Consider a mass $m$ at position $x(t)$ on a rough horizontal table attached to the origin by a spring with constant $k$ (restoring force $-kx$) and with a dry friction force $f$ $$\begin{cases} f=F, & \text{if $\dot x \lt 0$}\\ -F \le f \le F, & \text{if $\dot x =0$}\\ f=-F,& \text{if $\dot x \gt 0$} \end{cases} $$ a). What is the range of $x$ where the mass can rest?

b) Show that if the mass moves, the maximum excursion decreases by $\frac{2F}{k}$ per half cycle.

c) Discuss the motion

I nearly completed the question, but feel hard to answer part b due to no initial conditions given.

For part b, I started by noting $$m\frac{\mathrm{d}^2x}{\mathrm{d}t^2}=f-kx$$ Letting $y = \frac{f}{k}-x$, I get $$m{d^2y \over dt^2}=-ky$$ thus $y = A\cos (\omega t+\phi)$ where $A$ and $\phi$ depend on initial conditions and $\omega= \sqrt \frac{k}{m}$, hence I deduce $$x= \frac{f}{k}+A\cos(\omega t+\phi)$$

If I assume $x(0) \gt 0$, then $$x= \frac{F}{k}+A\cos(\omega t+\phi)$$ and hence $$x_{\text{max}_1}= \frac{F}{k}+A$$

In the other half cycle, $\dot x \gt 0$ so $$x= -\frac{F}{k}+A\cos(\omega t+\phi)$$ and $$x_{\text{max}_2}= {-F \over k}+A\cos(\omega t+\phi)$$ the difference between the previous $x_{\text{max}_1}$ and this $x_{\text{max}_2}$ is $\frac{2F}{k}$. $\square$

From the previous calculation I feel I have roughly completed the question, but I failed to demonstrate that between $x_{\text{max}_1}$ and $x_{\text{max}_2}$ the mass have moved a half cycle, although intuitively I feel that since it initially move backward($\dot x \lt 0$), when it starts to move forward ($\dot x \gt 0$) and achieve maximum displacement again, it should have completed half cycle.

Could someone help me to make the situation clear please?

Edit: On reading the hint given below, I tried to find the solution, here is my effort

First case: when $\dot x \gt 0$, $m \ddot x = -kx-F$. On substituting $y=x+{F\over k}$, I obtained $x_{(+)}=A\cos(\omega t+\theta _1)-{F\over k}$, where $A \ \text{and} \ \theta_1$ depends on initial conditions.

Second case: when $\dot x \lt 0$, $m \ddot x=-kx+F$. Using a similar method I obtained $x_{(-)}=B\cos(\omega t + \theta_2)+{F\over k}$.

However I find it hard to link $x_{(+)} \ \text{with} \ x_{(-)}$. How should I control the four arbitrary constants $A,B,\theta_1 \ \text{and} \ \theta_2$?

  • $\begingroup$ You've gotten so bound up in the mathematics of the problem that you're not thinking about the system. What kind of behavior does it exhibit if $F=0$ (that is we remove the friction from the system)? How---other than just slapping together a differential equation---can you analyze such systems? In what ways does the reduction of maximum excursion related to the frictional force, assuming the kind of system we have when there is no friction (I know, but start by assuming the friction is small)? And so on. $\endgroup$ Commented Jul 29, 2015 at 1:23
  • $\begingroup$ Hints: 1. Show that a constant branch $x(t)={\rm const}$ must have $|x|\leq \frac{F}{k}$. 2. Find the most general non-constant branches $x_+(t)$ and $x_-(t)$, when the mass moves forward $\dot{x}>0$ and backward $\dot{x}<0$, respectively. 3. Conclude that each non-constant branch is a shifted harmonic motion. 4. Now connect the three types of branches so that (among other things) the motion is continuous at the turning points $\dot{x}=0$. 5. Compare the positions of successive turning points. $\endgroup$
    – Qmechanic
    Commented Aug 1, 2015 at 14:50
  • $\begingroup$ @Qmechanic But how to decide the initial conditions? I have typed my efforts in my question(see edit). How can I progress from getting two branches? $\endgroup$
    – Rescy_
    Commented Aug 3, 2015 at 3:10
  • $\begingroup$ For the non-constant case, you may assume without loss of generality by shifting the $t$-axis that $x(t=0)=x_{\rm max}>\frac{F}{k}$ is at a turning point. $\endgroup$
    – Qmechanic
    Commented Aug 3, 2015 at 9:39

