I didn't think so, but I'm looking at a course problem,
A string of length $L$ and line density $\rho$ is under a tension $\rho c^2$. Its transverse displacement is $y(x,t)$. The string is fastened to rings at $x=0$ and $x=L$. The masses of the rings can be neglected, and they slide on two smooth fixed wires at $x=0$ and $x=L$, one ring being on each wire; the wires are parallel to $0y$. By considering the equations of motion of the rings, show that $$\frac{\partial y}{\partial x}=0$$ at $x=0$ and $x=L$ for all $t$. By seeking separable solutions, find the normal modes of vibration.
The given solution begins,
Consider either ring. Since each has zero mass, \begin{align} & \text{Newton’s Second Law} \\ \implies & x\ddot{y} = \text{vertical component of force} = \pm F\sin{\psi} \\ \implies & \sin{\psi}=0 \text{ at } x=0,L \end{align} but \begin{align} &\psi\ll1 \\ \implies&\sin(\psi)\approx\tan(\psi)=\frac{\partial y}{\partial x} \\ \implies & \frac{\partial y}{\partial x}=0 \text{ at } x=0,L. \end{align}
Doesn't this reasoning imply that any object with negligible mass is immovable? Couldn't one equally validly reason that $$\ddot{y}=\frac{F\sin{\psi}}{0}=\infty$$?
I also don't really understand why the $x$ in $x\ddot{y}$ seems to be mass.
ETA: My interest is in finding the problem in the argument rather than answering the title question experimentally.
