Solution

If $v=\sqrt{F/\rho}$ (or $F=\rho v^2$), then the wave moves at the natural speed of propagation in the string, and there will be no force exerted by the rings on the string. If $v\ne\sqrt{F/\rho}$, then additional tension $f=\rho v^2-F$ must be created in the string in order to allow wave propagation at the speed $v$. These additional tension (shown in green) are reactions to the forces that the rings exert on the string. At Point 1 and 3, we have $F_1=2f\sin(\alpha/2)$, while at Point 2 we have $F_2=2f\sin\alpha$. For a small $\alpha$, we have

$$F_1\approx(\rho v^2-F)\cdot\frac{b}{L}\qquad\mbox{and}\qquad F_2\approx2(\rho v^2-F)\cdot\frac{b}{L},$$

where $\vec F_1$ points almost downward and $\vec F_2$ points upward when $\rho v^2>F$ and vice versa.