2 Answers 2


Let us assume the initial conditions are $x(0)=A$ and $\dot x(0)=0$. The equation of motion as long as $\dot x(t)<0$ is given by $$m\ddot x=-kx+F$$

What happens at $t=0$? We should distinguish $F < kA$ and $F \geq kA$. In the case $F > kA$, suppose that the mass starts to move with $\dot x(0)<0$, the friction will immediately start pulling the mass to higher $x$. As soon as $\dot x(0)>0$, the friction force will start pulling again to smaller $x$ and so on. The net effect of this continuous change of sense of the friction force will be that the mass will be at rest, given that the spring repelling force is dominated by the friction force. This reasoning is also true for $F=kA$. Conclusion, the mass is at rest for $F \geq kA$.

What about the case where $F < kA$? in this case the friction is not dominating the repelling force of the spring. We should solve the equation piecewise, considering every time a time interval where $\dot x(0)$ and hence $f$ do not change sign.

The solution for the first phase is $$ x(t) = (A-\frac{F}{k})cos(\omega t) + F/k $$ with $\omega^2 = k/m$. This first phase comes to an end when $\dot x (t)=0$. This happens at $\omega t=\pi$. At that moment $x=-A+2F/k$. The equation of motion then becomes $$m\ddot x=-kx-F, (\pi < \omega t < 2\pi ) $$ which, taking into account the initial conditions, has as solution $$ x(t) = (A-\frac{3F}{k})cos(\omega t) - F/k $$. At $\omega t = 2\pi$, we get $$x=A-\frac{4F}{k}, \dot x = 0$$. We are now back in the same situation as in the very beginning, except that $A$ has been replaced by $$A-\frac{4F}{k}$$. So, what we see is that the amplitude diminishes with $2F/k$ over a period of time $\pi/\omega$. Phase $n$ will extend from $\omega t=(n-1)\pi$ to $\omega t=n\pi$ and the equation of motion will be $$ x(t) = (A-\frac{(2n-1)F}{k})cos(\omega t) - (-1)^n\frac {F}{k} $$

This process of dissipating energy and decreasing amplitude goes on until at a certain point the friction force starts dominating the repelling force of the spring, which means that the body will come to rest (cf. beginning of this answer). This will happen at the end of phase $n$ where $n$ is given by $$F>k(A-\frac{2nF}{k})$$ of or the lowest $n$ for which $$n>\frac{1}{2}(\frac{Ak}{F}-1)$$ General conclusion: as long as the friction is not dominating the repelling force of the spring, i.e. as long as the amplitude is high enough, i.e. $>F/k$, the system will dissipate such that every $\Delta t=\pi/\omega$, the amplitude decreases with $2F/k$. At a certain point, the amplitude has been reduced up to the point where the friction starts to dominate and then the body comes to rest.

The picture below shows the evolution of the position of the mass as a function of time for the case $A=13F/(2k)$.

enter image description here


Here is a method using energy.

Assume the mass starts a distance $A$ from its equilibrium point and it moves past the equilibrium point a distance $B$ before turning around.

The initial spring energy is equal to the final spring energy plus the energy lost due to the work by friction:

\begin{align} \frac12 kA^2 &= \frac12 kB^2 + FA + FB \\ 0 &= \frac12 kB^2 + FA + FB -\frac12 kA^2 \, . \end{align}

Then apply the quadratic formula to get $B$ in terms of $A$. One root is

$$B = A -2F/k \, .$$

The other root is $B = -A$, which simply means the mass does not move at all.


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