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Two masses attached by a spring

  1. Jul 22, 2009 #1
    I am studying the problem of two masses, m1 and m2 (not equal) attached by a spring of spring constant k on a frictionless surface set into oscillatory motion (unstretched spring is of length l, x1 < x2). When I first attacked this problem, I let the origin be the center of mass, calculated the reduced mass, and was able to get the frequency, etc. While this yielded the correct answer, I have now been asked to solve it setting up the equations of motion. I deduce that they are

    m1a1=-k(x1-x2+l)
    m2a2=-k(x2-x1-l)

    Can someone help me understand the notion of l? That is, for mass 1 is l positive because it is in the positive direction and vice-versa? I can solve it mathematically from there, but it makes me uncomfortable not fully understanding how the equations got set up.

    Thanks in advance!
     
  2. jcsd
  3. Jul 24, 2009 #2

    jmb

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    The best way to see it is like this.

    Write down an expression for the extension (i.e. difference from original length) of the spring as a function of [tex]x_1[/tex] and [tex]x_2[/tex] (draw a diagram if you need).

    Consider in which direction each of [tex]m_1[/tex] and [tex]m_2[/tex] would be accelerated by a positive extension. This should make everything clear...

    ...let me know if it doesn't!
     
  4. Jul 24, 2009 #3
    Hi,

    Thanks for replying. I have been looking at this problem from many different angles trying to determine why the prof included the value l. I found a great website that shows a much more complicated problem, but the process is basically the same:

    http://www.efunda.com/formulae/vibrations/mdof_eom.cfm

    I have concluded that the length l does not need to be included in the equations of motion, particularly since the time derivative of it would be zero and it falls out anyway. The only other trick I have noticed with these problems is that you have to take care of the signs and not just knee-jerk a -k.

    *sigh* Maybe I should become a psychology major!
     
  5. Jul 25, 2009 #4

    jmb

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    The reason the example you cite does not include spring lengths is because it is assuming that the springs are all at their natural lengths when the displacement variables of the various masses ([tex]x_i[/tex]) are all zero. In other words, the positions of the various masses are all measured from different origins.

    In the problem set out by your professor, [tex]x_1[/tex] and [tex]x_2[/tex] are measured from the same origin. Thus the spring is not at its equilibrium length when [tex]x_1=x_2=0[/tex]. Thus, using this coordinate system you do need to have [tex]l[/tex] in the equations.
     
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