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?
I think using energy might be easier. I will assume that in part b. the block is released from rest. Let's say it 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:
$\frac12 kA^2 = \frac12 kB^2 + FA + FB $
$0 = \frac12 kB^2 + FA + FB -\frac12 kA^2$
Then apply the quadratic formula to get $B$ in terms of $A$. One root is
$B = A -2F/k$
The other root is $B= -A$. I'm not sure of the physical significance of this second root.