# Different frictional forces- damped harmonic motion

What classifies as damped harmonic motion? All of the books/Web pages I have looked at about damped harmonic motion have used a damping force that is proportional in magnitude to the velocity, even if it is not appropriate for a particular problem. For example the equation is generally derived with a mass on a spring situation with friction between the mass and the floor, however this friction should be constant and independent of the velocity.

I tried to find a solution myself to the constant friction problem (although I had to restrict myself to considering only half a cycle because otherwise the force would be in the wrong direction. I am not too familiar with solving differential equations (although this is quite a simple one!) And the solution I got to $m\ddot x +kx +F=0$

Was

$x=Acos (\omega t +\phi ) -\frac {F}{k}$

Which is clearly wrong as then the amplitude isn't decaying.

But I guess my main question is: is damped harmonic motion only for resistive forces proportional to the velocity?

• Why would you include $k\dot{x}$ if you want a constant velocity problem? Do you mean a constant friction problem? And is F supposed to be a linear restoring force? – Bill N Dec 23 '16 at 20:35
• @BillN Thanks for pointing out the typo. Yes, I meant to write friction there. – Meep Dec 23 '16 at 21:04
• And $kx$, not $k\dot{x}$ ?? – Bill N Dec 23 '16 at 21:11
• your term $F$ represents the resultant of the applied forces to the object's centre of mass. The friction force that is of our interest does depend on the velocity, because when the solid is at rest the solid is in mechanical equilibrium and $F=0$ . – HBR Dec 24 '16 at 1:29
• @BillN yes, apologies. – Meep Dec 25 '16 at 11:09

Let us rename your parameters in order to write the equation in a more usual form: $$m\ddot{x}+c\dot{x}+F(x)=0 \qquad m>0,c\geq 0$$ For a suitable restoring force $F(x)$ to force the system to exert a harmonic motion.

For the sake of simpicity let the system exert small oscillations and therefore the function $F(x)$ can be expanded close to its minimum $F(x)\approx kx$ with $k> 0$. Let us also divide your equation by the mass $m$, defining two new quantities: $$\gamma=\frac{c}{m}\qquad \omega^2 = \frac{k}{m}$$ Hence $$\ddot{x}+\gamma\dot{x}+\omega^2x=0\tag1$$

Multiplying $(1)$ by $\dot{x}$ we have $$\frac{1}{2}\frac{d}{dt}\left(\dot{x}^2+\omega^2x^2\right)=-\gamma\dot{x}\dot{x}\tag2$$ We can say that if $\gamma$ is zero the energy of the system $$E=\frac{1}{2}\left(\dot{x}^2+\omega^2x^2\right)$$ is conserved. Note that the RHS of $(2)$ is strictly negative throughout the motion.

It is clear that $(2)$ is the generalisation of the equation $(1)$ you proposed. Answering your question, any function that is strictly positive once multiplied by $\dot{x}$, throughout the motion may serve for this purpose.

Therefore if the energy $E$ must decrease throughout the motion it is mandatory that the friction coefficient be an even function in $\dot{x}$ $$\gamma = \gamma_0+\gamma_2\dot{x}^2+\gamma_4\dot{x}^4+...$$

Hope this helps

P.S. The solution you gave for constant $F$ is wrong, if $F=constant$, changing variables to $x=y-F/kt$ results that y must verify $$\frac{d}{dt}\left(m\dot{y}+ky\right) = 0$$ Solving for $y$ we have $$m\dot{y}+ky=constant \rightarrow y=c_0 + c_1\exp{(-k/mt)}$$ Being $$x = c_0 + c_1\exp{(-k/mt)}-F/kt$$