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!
Best Answer
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 $x'(0)>0$.