Plucked string eigenvalues/harmonic frequencies: integer multiples (or not) I'm trying to derive a model of a plucked string from Newton's second law. My derivation results in $$ω_n = C\cdot\sqrt{n},\, n=1,2,3\dots\text{integer}$$  I think it should be $$ω_n = C\cdot n,\, n=1,2,3\dots\text{integer}$$
I started with beads each having mass $m$ evenly spaced on a string with string tension $T$. Bead $n$ is displaced upward by a distance $y$. The forces on bead $n$ seem to me to be:
$$F_n = m_n  \ddot y_n =-T \sin⁡(\theta_{n-1,n}) -T \sin⁡(\theta_{n,n+1}) $$
$$\sin⁡(\theta_{n-1,n}) \approx (y_n - y_{n-1})/d$$
$$\sin⁡(\theta_{n,n+1}) \approx (y_n - y_{n+1})/d/d$$
Making substitutions and rearranging: $$\ddot y_n+\frac{T}{dm_n}(-y_{n-1} + 2y_n -y_{n+1})=0$$
Now let $$y_n = A_n \cdot e^{i\omega t}, \implies \dot y_n = i\omega \cdot y_n,    \ddot y_n = -w^2 \cdot y_n$$
Also define $$C ≡ \sqrt{\frac{T}{d\cdot m}}
$$ Now
$$\omega^2 + C^2\cdot(-y_{n-1} + 2y_n -y_{n+1})=0$$ Expanding in matrix form: $$\begin{bmatrix} A\end{bmatrix}  \vec y_n = \frac{\omega^2}{C^2} \vec y_n$$
Where $$|A| \equiv 
 \begin{bmatrix} 
 2 & -1 & 0 & 0 & 0 & 0\\ 
 -1 & 2 & -1 & 0 & 0 & 0\\
 0 & -1 & 2 & -1 & 0 & 0\\
 0 & 0 & -1 & 2 & -1 & 0\\
 0 & 0 & 0 & -1 & 2 & -1\\
 0 & 0 & 0 & 0 & -1 & 2 &\\
 \end{bmatrix}$$ An on-line calculator shows the eigenvalues of $|A|$ to be: $1, 2, 3, 4\dots$
So my result would seem to be $$ω_n = C \cdot\sqrt{n},\, n=1,2,3\dots\text{integer}$$ What's wrong?
 A: Why would you think $\omega_n=n{\omega_0}$ or $\sqrt{n}\omega_0$?
The "quantum number" $n$ here is momentum $k$ since this is obviously a translationally invariant system. With some Fourier transformations, you can get the energy $\omega_n=2\omega_0\sin(k)$, where $k=\pi/(2N)\cdot i$ and $i=-N,\cdots,N$, $N$ for the total number of lattice site. Small deviations in $k$ if the boundary is open.
This dispersion agrees with the eigenvalues of your tridiagonal matrix.
A: The answers posted by previous experts leave the fundamental two question unanswered:

*

*Why don't the eigenvalues of my uniform, tri-diagonal, symmetric (Toeplitz) harmonic, corresponding to the known characteristics of a real guitar string?

*Why don't the eigenvalues for progressively larger matrices converge to the continuum solution of Lagrangian analysis?

Help can be found in "Almost-dispersionless pulse transport in long quasi uniform spring-mass chains ..." R. Via, August 2, 2018. See equations (38)- (41).
The eigenvalues of matrix $(A)$ defined in my question (as noted in the previous answers and confirmed in the referenced paper) are:
$$\lambda_n = 2-2\cdot \cos(k_n)$$
where:
$$k_n ≝ \left(\frac{ n\,\pi}{N+1}\right)$$
So:
$$\frac{\omega_n^2}{C^2}= 2-2 \cdot \cos \left(k_n\right)$$
$$\omega_n = C\, \sqrt { 2-2\cdot \cos( k_n ) } = 2\, C \cdot \sin { \left( \frac{k_n}{2} \right) }  = 2 \, C \cdot \sin { \left(\frac{ n\cdot \pi}{2\cdot (N+1)}\right) } $$
My first remaining question is answered by a key sentence in the reference paper is:

For $k\ll 1$ (i.e. $n\ll N$) they [the $\omega_n$] are almost linear in $k$ and equally spaced in $n$.

For $n\ll N$
$$\sin { \left(\frac{ n\cdot \pi}{2\cdot (N+1)}\right) } \approx { \left(\frac{ n\cdot \pi}{2\cdot (N+1)}\right) }$$
So in this particular case:
$$\omega_n \approx {\left(\frac{C\cdot \pi}{N+1}\right) n}$$
But as noted in my question:
$$C ≝  \sqrt{\frac{T}{d\cdot m}}$$
Holding the length of the string as a constant $L$ and the total mass of the string as constant $M$:
$$ m = \frac{M}{N},\, d = \frac{M}{N+1}$$
For very large $N$, i.e. as the chain of masses becomes a continuous string:
$$C \approx \sqrt{\frac{T}{L\cdot M}}\cdot N$$
We can now define:
$$\omega_0 ≝\sqrt{\frac{T}{L\cdot M}}\cdot \frac{N}{N}\cdot \pi = \pi\,\sqrt{\frac{T}{L\cdot M}}$$
Finally:
$$\omega_n \approx \omega_0 \cdot n,\, n=1,2,3\dots\text{integer}$$
which is the classical answer and resolves my last remaining question.
