I’m having some trouble predicting the behavior of ODE’s using the mass-spring analogy. For example, consider the second order IVP listed below: $$y’’ – \space 6y’ + 8y = 0, \space \space \space \space y(0) = 2, \space \space y’(0) = -8$$ Now I believe that the initial condition of $y(0) = 2$ indicates that the original displacement of the spring from equilibrium is 2 units, but I could be wrong on that. I have no idea what $y’(0) = -8$ means intuitively, so that’s the first thing I’m hoping to get cleared up (ie interpreting the initial conditions).

Next, I understand that each coefficient means different things. From the equation above I see that there is an inertial mass of 1, a damping factor of -6, and a stiffness factor of +8. Since the damping factor is negative I think that the spring will not converge, but outside of that I’m not sure how to interpret the other coefficients. So as another question, how do you know when the solution will oscillate with increasing distance, or go off to positive/negative infinity, or simply oscillate, etc.

I'm really seeking advice on interpreting what the behavior of the spring will be for all 8 different scenarios when the coefficients are positive and negative. I’m already planning to plug in some examples using wolfram alpha to see what’s going on, but getting some deeper understanding from the forum will definitely help. Any tips and help with this would be greatly appreciated.

Edit: More specifically, I'm interested in how the solution behaves graphically over time.

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    $\begingroup$ $y'(0)$ is the initial velocity of the mass. $\endgroup$ – copper.hat Feb 24 '13 at 23:19
  • $\begingroup$ So since $y'(0) = -8$ does this mean the spring is currently recoiling with a velocity of 8 units? Does the negative sign imply backwards motion? $\endgroup$ – Amateur Math Guy Feb 24 '13 at 23:23
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    $\begingroup$ It means that initially it is recoiling with a speed of 8 units/sec in the negative direction. $\endgroup$ – copper.hat Feb 24 '13 at 23:24
  • $\begingroup$ Alright thanks! Any thoughts on the second part? $\endgroup$ – Amateur Math Guy Feb 24 '13 at 23:30

You are correct that the natural interpretation is that the mass is $1$ and the restoring force is $8$, but the equation could be multiplied by any constant. This could equally describe a mass of $10$ and a restoring force of $80$. In either case, the natural frequency, which it would vibrate at without friction is $\sqrt 8$. The damping factor being negative is not normal for friction. If isolate the acceleration we have $y''=6y'-8y$ so the acceleration due to friction is in the same direction as the velocity.

As mentioned above, multiplying by a constant doesn't change the behavior, even if the constant is negative. You only have four sign patterns to consider. I would suggest working with the equation in the form $y''=ky'+my$ as it shows clearly the sign of the acceleration from each term. As you say, $k$ should be less than zero, as we expect friction to lead to acceleration opposite to the velocity. Otherwise something is adding energy to the system. We also expect $m$ to be less than zero as we want a restoring force-we want the acceleration opposite in sign to the displacement.

Added: If you start with a second order equation like this, $y''+by'+cy=0$ you can learn a lot by factoring the similar quadratic equation $t^2+bt+c=0$. If the roots are $r,s$ the solution (unless it is a perfect square) is $A\exp(rt) + B\exp(st)$. If the roots are real, there will be no oscillatory behavior. If the roots are complex the $it$ in the exponential gives the oscillation.

  • $\begingroup$ What would this mean for the solution graphical. That is, when you say that the acceleration due to friction is in the same direction as the velocity, what does this mean for the solution over time? I've made this distinction more clear in the question $\endgroup$ – Amateur Math Guy Feb 24 '13 at 23:50
  • $\begingroup$ @AmateurMathGuy: If the acceleration due to friction is in the direction of the velocity, the velocity will grow exponentially with time because (ignoring the spring for now) you have $y''=ky'$ which is solved by $y'=A\exp(kt)$. No oscillation at all. $\endgroup$ – Ross Millikan Feb 24 '13 at 23:57
  • $\begingroup$ I've already upvoted your answer, but I have just one last question. So I inputted the original equation into wolfram alpha with the current boundary conditions and it did not oscillate but then when I changed the equation to $y'' - 2y' + 8y$ with the same intial conditions and I got some oscillation. Do you know what's going on here or have I just done something wrong in wolfram alpha? $\endgroup$ – Amateur Math Guy Feb 25 '13 at 0:19
  • $\begingroup$ @AmateurMathGuy: Alpha gives me exponential solutions in both cases, but if I change the sign on the $y'$ term and solve $y''+2y'+8y=0$ it gives nice oscillations. wolframalpha.com/input/… $\endgroup$ – Ross Millikan Feb 25 '13 at 0:30
  • $\begingroup$ Ok, thanks for the help $\endgroup$ – Amateur Math Guy Feb 25 '13 at 0:35

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