Here's the problem I have to solve. I can do the math just fine once I get the IVP set up, but getting it set up is what I don't know how to do.

A 2 kg (20 N) mass is attached to a spring, thereby stretching it 0.5 m beyond its undisturbed length. The system is then suspended horizontally in a hydraulic fluid that provides a resistive force of $40\sqrt{5}$ N for every 5 m/s of velocity. Assume no external forces act on this system.

(a) Find a general solution for the equation modeling the horizontal displacement $x$ of the mass from its equilibrium position $t$ seconds after it is set in motion.

(b) Suppose the mass is set in motion from equilibrium with an initial rightward velocity of 2 m/s. Find the particular displacement model given these initial conditions.

My attempt.

(a) Supposedly this system is modeled using the ODE $$mx''+bx'+kx=F_{\text{ext}}(t)$$ where $m$ is inertia (? or maybe mass of the spring? or both?), $b$ is the damping coefficient, $k$ is stiffness of the spring, and $F_{\text{ext}}$ is the external force function. So I would use $m=2$ and $F_{\text{ext}}(t)=0$. I think (but am not sure) that we find $b$ by taking $b=(40\sqrt{5})/5$ and $k$ by taking $k=20/0.5$. Is that correct? Once I have $m,b,k,F_{\text{ext}}$, I can solve it easily.

(b) Clearly $x'(0)=2$---or is it negative 2? But I am confused about the system setup. I take the 0.5 m stretching to be vertical. In its horizontal position, equilibrium is just 0 m, right? So that would mean $x(0)=0$. Is that correct? Again, once I have the IVP I can take it from there.

Thanks guys!

  • $\begingroup$ Read the wording of the problem carefully. $x$ is taken to be the distance relative to equilibrium. This means $x=0$ is the equilibrium point, regardless of direction. $\endgroup$
    – Dylan
    Mar 6, 2019 at 18:09

1 Answer 1


The vertical part is just to calculate the spring coefficient, as you did for part (a). You've got those numbers right. For part (b), you can choose $x(0)=0$ and you can choose $x'(0)>0$. But if you choose $x'(0)<0$ you should still get a very similar equation, where the solution for $x$ position is the mirror image with respect to the equilibrium position when compared to the case $x'(0)>0$.

  • $\begingroup$ Thank you good sir! $\endgroup$
    – Ben W
    Mar 7, 2019 at 10:24

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