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Springs Q

  1. Dec 22, 2005 #1
    Here's another Q I'm stuck on:

    "Two particles P1 and P2 of equal mass m are connected to springs as in the diagram below.
    * p1
    * p2

    The springs are identical, each having natural length l and modulus [tex]\lambda[/tex]. The fixed point Q is at a height L vertically above the fixed point O (L > 3l). Take the origin of coordinates at O and the z-axis vertically upwards and let c be the z-coordinate of the centre of mass of the particles.

    The particles are in motion. Suppose that the particles remain on the z-axis throughout the motion. Show that c satisfies the equation

    [tex]\ddot{c} = ( \frac {\lambda L} {2lm} - g ) - \frac { \lambda} {lm} c[/tex]

    [You should assume that acceleration due to gravity, g, is constant and that the springs obey Hooke's law throughout the motion.]"

    I don't know where to begin on this one.
  2. jcsd
  3. Dec 24, 2005 #2


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    Homework Helper

    First draw free-body diagrams and find the location of c at equilibrium.
    Notice that if g were 0 , c would be L/2 .

    You could treat P1 + P2 + MiddleSpring as one object, but
    you might want to treat P1 and P2 separately, at z1 and z2 ...
    (c = (z1 + z2)/2 ; c's acceleration is the average of az1 and az2).

    displace P1 from its equilibruim by dz1 , displace P2 from equilib by dz2 .
    calculate the acceleration of each , then average them.
    If you keep your signs straight, you'll see that MiddleSpring causes no acceleration of the center-of-mass since it pulls oppositely on P1 and P2.
  4. Dec 26, 2005 #3
    Hmm, I'm not really too confident as to how to lay out my solution. I'm assuming that the springs have been pulled away from the origin.
    Using your method said above, so far I have:

    For particle 1

    [tex] m \ddot{z}_1 = T_1 - T_2 - mg [/tex]

    [tex] m \ddot{z}_1= \frac {\lambda} {l} ( dz_1 - l) - \frac {\lambda} {l} ( dz_1 - l) - mg[/tex]

    And particle 2

    [tex] m \ddot{z}_2 = T_2 - T_3 - mg [/tex]

    [tex] m \ddot{z}_2= \frac {\lambda} {l} ( dz_2 - l) - \frac {\lambda} {l} ( dz_2 - l) - mg[/tex]

    Am I on the right lines or is this completely wrong?
    Last edited: Dec 26, 2005
  5. Dec 27, 2005 #4


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    Homework Helper

    The centre of mass of the system is c where [tex]c = (z_1 + z_2)/2.[/tex]
    Also, [tex]2\ddot c = (\ddot z_1 + \ddot z_2)[/tex]

    [tex]m\ddot z_1 =T_1 - T_2 - mg[/tex]
    [tex]m\ddot z_2 = T_2 - T_3 - mg[/tex]

    [tex]m(\ddot z_1 + \ddot z_2) = T_1 - T_3 - 2mg[/tex]
    [tex]2m\ddot c = T_1 - T_3 - 2mg, \mbox{ putting }2\ddot c = (\ddot z_1 + \ddot z_2)[/tex]

    [tex]\mbox{Expansion in Spring1: } T_1 = \frac{\lambda}{l}(l_1-l) = \frac{\lambda}{l}
    (-z_1 -l)\ <-- \ (z_1 \mbox{ is -ve})[/tex]
    [tex]\mbox{Expansion in Spring2: } T_2 = \frac{\lambda}{l}(l_2-l) = \frac{\lambda}{l}(-(z_2 - z_1) - l)\ <-- \mbox{ don't acually need this!}[/tex]
    [tex]\mbox{Expansion in Spring3: } T_3 = \frac{\lambda}{l}(l_3-l) = \frac{\lambda}{l}((-L + z_2) - l)[/tex]

    [tex]T_1 - T_3 = \frac{\lambda}{l}(-z_1 - l) - \frac{\lambda}{l}((-L + z_2) - l)[/tex]
    [tex]T_1 - T_3 = \frac{\lambda}{l}(-z_1 - l + L - z_2 + l)[/tex]
    [tex]T_1 - T_3 = \frac{\lambda}{l}(L - (z_1 + z_2))[/tex]
    [tex]T_1 - T_3 = \frac{\lambda}{l}(L - 2c)[/tex]

    [tex]\mbox{Substituting for } T_1 - T_3,[/tex]

    [tex]2m\ddot c = \frac{\lambda}{l}(L - 2c) - 2mg[/tex]
    [tex]\ddot c = (\frac{\lambda L}{2lm} - g) - \frac{\lambda}{lm}\cdot c[/tex]

    I've had to assume that, as a measured dimension, L is a vector quantity and hence negative, rather than a (positive) scalar quantity.

    http://img322.imageshack.us/img322/738/eldavidas4zn.th.jpg [Broken]

    Attached Files:

    Last edited by a moderator: May 2, 2017
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