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The equation of motion for a Harmonic oscillator with linear drag is, by Newton's second law, $$m \ddot{x}(t)+b \dot{x}(t)+kx(t)=0.\tag{1}$$ This is a linear differential equation that can be solved analytically by well known techniques and this system is studied in all undergraduate classical mechanics courses. My question is, what happens when we cannot longer assume that drag is linear and we have to use some other drag force, in fluids of low viscosity, like air or water?

In the case of quadratic drag, the equation of motion changes to $$m \ddot{x}(t)+c\dot{x}(t)|\dot{x}(t)|+kx(t)=0,\tag{2}$$ where $c$ is the drag coefficient for quadratic drag. This case is rarely treated in textbooks and I assume that no closed-form analytical solution to this nonlinear differential equation is possible, but what can be said about the motion and/or behaviour of the system? How can at least be solved numerically? Also, is there a Lagrangian formulation for this system?

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    $\begingroup$ The fact that the damping term is always positive (assuming $c$ positive too) while $x(t)$ can be positive and negative is an issue (on a physical point of view of course, not a mathematical one)... You might need to define two equations: one for $\dot{x}>0$ and one for $\dot{x}<0$ so the dampig force will always oppose the movement. $\endgroup$ – Syrocco Dec 18 '19 at 2:08
  • $\begingroup$ @Syrocco: Added an absolute value. $\endgroup$ – Qmechanic Dec 18 '19 at 2:19
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    $\begingroup$ How can at least be solved numerically? Any ordinary differential equation can be solved numerically, so what do you mean? There isn’t anything about this equation that makes standard numerical methods inapplicable. $\endgroup$ – G. Smith Dec 18 '19 at 3:00
  • $\begingroup$ @G.Smith What I mean is that I have problems and conceptual difficulties in solving numerically a differential equation. $\endgroup$ – Don Al Dec 18 '19 at 4:24
  • $\begingroup$ If you have problems with numerical solution concepts then you may want to read up on them, try Numerical Recipes, or a text on ODE methods. $\endgroup$ – ggcg Dec 18 '19 at 16:35
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Depending on the boundary conditions of the problem, this can have a closed form analytical solution. You can proceed as follows.

Check that the above equation with quadratic drag can be rewritten as: $$mv\frac{dv}{dx}+cv^2+kx=0$$ $$\implies \frac{m}{2}\frac{d}{dx}(v^2)+cv^2+kx=0$$

This is a linear ODE in $v^2$ and you use an integrating factor of $\exp{\left(\frac{2cx}{m}\right)}$.

So you get

$$v=\frac{dx}{dt}=v(x=x_1)+\sqrt{\int_{x_1}^{x} dx \frac{2kx}{m} \exp{\left(\frac{2cx}{m}\right)}}$$

And after this, again depending on the boundary conditions, you can solve further or predict facts about the motion from this $\frac{dx}{dt}$ vs. $x$ equations, which is called a phase space plot in NLD.

This is how you can work through this purely via classical methods, without any sort of numerical analysis.

Hope this helps.

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  • $\begingroup$ It might be worth noting that you will have to switch the sign of the drag coefficient c "by hand" when ever the velocity reaches zero. $\endgroup$ – Luke Pritchett Dec 18 '19 at 16:40

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