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Two masses hanging by springs

  1. May 18, 2008 #1
    1. The problem statement, all variables and given/known data
    A mass [tex]M_1[/tex] is suspended by a spring with constant [tex]k_1[/tex]. A second mass [tex]M_2[/tex] is suspended from the first by a spring with constant [tex]k_2[/tex]. The equilibrium height of the first is [tex]y_1[/tex] and the equilibrium height of the second is [tex]y_2[/tex].

    I just want to write down the Lagrangian for now.


    2. Relevant equations
    [tex]L=T-V[/tex] (i.e. Lagrangian is Kinetic energy minus Potential energy).


    3. The attempt at a solution
    I'm using as generalized coordinates the displacements of the two masses from equilibrium [tex]\delta_1[/tex] and [tex]\delta_2[/tex]
    I'm pretty sure that the kinetic energy parts are easy
    [tex]T_1 = {1\over 2}M_1 \dot\delta_1^2[/tex] and [tex]T_2 = {1\over 2}M_2 \dot\delta_2^2[/tex]
    Now for the potential energy for the second mass I think is
    [tex]V_2 = -M_2 g (y_2+\delta_1+\delta_2) + {1\over 2}k_2(\delta_2-\delta_1)^2[/tex].
    The first term is the gravitational part (taking the zero to be at the ceiling where the first spring is anchored) and the second term is the spring part.
    But for the first mass, I'm tempted to write
    [tex]V_1 = -M_1 g (y_1+\delta_1) + {1\over 2}k_1 \delta_1^2[/tex].
    I think the spring part is wrong because [tex]M_1[/tex] is attached to spring 2, so there should be some contribution involving [tex]k_2[/tex]. This leads me to want to add another term like [tex]{1\over 2}k_2(\delta_2-\delta_1)^2[/tex], but I don't know if it should be positive or negative, and also I feel like I am 'double counting' the spring term in [tex]V_2[/tex]. For example, if spring 1 is streched and spring 2 is compressed, I expect [tex]V_1[/tex] to be increased by the compression of spring 2. I hope this is clear.
     
  2. jcsd
  3. May 18, 2008 #2
    I've almost convinced myself that I should have
    [tex]V1=-M_1 g(y_1+\delta_1)+{1\over 2}k_1\delta_1^2\pm{1\over 2}k_2(\delta_2-\delta_1)^2[/tex].
    The sign is fuzzy to me, but thinking about the force (saying that down is negative), I think I should keep the minus. If this is correct, then I can get the equations of motion, which brings me to my real question: How do I find the normal mode frequencies? From similar examples, I've tried putting [tex]\delta_j = A_j e^{i\omega t}[/tex] (i.e. same frequency for both displacements (gen. coords)), into the equations of motion, then finding [tex]\omega[/tex] such that the solution exists. But, the gravitational terms don't allow us to cancel off the exponentials, which usually happens. I hope you can follow.
     
  4. May 18, 2008 #3
    Sorry. It should just be
    [tex]V_1=-M_1 g(y_1+\delta_1)+{1\over 2}k_1\delta_1^2[/tex].
    But I still don't know how to find the normal mode frequencies.
     
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