 Solution

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 intoLet $z=k\ell$ and denote the mass's displacement amplitude by $X_0$. Continuity and the mass equation becomeThe two possibilities must be separated before dividing by $\sin z$.

Antisymmetric modes. If $\sin z=0$, then $z=n\pi$, $n=1,2,\dots$, and $X_0=0$. The force equation gives $A+B=0$, since $\cos z\ne 0$. Thus $X(-x)=-X(x)$ and the mass stays at rest. These allowed angular frequencies satisfyThe restoring forces on the mass cancel even though the string vibrates.

Symmetric modes. If $\sin z\ne 0$, continuity forces $A=B$. Cancelling the nonzero amplitude in the mass equation and using ${\omega }^2=(T/\rho )k^2$ givesThese 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.

For the antisymmetric family, the graph of $\sin z$ crosses zero at every positive multiple of $\pi$. For the symmetric family, on each interval $(n\pi ,n\pi +\pi /2)$, $n=0,1,\dots$, 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 givefor a nonzero real mode. Thus no nonzero static or exponentially growing mode has been omitted.