Plucked string eigenvalues/harmonic frequencies: integer multiples (or not)

Question

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?

Answer 1

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.

Answer 2

The answers posted by previous experts leave the fundamental two question unanswered:

1. Why don't the eigenvalues of my uniform, tri-diagonal, symmetric (Toeplitz) harmonic, corresponding to the known characteristics of a real guitar string?
2. 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.

Answer 3

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The results of OP are wrong. More precisely if \begin{equation} \mathrm\Xi\e \begin{bmatrix} \hp\m 2&\m1&\hp\m 0&\cdots&\hp\m 0&\hp\m 0\hp\m\\ \m1 &\hp\m 2&\m1&\cdots&\hp\m 0&\hp\m 0\hp\m\\ \hp\m 0&\m1&\hp\m 2&\cdots&\hp\m 0 &\hp\m 0\hp\m\\ \hp\m\vdots &\hp\m\vdots &\hp\m\vdots&\cdots &\hp\m \vdots&\hp\m\vdots\hp\m\\ \hp\m 0&\hp\m 0&\hp\m 0&\cdots&\hp\m 2&\m1\hp\m\\ \hp\m 0&\hp\m 0&\hp\m 0&\cdots&\m1&\hp\m 2\hp\m \end{bmatrix} \tag{01}\label{01} \end{equation} is a $\:n\times n\:$ symmetric Toeplitz matrix, its eigenvalues are not those of the OP \begin{equation} \lambda_\rho\e\rho\,, \qquad \rho\e1,2,3,\cdots n \qquad\qquad \texttt{(wrong)} \tag{02}\label{02} \end{equation} but \begin{equation} \xi_\rho\e 4\sin^{2}\blr{\rho\dfrac{\pi}{2\plr{n\p1}}}\e 2\Bigg(1\m\cos\left[ \rho\dfrac{\pi}{\plr{n\p1}} \right]\Bigg), \quad \rho\e 1,2,\cdots, n \tag{03}\label{03} \end{equation}

$\bl{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!}$

Counterexample 1 : Consider the $\:2\times 2\:$ symmetric Toeplitz matrix \begin{equation} \mathrm\Xi\e \begin{bmatrix} \hp\m 2&\m1\hp\m\vp\\ \m1 &\hp\m 2\hp\m\vp \end{bmatrix} \tag{04}\label{04} \end{equation} According to the OP results its eigenvalues are \begin{equation} \lambda_1\e 1\,, \qquad \lambda_2\e 2 \qquad\qquad \texttt{(wrong)} \tag{05}\label{05} \end{equation} instead of \begin{equation} \begin{split} \xi_1 & \e 4\sin^{2}\plr{\dfrac{\pi}{6}}\e 1 \\ \xi_2 & \e 4\sin^{2}\plr{\dfrac{\pi}{3}}\e 3 \end{split} \tag{06}\label{06} \end{equation} in agreement with equation \eqref{03}.

$\bl{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!}$

Counterexample 2 : Consider the $\:3\times 3\:$ symmetric Toeplitz matrix \begin{equation} \mathrm\Xi\e \begin{bmatrix} \hp\m 2&\m1&\hp\m0\hp\m\vp\\ \m 1 &\hp\m 2&\m1\hp\m\vp\\ \hp\m 0 &\m 1&\hp\m 2\hp\m\vp \end{bmatrix} \tag{07}\label{07} \end{equation} According to the OP results its eigenvalues are \begin{equation} \lambda_1\e 1\,, \qquad \lambda_2\e 2\,, \qquad \lambda_3\e 3 \qquad\qquad \texttt{(wrong)} \tag{08}\label{08} \end{equation} instead of \begin{equation} \begin{split} \xi_1 & \e 4\sin^{2}\plr{\,\dfrac{\pi}{8}\,}\e 2\Bigg[1\m\cos\left(\,\dfrac{\pi}{4}\,\right)\Bigg]\e 2\m\sqrt{2} \\ \xi_2 & \e 4\sin^{2}\plr{\,\dfrac{\pi}{4}\,}\e2\Bigg[1\m\cos\left(\,\dfrac{\pi}{2}\,\right)\Bigg]\e 2\\ \xi_3 & \e 4\sin^{2}\plr{\!\dfrac{3\pi}{8}\!}\e 2\Bigg[1\m\cos\left(\!\dfrac{3\pi}{4}\! \right)\!\Bigg]\e 2\p\sqrt{2} \\ \end{split} \tag{09}\label{09} \end{equation} in agreement with equation \eqref{03}.

Note that \begin{equation} \lambda_1\lambda_2\lambda_3\e 1\cdot 2 \cdot 3 \e 6 \bl\ne 4\e \det\plr{\mathrm\Xi} \tag{10}\label{10} \end{equation} while \begin{equation} \xi_1\xi_2\xi_3\e \plr{2\m\sqrt{2}}\cdot 2\cdot\plr{2\p\sqrt{2}} \e 4\e \det\plr{\mathrm\Xi} \tag{11}\label{11} \end{equation}

An other check is \begin{equation} \sum\limits_{\rho\e 1}^{\rho\e n}\lambda_\rho\e\sum\limits_{\rho\e 1}^{\rho\e n}\rho\e\dfrac{n\plr{n\p1}}{2}\bl\ne 2n\e \texttt{Trace}\plr{\mathrm\Xi} \tag{12}\label{12} \end{equation}