Following a course in dynamical systems I am studying a mass spring damper system. In the particular case it is a cart constrained to a fixed point by a spring whose oscillation is damped by a damper b. mass spring damper system

If I understand correctly the resulting position graph is the following (and input):

Position and input plot

My question is: why is there an oscillation in the forced response (when applying force i don't know if it's the correct translation) instead of an asymptotic approach to the equilibrium value? That is, if a cart is pulled with a constant force for a given time window, the cart would not reach its maximum distance (i.e. the maximum extension of the spring for the determined force applied) gradually and without oscillations (without go back and forth)? Shouldn't his trend be as follows?

enter image description here

I understand it could be an ideal case in which the spring has infinite extension, but at a constant input (for infinite time) at some point does the force of the spring equals the force of the input( $F_k=u$ )? Again, there would not still be an asymptotic approach to the maximum position (without oscillation) until the spring would be released?

What am I doing wrong?

P.S. I hope I was clear, it's my first post here and English is not my native language.


So assuming a magnetic monopole exists, this is how I am to intend my input ?: enter image description here

  • $\begingroup$ This is Simulink implementation?. put u=0 and initial conditions y(0)=0.3,$\dot y(0)=0$ and b=0 the simulation result y(t) should be sinus wave $\endgroup$
    – Eli
    Commented Oct 31, 2020 at 20:31
  • $\begingroup$ @Eli Yes I tried to use simulink to understand / explain myself better. I'm ok with the fact that leaving the cart free its position will be sinusoidal (the cart go back and forth). I don't understand why it is sinusoidal even when I pull it. $\endgroup$
    – Dabadack
    Commented Oct 31, 2020 at 23:17

1 Answer 1


Without actually solving it, I expect an oscillation during the constant force application. Remember that a constant force has only the effect of moving the equilibrium position (think of a vertical spring in a gravitational field). Thus, when you start applying the force, the spring will no longer be at equilibrium but "compressed" in an out of equilibrium position, and it will start oscillating in a damped way until it stops, or until you switch the force off, which will return the equilibrium point to its original place, thus starting a new oscillation.

If you want a more basic explanation, notice that $u-kx=ma$ (forget the damping for now), thus when $u=kx$ then $a=0$, but the cart still has a velocity, so it will move past the equilibrium point (in which $u=kx$), as $u$ remains constant the spring force grows and eventually decelerates the cart to the point in which $v=0$, and the acceleration is to the left, so the cart will return, etc

  • $\begingroup$ Ty for reply.I understand the sinusoidal movement when I release the spring, but not when it is compressed. I also see that by applying an input I shift the equilibrium point of the system, but I don't understand why the new equilibrium is reached in a "sinusoidal way". If I have a vertical spring in a gravitational field and compress it why should the spring oscillate during the compression? I mean, I am compressing it, the plot I posted above say that the spring is moving me in the opposite direction while I am applying the force/compression. Am I wrong to interpret the time-position graph? $\endgroup$
    – Dabadack
    Commented Oct 31, 2020 at 23:14
  • $\begingroup$ When you apply the force the spring equilibrium point will change. Thus its motion would be as if you compressed it and released it: a decaying sinusoidal $\endgroup$
    – user65081
    Commented Oct 31, 2020 at 23:29
  • $\begingroup$ that is, applying a constant force is equivalent to instantaneously compressing and releasing it $\endgroup$
    – user65081
    Commented Oct 31, 2020 at 23:31
  • $\begingroup$ But in this way what is the difference between applying a force for a time interval t = [0, 10] or t = [0, inf]? If applying continuous force means instantaneously compressing and releasing the mass then when I stop applying force what does that mean? "Move the point of equilibrium again", but at this point what do I mean by equilibrium point? Isn't the equilibrium point in this case the position of the cart that balances the forces? From the "real" point of view, constantly pulling the cart does not mean applying a force and instantly releasing it. So what does that graph describe? $\endgroup$
    – Dabadack
    Commented Oct 31, 2020 at 23:59
  • $\begingroup$ Are you saying that I have to understand the inputs of a system differently? Is the equilibrium point something more abstract? Thanks anyway you made me understand that I did not understand the concept of balance point. $\endgroup$
    – Dabadack
    Commented Nov 1, 2020 at 0:01

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