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Spring problem

  1. Feb 9, 2010 #1
    1. The problem statement, all variables and given/known data

    A pair of masses M1, M2 is suspended vertically by a pair of spring, with spring constant k1, k2. ( see the attachment for the picture)

    a.A downward force F is applied to bottom mass. Find the downward displacements d1 and d 2 of the equilibrium positions of the Mass M1and M2 due to the force. Note that effect of gravity is already taken into account in determing the equilibrium positions.


    b.At time t =0, the downward force is removed. What are the equation of motion and initial conditions that determine the displacements d1(t) and d2(t) for t greater than 0? You need not solve the equations.


    2. Relevant equations

    m2g=k2x2
    k2x2 + m2g+ m1g=k1x1
    then i added them, I get 2m2g +m1g=k1x1

    F=-kx

    3. The attempt at a solution

    m2g=k2x2
    k2x2 + m2g+ m1g=k1x1
    then i added them, I get 2m2g +m1g=k1x1.
    after this, I hav no idea how to solve this! please help!
     

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  2. jcsd
  3. Feb 10, 2010 #2
    Btw, this is a Caltech problem!
     
  4. Feb 10, 2010 #3

    jhae2.718

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    Gold Member

    For A, try creating separate free body diagrams for [itex]m_1[/itex] and [itex]m_2[/itex] and finding [itex]\Sigma F_y=F_{app}[/itex]. Then, solve each for [itex]d_1[/itex] and [itex]d_2[/itex].
     
  5. Feb 10, 2010 #4
    a) Both springs will experience force F. Since they are already in equilibrium you can "ignore" the masses.
    [tex] F=k_1d_1[/tex]
    [tex] F=k_2d_2[/tex]
    b) Simple SHM equations.
     
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