1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

3 Springs and 2 Mass System in a line

  1. Feb 5, 2014 #1
    1. The problem statement, all variables and given/known data
    I have 2 walls, 3 springs, and two masses. The masses and springs are not necessarily similar to the other(s).
    They are connected in a configuration: Wall-Spring1-Mass1-Spring2-Mass2-Spring3-Wall.
    I need to find equation for both masses, given that not all springs are in equilibrium when the system is set into motion.
    Assume that for this system, there is one scenario in which the system is in its equilibrium state. In this state, x_1 and x_2 (defined below) are 0.

    Springs constants: k_1, k_2, k_3
    Masses: m_1, m_2
    Displacement of mass 1 from equilibrium: x_1
    Displacement of mass 2 from equilibrium: x_2


    2. Relevant equations
    F = -kx
    U = 1/2 * kx^2
    K = 1/2 * mv^2
    E = U + K
    dE/dt = 0 (closed system)


    3. The attempt at a solution
    I first calculated the forces on each mass for a given position and found that
    F_1 = -k_1 * x_1 - k_2 * (x_1 - x_2)
    F_2 = -k_3 * x_2 - k_2 * (x_2 - x_1)
    Also, E = 1/2(∑kx^2 + ∑mv^2)

    (' denote derivative with respect to time)
    and dE/dT = k_1*x_1*x_1' + k_3*x_2*x_2' + k_2*(x_1 - x_2)(x_1' - x_2') + m_1*x_1'*x_1'' + m_2*x_2'*x_2'' = 0

    But I'm not sure what to do from here.
    First, are my equations for force correct? The middle spring (#2) applies the force F=-kx to two objects and I have only ever dealt with springs fixed on one end so I'm not sure.
    Also the dE/dT = 0 is a nonlinear differential equation and I have only ever dealt with linear differential equations so I don't know how to solve it.

    Would this be better in Physics hw help? I posted here since it is hw for a DE course.
     
    Last edited: Feb 5, 2014
  2. jcsd
  3. Feb 6, 2014 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You've started correctly, but I would not use "energy" here. Directly from "F= ma" you have the two equations for the motion of the two objects
    [tex]m_1\frac{d^2x_1}{dt^2}= -k_1x_1- k_2(x_1- x_2)= -(k_1+ k_2)x_1+ k_2x_2[/tex]
    [tex]m_2\frac{d^2x_2}{dt^2}= -k_3x_2- k_2(x_2- x_1)= k_2x_1- (k_2+ k_3)x_3[/tex]

    Depending upon how you want to treat it, that can be reduced to a single fourth order equation or to four first order equations.
     
  4. Feb 6, 2014 #3
    I have only ever worked with single second-order linear differential equations and I'm not sure how to approach reducing. I tried taking the derivative of the second equation with respect to time twice and substituting [tex]\frac{d^2x_1}{dt^2} = -\frac{k_1}{m_1}x_1 - \frac{k_2}{m_1}(x_1 - x_2)[/tex] and get (after substituting, distributing minus signs, and multiplying both sides by [itex]m_1[/itex]) [tex]m_1m_2\frac{d^4x_2}{dt^4} = -m_1(k_2 + k_3)\frac{d^2x}{dt^2} - k_2(k_1 + k_2)x_1 + k_x^2x_2[/tex] but there is still the [itex]x_1[/itex] term that I'm not sure how to get rid of.

    edit; Ok, I substituted [itex]x_1[/itex] from the original equation for [itex]m_2\frac{d^2x_2}{dt^2}[/itex] and got fourth order differential equations for both x_1 and x_2. Unsurprisingly, the coefficients were the same. I got
    [tex]m_1m_2\frac{d^4x}{dt^4} + (m_1k_2 + m_1k_3 + m_2k_1 + m_2k_2)\frac{d^2x}{dt^2} + (k_1k_2 + k_1k_3 + k_2k_3)x = 0[/tex] in which [itex]x[/itex] is either [itex]x_1 or x_2[/itex].

    Assuming solution is [itex]e^{\lambda t}[/itex] I get four lambdas so my solution is [tex]
    x = c_1e^{\lambda_1 t} + c_2e^{\lambda_2 t} + c_3e^{\lambda_3 t} + c_4e^{\lambda_4 t}[/tex] but I only have two initials equations for each variable:
    [tex]x(0) = x_i \quad and \quad x'(0) = 0[/tex]
    so I have four variables and two equations which is not enough for me to find the coefficients
     
    Last edited: Feb 6, 2014
  5. Feb 6, 2014 #4

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    There are two masses. The motions of each are controlled by the same four unknowns. For each mass you have an initial position and an initial velocity: four equations.
     
  6. Feb 6, 2014 #5
    The coefficients on each exponent are not necessarily the same for [itex]x_1[/itex] and [itex]x_2[/itex] right?
     
    Last edited: Feb 6, 2014
  7. Feb 6, 2014 #6

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Right, but they must be related. What happens if you substitute your generic x1 solution back into the original two differential equations?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: 3 Springs and 2 Mass System in a line
Loading...