Solution

The speeds of propagation of the waveform to the left and the right are $c_1=\sqrt{F_\parallel/\rho_1}$ and $c_2=\sqrt{F_\parallel/\rho_2}$, respectively. Consequently, the mass of the portion of the rope that has transverse momentum increases linearly with time, and its transverse speed $u$ must be constant, because the transverse force is constant. After time $t$, transverse momentum of the waveform is

$$P_\perp=\left(\rho_1\sqrt{(c_1t)^2+(ut)^2}+\rho_2\sqrt{(c_2t)^2+(ut)^2}\right)u.$$

As a result,

$$F_\perp=P_\perp/t=\rho_1c_1^2\cdot\frac{u}{c_1}\sqrt{1+\left(\frac{u}{c_1}\right)^2}+\rho_2c_2^2\cdot\frac{u}{c_2}\sqrt{1+\left(\frac{u}{c_2}\right)^2}.$$

Under the assumption that $F_\perp\ll F_\parallel=\rho_1c_1^2=\rho_2c_2^2$, it follows that $u\ll c_1,c_2$, and we have the approximation

$$F_\perp\approx F_\parallel\left(\frac{u}{c_1}+\frac{u}{c_2}\right).$$

Therefore,

$$u\approx\frac{c_1c_2}{c_1+c_2}\cdot\frac{F_\perp}{F_\parallel}.$$