Take two bodies of masses m and M attached by a spring of constant K on a smooth horizontal surface. The system is at rest. A constant force F acts on body M, horizontally. To study the motion of the bodies I wrote the net forces acting on the 2 bodies.
My goal was to find minimum and maximum seperation of the bodies. For that, i took the condition that at minimum or maximum seperation, the relative velocity of one body with respect to the other is zero
I took the relative velocities between the bodies to be zero, when they are at maximum and minimum separation. My issue is that::
-I got minimum seperation to be 0. This makes no sense. My only idea is that the condition that both bodies have same velocity at minimum seperation is wrong. Why so? How to calculate minimum seperation between the bodies?
Best Answer
This is my solution. Place $m$ to the left described by the coordinate $x_1$ and $M$ to the right, described by the coordinate $x_2$, both coordinates increasing to the right. Take $\vec{F}$ to be acting upon $M$ to the right and $\eta_0$ to be the natural separation of the spring. Finally, take $x$ to be the center of mass of the system and $\eta$ be the separation between both masses, such that the spring force clearly has magnitude $\|\vec{F}_s\|=k|\eta-l|$.
First, we analyze the motion of the center of mass. Since the constant force $\vec{F}$ is the only external force, we have $$(m+M)\ddot{x}=F\implies \ddot{x}=\frac{F}{m+M}$$
Then, lets write the coordinates $x_1$ and $x_2$ in terms of $x$ and $\eta$. From our definitions we have $$(m+M)x=mx_1+Mx_2$$ $$\eta=x_2-x_1$$ Let's focus our attention on $x_1$ (if you do the same procedure with $x_2$ you will get the same answer). This system of equations can be inverted to yield $$(m+M)x_1=(m+M)x-M\eta$$ Differentiating twice we obtain $$(m+M)\ddot{x}_1=(m+M)\ddot{x}-M\ddot{\eta}=F-M\ddot{\eta}$$ Now, focusing our attention on mass $m$, we have newton's second law $$m\ddot{x}_1=\frac{m}{m+M}(F-M\ddot{\eta})=k(\eta-l)$$ Solving for $\eta$ one arrives to $$\ddot{\eta}+\frac{m+M}{mM}k\eta=\frac{F}{M}+\frac{m+M}{mM}kl$$ By defining $\omega=\sqrt{k\frac{m+M}{mM}}$, the solution to this differential equation is $$\eta(t)=A\cos(\omega t+\phi)+\frac{F}{M\omega^2}+l$$It is easy to see that $$\dot{\eta}(t)=-A\omega\sin(\omega t + \phi)$$ Since the system starts at rest, $$\dot{\eta}(0)=0=-A\omega\sin(\phi)\implies\phi=k\pi,\quad k\in \mathbb{Z}$$ and $$\eta(0)=l=A\cos(k\pi)+\frac{F}{M\omega^2}+l\implies A=\pm\frac{F}{m\omega^2}$$ Therefore, the maximum and minimum are $$\eta_{\text{max}}=\frac{2|F|}{M\omega^2}+l$$ $$\eta_{\text{min}}=l$$ I hope this is useful!