a##.

    ##u = r' _a - r' _b - L \rightarrow \ddot{u} = \ddot{r}'_a - \ddot{r}'_b ## then ##m\ddot{u}+2ku=0## solving for the boundary conditions (##u(0)=\frac{L}{2}## and ##\dot{u}(0)=0##) we have

    $$u(t)=\frac{L}{2}\cos\left( \sqrt{\frac{2k}{m}}t\right)$$

    At this point I'm a little bit confused. We need to find velocity of ##R##, so I think we can fiddle with ##u=r'_a - r'_b - L##.

    Edit: Because we can rewrite this as ##\dot{u}(t)=-\omega \frac{L}{2} \sin (\omega t)##, we can write ##\dot{r}_a - \dot{r} _b = -\omega \frac{L}{2} \sin (\omega t)##, if we multiply the entire equation by ##\frac{1}{2}##, we would have ##\dot{R} = \frac{1}{2}(\dot{r}_a - \dot{r} _b ) = -\omega \frac{L}{4} \sin (\omega t)##? The only thing that's really weird is that this is negative, which can't be right.

    Edit: Edit: Forget the top edit, I didn't realize that ##R = \frac{1}{2}(r_a + r_b )##


    Edit:Edit:Edit: I tried another method. Because ##\cdot{r}' _a - \dot{r}'_b = \dot{u}(t)## and ##\cdot{r}' _a = -\dot{r}'_b ##,this implies that ##\dot{r}'_a = -\omega \frac{L}{4} \sin (\omega t)## and ##\dot{r}'_b = \omega \frac{L}{4} \sin (\omega t)##. Now that we have two expressions, we can plug these into ##\dot{R} = \frac{1}{2}(\dot{r}'_a + \dot{r}'_b ) = 0## This result also seems counterintuitive The more I try, the more confused I become
     
    Last edited: Dec 11, 2013
  2. jcsd
  3. Dec 11, 2013 #2

    haruspex

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    I would split the problem into two phases. In the first phase, block A is against the wall and there is a force from the wall onto A. Later on, the system will move away from the wall, never to return. Find out when that happens.
     
  4. Dec 11, 2013 #3
    I was thinking that a valuable piece of information might be that the force on A and B is ##k\frac{L}{2}##, but I can't figure out how to use it. These forces from the spring oppose each other.

    I can't understand why we should need to do anymore work on this, we have the EoM, it seems to me that it's just a matter of subbing in the right values.

    With ##\ddot{u}(t)##, we have information on what all the other variables can be.

    What I can think of physically, is the spring will push block B out past it's rest length, at which point A will have a force pulling it away from the wall.

    ##\dot{u}(t) = \dot{r}' _a - \dot{r}'_b## and ##\dot{r}' _a = -\dot{r}'_b## must give us the velocities of each, it's just a matter of solving them. I tried this method in my last edit, and it didn't seem to work.

    Do you see what I'm saying? You may very well have suggested the correct way to look at this problem, but I'm having a difficult time seeing why my approach won't work.
     
  5. Dec 11, 2013 #4

    haruspex

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    I confess I've not tried to follow your working so far, but my point is that the circumstances will undergo a distinct change when block A loses contact with the wall. Up until that point, it will be as though the spring were fixed to the wall, and block A might as well not exist. After that point, there is no external force on the system. I don't see anywhere that you've modelled that, which makes me doubt your approach.
     
  6. Dec 12, 2013 #5
    I got the answer through conservation of momentum, but I think it was a little bit of an underhanded tactic. But it's all good, thanks for the help, you made me realize I could "cheat". I could have gotten an answer with 1/4 the work involved.
     
  7. Dec 12, 2013 #6

    haruspex

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    OK, that's good.
    Fwiw, it looks to me that the answer must be like
    (L/2)(1-cos(γt)/2) for t < π/(2γ)
    L/2 + v(t-π/(2γ)) for larger t.
     
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