# horizontal spring - mass system with damping

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!

• 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. – Dylan Mar 6 at 18:09

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 $$x'(0)>0$$.