# Non-conservative force, but equation of harmonic motion

A small disc is projected on a horizontal floor with speed $$u$$. Coefficient of friction between the disc and floor varies as $$\mu = \mu_0+ kx$$, where $$x$$ is the distance covered. Find distance slid by the disc on the floor.

I have correctly solved this by using two methods :

1. By integrating the kinematic equation $$\displaystyle a=v\frac{dv}{dx}=-(\mu_0+kx)g$$ and
2. by considering the work done by frictional force ($$\int_0^x (\mu_0+kx)mgdx$$) to dissipate the initial kinetic energy of the disc.

Now I notice that the equation of motion, $$a=-(\mu_0 +kx)g$$, is kinda SHM-like, but with a non-conservative force instead. I conjecture that as soon as the disc reaches $$v=0$$, the force of friction, and hence the acceleration, drop to 0. Now, I think that this is pretty similar (except the $$a=0$$ at $$v=0$$ part; in SHM, from my knowledge, $$a$$ and $$v$$ are separated by a phase angle of $$\frac π2$$) to the first $$\frac 14$$ of an SHM cycle, i.e. starting from the mean position till it reaches the amplitude.

Now my calculations for the third method: The maximum speed obtained is at mean position: $$v_{max}=u=A\omega$$ where $$\omega$$ is the angular frequency of oscillation. Also, $$\omega ^2=kg$$ and so maximum distance travelled is equal to the amplitude, i.e, when velocity becomes 0. So $$A=\frac{u}{\sqrt{kg}}.$$ However, this is not the correct answer.

What’s my error in concept?

EDIT: I understood where I got confused, thanks to user @J.Murray. But I’d like to see this question solved (or get the easily solvable equation $$kgx^2+2\mu_0gx-u^2=0$$) with concepts borrowed from SHM, and not directly the two methods I have listed above. No big deal if it is complicated.

• You're right that the equation of motion is kinda SHM-like, but your analysis seems to proceed as though it is exactly SHM. What happened to the $\mu_0$ term? Jun 25, 2022 at 17:58
• Uh… @J.Murray why would that affect any calculations? It doesn't change $\omega$ Jun 25, 2022 at 18:01
• Ah….I see… At the expected mean position the acceleration is not 0 (as it should be in SHM). Or is it? Because we start from there. So at x=0, is a=0? @J.Murray can you please see if the question be solved via some SHM-concepts? Jun 25, 2022 at 18:03
• When $x=0$, we have $a = -(\mu_0 + k\cdot 0 )g = -\mu_0 g\neq 0$. The equilibrium point will be at $x= -k/\mu_0$. You could use some concepts from SHM, but it would not be quite as simple as just doing the integral. Jun 25, 2022 at 20:03
• @J.Murray I may come off as annoying, but pleaseeee confirm that even when the disc has started from a point (say A) with nonzero initial speed, there may be acceleration at A? I just feel silly asking this, but please confirm this for me. Jun 25, 2022 at 20:15

Consider a mass dangling from a vertical spring. The equation of motion is

$$ma = mg -k(x-x_0)$$

where $$x_0$$ is the position of the end of the unstretched spring, and I’m using a downward-positive coordinate system. This system has an equilibrium at

$$x-x_0= \frac{mg}k$$

The potential energy for a mass on a vertical spring is

$$U= -mgx +\frac12 k(x-x_0)^2 =\frac12 k \left( x-\left(x_0+\frac{mg}k\right) \right)^2 + \text{constant}$$

That is, the potential energy is the sum of a linear gravitational term and a quadratic term from the spring. However, you may recall from algebra that a quadratic function plus a linear function is just a quadratic function with the same curvature but a different minimum. If you enjoy algebra and completing the square, you can find the second equality.

So oscillations of a dangling spring about its effective equilibrium are described by simple harmonic motion with the same frequency $$\omega^2=k/m$$ as the free spring.

$$ma=-m(\mu+kx)g$$
has exactly the constant-plus-linear form of a dangling spring, so you can find the stopping position by treating the stopping process as a partial oscillation, from $$x=0$$ (with nonzero initial velocity) to the oscillation’s turning point.
$$x'' + \beta ^2 x + C$$ where $$\beta^2 = gk$$ and $$C = \mu_0 g$$
The equation is for a SHM plus a constant. This means the solution will be the sum of the SHM solution and the particular solution for this equation. When the math is done, the solution takes the general form: $$x = Asin(\beta t) + Bcos(\beta t) - \frac C {\beta^2}$$ and $$x' = A\beta cos(\beta t) - B\beta sin(\beta t)$$