= 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 select
$$
Y_1(x)=A\sin k_1x,\qquad
Y_2(x)=B\cos[k_2(4L-x)].
$$
Let $q=k_1L$. Since $k_2(4L-L)=q$, continuity of displacement and of transverse force at the junction gives
$$
A\sin q=B\cos q,\qquad
Ak_1\cos q=Bk_2\sin q.
$$
Equal tension makes the second condition continuity of slope. A nontrivial pair of amplitudes requires the coefficient <determinant> to vanish:
$$
k_1\cos^2q-k_2\sin^2q=0.
$$
For positive frequency, $k_2/k_1=1/3$, so
$$
\boxed{\sin^2(k_1L)=\frac34,\qquad\sin(k_1L)=\pm\frac{\sqrt3}{2}.}
$$
Neither $\sin q=0$ 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 $n\ge0$. 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.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2017-resit-composite-string-modes.png]
{title=The first three modes of a fixed-free string with a density junction}
{description=Original eigenfunctions for the two segment densities. Displacement and slope are continuous at x equal to L; the lighter segment has three times the local wavelength.}
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