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Spring with 2 masses

  1. Dec 24, 2009 #1
    1. The problem statement, all variables and given/known data
    A spring of constant k connects two masses, m1 and m2. The system is constrained
    to longitudinal oscillations (in the direction of the line connecting the masses).
    (a) Find the equivalent 1D problem and solve the equations of motion.
    (b) Use the solution to the previous point to obtain the solution of the original

    2. Relevant equations
    I solved point a. I got that the lagrangian is :(0.5*m1*v1^2)+0.5*m2*v2^2)-0.5*k*(x1+x2-lo)
    Is that correct?
    And the solution of the NDE are without cos,sin because x1''=-k/m1 for example.
    was I wrong?

    Now for point b, I should you coordinate x1,y1,x2,y2? beacause the masses can also go up and down during their motion?
    How will the lagrangian will look like?
    Please help me.
  2. jcsd
  3. Dec 26, 2009 #2
    Now I could be wrong on this, but I would think an equivalent 1D problem would be a simple harmonic oscillator. If you place the origin (and your reference point) on the left mass, it will only appear that the right mass is moving. The Lagrangian for that would be


    because you only have 1 mass moving with 1 direction, hence no subscripts to distinguish particles.

    For solving this actual problem, you won't need to consider the [itex]y[/itex] axis at all because you are told the motion is longitudinal only. What you should do is consider the unstretched length of the spring, call it [itex]\ell[/itex] and then use some reference point that is some distance, call it [itex]x_0[/itex] to the left of the left mass. Hopefully this little picture can help describe what I'm talking about:

  4. Dec 26, 2009 #3
    But from the question , How can I be sure that only one mass is moving and not both of them? and regrading to the actual problem , in the 1D problem I also have some stretched length of the the spring , so What's the difference between the two problem?
  5. Dec 26, 2009 #4
    If we call the left mass [itex]m_1[/itex] and the second mass is [itex]m_2[/itex], then when we place our origin on [itex]m_1[/itex], its position is [itex]x_{m_1}=0[/itex]; its time-derivative would also be zero: [itex]\dot{x}_{m_1}=d(x_{m_1})/dt=d(0)/dt=0[/itex].

    In the second scenario, your origin is not at [itex]m_1[/itex], but somewhere to the left of it. So then its position would be [itex]x_{m_1}=x_0+ x_1[/itex] so that its time-derivative would be [itex]d(x_{m_1})/dt=dx_1/dt=\dot{x}_1[/itex] because [itex]x_0[/itex] is a constant. Your Lagrangian in this case would then be something like


    As you can see, there is a difference in the Lagrangian, so there will be a difference in equations of motion. You should find http://en.wikipedia.org/wiki/Normal_mode" [Broken] of oscillation with this problem (I think there will be only 2 modes of oscillation with this set up).
    Last edited by a moderator: May 4, 2017
  6. Dec 26, 2009 #5
  7. Dec 26, 2009 #6
    Now another thing I don't understand is why you took the 1D problem as the simple harmonic oscillator?
  8. Dec 26, 2009 #7
    If you let [itex]x_0=x_1=0[/itex], then the Lagrangian I wrote in post #4 reduces to the Lagrangian in post #2
  9. Dec 26, 2009 #8
    and still, the solution to the motion equations is with exp, and not sin and cos... and also not imaginary soultion
  10. Dec 26, 2009 #9
    Whether or not an imaginary solution exists depends on how the potential is defined. But you should recognize that, through the Euler relations,

    \exp\left[\pm ix\right]=\cos\left[x\right]\pm i\sin\left[x\right]

    So either the exponential or the sine/cosine answer is correct. You would only be able to definitively decided which would give the right answer when you have initial conditions.
  11. Dec 26, 2009 #10
  12. Dec 27, 2009 #11


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    Homework Helper

    This is a two-body problem that can be reduced to one-body problem by appropriate choice of coordinates. As this is a closed system, the centre of mass does nor accelerate. So we choose Xc, the position of centre of mass as one of the new coordinates.
    The potential energy of the system depends on the change of length of the spring. The spring connects the two masses, their distance is equal to the length of the spring, x2-x1=L. The potential energy is PE=1/2k(L-L0)2. It is obvious to choose x2-x1-L0 = u as the other new coordinate.

    Now we have to write the KE with the new variables. It is easy to show that



    The time derivatives are:

    [tex]\dot x_1=\dot X_c-\frac{m_2}{m_1+m_2}\dot u [/tex]

    [tex]\dot x_2=\dot X_c+\frac{m_1}{m_1+m_2}\dot u [/tex]

    The expression for the KE can be obtained in terms of the time derivatives of the new variables:

    [tex]KE=\frac{1}{2} (m_1\dot x_1^2+m2 \dot x_2^2)= \frac{1}{2} (m_1+m_2)\dot X_c^2+\frac{1}{2} \frac{m_1*m_2}{m_1+m_2} \dot u^2

    [tex]\frac{m_1*m_2}{m_1+m_2}=\mu [/tex]

    is called the reduced mass of the system.

    The Lagrangian of the system :

    [tex]L=KE-PE = \frac{1}{2} (m_1+m_2)\dot X_c^2+\frac{1}{2} *\mu \dot u^2 -\frac{1}{2}k* u^2

    The Lagrange equations of motion:

    [tex](m1+m2)\frac{d^2 X_c}{dt^2}=0 [/tex]

    the centre of mass stays in rest or travels with its original velocity,

    [tex]\mu\frac{d^2 u}{dt^2}+k*u=0 [/tex]

    the variable u performs a simple harmonic motion [tex]u=A\sin(\omega t +\alpha) [/tex]

    with the angular frequency

    [tex] \omega=\sqrt{\frac{k}{\mu}}[/tex]

    Inserting Xc and u back into the expression for x1 and x2, we obtain the solution of the original problem: The two masses will oscillate around the CM with the same frequency and phase.

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