Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Double Spring Mass System - Maximizing Velocity

  1. Aug 11, 2008 #1
    Looking for some pointers on how to approach this problem:

    I'm considering a system like the following - two springs and two masses connected in sequence:


    I would like to find the following:

    Given m1, m2, k1, k2. The springs are compressed each by a certain amount (x1start and x2start) and the velocities are set to 0. The springs can be released at any point in time. Once a spring is released and it stretches so that its stretch reaches a peak, it is clamped at that position. Once both springs reach their peak stretch the simulation is ended. Here is what I'm trying to find:

    Assuming the left spring is released at t=0. At what time should the right spring be released to maximize the peak velocity of the right block.

    I could iterate numerically using Runge-Kutta and try stepping up the release time of the right spring and record peak velocity of the right block but that seems very brute forced. I would like to find a better solution for this if possible.

    As a side question, I was wondering if there was an analytical solution to this system?

    Thanks for any pointers.
  2. jcsd
  3. Aug 12, 2008 #2
    As a side note: If you are looking at "optimized" velocity acceleration using a spring system, consider going "out-of-the-box"

    I have worked on a similar problem, and concluded through my work that a "conical" spring works extremely well, especially if one attaches "tension stage" kevlar filaments in a certain way. I can not go into more detail, as such a design is nothing short of a powerful gun without using explosives.

    Your on the right track, but allow yourself to be creative.
  4. Aug 13, 2008 #3
    You can solve the system that you linked to analytically. I would simplify the system by making x_1 and x_2 measured from the equilibrium points of the masses (0 for both when the whole system is in equilibrium.. both springs at their rest lengths). Then the equations are:
    [tex]m_1 \ddot{x_1}+(k_1+k_2)x_2-k_2 x_2=0[/tex]
    [tex]m_2 \ddot{x_2}+k_2 x_2-k_2 x_2=0[/tex]

    To see how to go about an analytic solution you can look in Marion Thornton's Classical Dynamics of Particles and Systems chapter 12 (coupled oscillations) though the example they give is simpler because it has a lot of symmetry. The solution is basically the same, but your frequencies will be much messier.

    Basically you guess a solution of the form:
    [tex] x_1(t)=B_1 e^{i \omega t}[/tex]
    [tex] x_2(t)=B_2 e^{i \omega t}[/tex]
    plug them into the equations above and get a determinant of the coefficients of B_1 and B_2. You'll actually be getting four frequencies (two each of the same magnitude, but different signs) because of a square root. You end up with a final solution of the form:
    [tex]x_1(t)=B_1^{+} e^{i \omega_1 t}+B_1^{-} e^{-i \omega_1 t} +B_2^{+} e^{i \omega_2 t} +B_2^{-} e^{-i \omega_2 t}[/tex]
    [tex]x_2(t)=-B_1^{+} e^{i \omega_1 t}-B_1^{-} e^{-i \omega_1 t} +B_2^{+} e^{i \omega_2 t} +B_2^{-} e^{-i \omega_2 t}[/tex]
    where the B coefficients are determined by the initial conditions and the frequencies are just the two magnitudes of the four total frequencies (if I said that correctly). Anyway, there you go. You can also find a set of normal coordinates, but I don't know if that will really help. Once m_1 is stopped, the other mass will just behave like a simple oscillator so it should be pretty simple from there.
  5. Aug 13, 2008 #4

    It look that after you reduced the system of linear second order DE to a system of linear first order DE, your system of equations can be written as a linear matrix equation

    [tex]Y' = AY + F[/tex]

    [tex]A=\left(\begin{array}{cccc} 0& 0& 1&0\\ 0&0&0&1\\
    -\frac{(k_1 + k_2)}{m_1}& \frac{k_2}{m_1}&0&0\\
    \frac{k_2}{m_2}& -\frac{k_2}{m_2}&0&0\end{array} \right) ,

    Y=\left(\begin{array}{c} x_1\\ x_2\\ v_1\\ v_2\end{array}\right)
    \mbox{ and }
    F=\left(\begin{array}{c} 0\\ 0\\ \frac{k_1R_1}{m_1} - \frac{k_2(w_1+R_2)}{m_1}\\ \frac{k_2(w_1+R_2)}{m_2}\end{array}\right) [/tex]

    if I copied correctly.

    I think the matrix equation can be solve for Y by diagonalizing the matrix A. But haven't try yet.
  6. Aug 13, 2008 #5


    Staff: Mentor

    I would think that you could answer the question using energy conservation much easier than actually solving the differential equation. The maximum velocity will be achieved if, in the end, the two springs are uncompressed (and unstretched), and the other mass is stationary.
  7. Aug 14, 2008 #6
    I'm not that sure how easy this problem can be solve using the conservation of mechanical energy. I thought from the conservation law you only get some relations involving velocities and displacements. The original problem post by ejhong has time factor.

    The equation of motion is a system of coupled DE. I think it is only a hypothesis that the right block attain maximum speed when the two springs are uncompressed. How do we know that this situation is really a particular solution to the problem? The displacements of the two masses are not independent of each other.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook