= Solution
{c}
Use <separation of variables> to seek a <normal mode> $y(x,t)=X(x)\cos(\omega t+\delta)$, with <angular frequency> $\omega>0$ and <wavenumber> $k=\omega/c$. The endpoint <boundary conditions> are built into
$$
X(x)=\begin{cases}
A\sin(k(x+\ell)),&-\ell\leq x\leq0,\\
B\sin(k(\ell-x)),&0\leq x\leq\ell.
\end{cases}
$$
Let $z=k\ell$ and denote the mass's displacement amplitude by $X_0$. Continuity and the mass equation become
$$
A\sin z=B\sin z=X_0,\qquad
Tk(A+B)\cos z=m\omega^2X_0.
$$
The two possibilities must be separated before dividing by $\sin z$.
\b[Antisymmetric modes.] If $\sin z=0$, then $z=n\pi$, $n=1,2,\ldots$, and $X_0=0$. The force equation gives $A+B=0$, since $\cos z\ne0$. Thus $X(-x)=-X(x)$ and the mass stays at rest. These allowed <angular frequencies> satisfy
$$
\boxed{\sin(k\ell)=0,\qquad k\ell=n\pi\ (n\geq1),\qquad\omega_n^{\mathrm{odd}}=\frac{cn\pi}{\ell}.}
$$
The restoring forces on the mass cancel even though the string vibrates.
\b[Symmetric modes.] If $\sin z\ne0$, continuity forces $A=B$. Cancelling the nonzero amplitude in the mass equation and using $\omega^2=(T/\rho)k^2$ gives
$$
2Tk\cos z=m\frac T\rho k^2\sin z,
\qquad
\boxed{\cot(k\ell)=\frac{mk}{2\rho}.}
$$
These modes have $X(-x)=X(x)$ and move the mass. In dimensionless form their equation is $\cot z=\alpha z$, where $\alpha=m/(2\rho\ell)>0$. This is the spectral splitting for a <string with a central point mass>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-oxford/2023/prelims/mathematics-v-string-frequencies.png]
{title=Antisymmetric zeros and symmetric frequency intersections}
For the antisymmetric family, the graph of $\sin z$ crosses zero at every positive multiple of $\pi$. For the symmetric family, on each interval $(n\pi,n\pi+\pi/2)$, $n=0,1,\ldots$, the graph of $\cot z$ decreases continuously from $+\infty$ to zero while $\alpha z$ increases. By the <intermediate value theorem> and strict <monotonicity>, there is exactly one intersection in each interval. On the remaining half-intervals $\cot z<0<\alpha z$, so there are no further positive roots. This proves graphically and analytically that both families contain infinitely many <normal modes>. The figure uses $\alpha=1$ for illustration.
The algebraic case $k=0$ is not a mode: a zero-frequency profile would be linear on each half, continuous and zero at the endpoints; the zero-acceleration mass equation also makes its slopes agree, forcing the entire profile to be zero. More generally, <integration by parts> on the two halves and the matching condition give
$$
\omega^2\left(\rho\int_{-\ell}^{\ell}X^2\,dx+mX_0^2\right)
=T\int_{-\ell}^{\ell}(X')^2\,dx>0
$$
for a nonzero real mode. Thus no nonzero static or exponentially growing mode has been omitted.
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