 Solution

At common tension, the lighter segment has three times the wave speed, so its wavenumber at the same frequency is $k_2=k_1/3$. The fixed left end and free right end selectLet $q=k_{1}L$. Since $k_2(4L-L)=q$, continuity of displacement and of transverse force at the junction givesEqual tension makes the second condition continuity of slope. A nontrivial pair of amplitudes requires the coefficient determinant to vanish:For positive frequency, $k_2/k_1=1/3$, soNeither $\sin q=0$ nor $\cos q=0$ gives a nonzero mode, as direct substitution into the two matching equations confirms. The positive roots are $q=\pi /3+n\pi$ or $q=2\pi /3+n\pi$ for $n\ge 0$. The first modes in the figure show the changing wavelength across the junction while retaining continuous displacement and slope. The zero-frequency limit must be solved separately; the fixed and free boundary conditions allow only the trivial static solution.