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In Boas' Mathematical Methods there is a section on linear algebra in which it is stated that we can write the eigenvalue equation for a set of springs using the kinetic energy and the potential energy where $$ V = \frac{1}{2}kr^{T}Vr $$ and $$T = \frac{1}{2}m\dot r^{T}V\dot r $$

Then it's stated that we can write the equations of motion as $$\lambda T r = Vr$$ where $$\lambda = \frac{mw^{2}}{k} $$

My question is, what is the logic of setting up an eigenvalue equation in which we set the potential energy equal to the kinetic energy times an eigenvalue factor?

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enter image description here

On the top of Figure we have $\:n+1\:$ ideal springs and $\:n\:$ particles in equilibrium. The constants of the springs are $\:k_{\rho}\: (\rho=1,2,\cdots, n+1) \:$ with equilibrium lengths $\:\ell_{\rho}\:(\rho=1,2,\cdots, n+1 )\:$ and the particle masses $\:m_{\rho}\:(\rho=1,2,\cdots, n)$. Disturbing the system from this equilibrium, the equation of motion of the particle $\:m_{\rho}\:$ is

\begin{equation} m_{\rho}\ddot{x}_{\rho} =-k_{\rho}\left(x_{\rho}-x_{\rho-1} \right)+k_{\rho+1}\left(x_{\rho+1}-x_{\rho} \right) \tag{01} \end{equation}

where $\:x_{\rho}(t)\:$ the displacement of this particle from its equilibrium position, see in the middle of above Figure. We set $\:x_{0}(t)=0\:$ and $\:x_{n+1}(t)=0\:$ for the extreme fixed points A and B respectively.

Equation (01) may be written as \begin{equation} m_{\rho}\ddot{x}_{\rho}-k_{\rho}x_{\rho-1} +\left(k_{\rho}+k_{\rho+1} \right)x_{\rho}-k_{\rho+1}x_{\rho+1}=0 \tag{02} \end{equation}

or

\begin{equation} \mathrm{M}\ddot{\mathbf{x}}+\mathrm{K}\mathbf{x}=0 \tag{03} \end{equation}

where

\begin{equation} \mathbf{x}= \begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ \vdots\\ x_{n-1}\\ x_{n} \end{bmatrix} \in \mathbb{R}^{n} \tag{04} \end{equation}

$\:\mathrm{M}\:$ the $\:n \times n\:$ diagonal matrix

\begin{equation} \mathrm{M}= \begin{bmatrix} m_{1} & 0 & 0 & \cdots & 0 & 0 \\ 0 & m_{2} & 0 & \cdots & 0 & 0 \\ 0 & 0 & m_{3} & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & m_{n-1} & 0 \\ 0 & 0 & 0 & \cdots & 0 & m_{n} \end{bmatrix} \tag{05} \end{equation} and $\:\mathrm{K}\:$ the $\:n \times n\:$ tridiagonal symmetric matrix

\begin{equation} \mathrm{K}= \begin{bmatrix} (k_{1}+k_{2}) & -k_{2} & 0 & \cdots & 0 & 0 \\ -k_{2} & (k_{2}+k_{3}) & -k_{3} & \cdots & 0 & 0 \\ 0 & -k_{3} & (k_{3}+k_{4}) & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & (k_{n-1}+k_{n}) & -k_{n} \\ 0 & 0 & 0 & \cdots & -k_{n} & (k_{n}+k_{n+1}) \end{bmatrix} \tag{06} \end{equation}

Equation (03) yields

\begin{equation} \ddot{\mathbf{x}}+\left(\mathrm{M}^{-1}\mathrm{K}\right)\mathbf{x}=0 \tag{08} \end{equation}

or

\begin{equation} \ddot{\mathbf{x}}+\mathrm{S}\mathbf{x}=0, \qquad \mathrm{S}\equiv \mathrm{M}^{-1}\mathrm{K} \tag{09} \end{equation}

Now if $\:\mathrm{S}=\mathrm{M}^{-1}\mathrm{K}\:$ is diagonalizable with eigenvalues $\:\lambda_{\rho}\:(\rho=1,2,\cdots, n)$ and $\:\mathrm{P}\:$ a invertible matrix which diagonalizes it then

\begin{equation} \mathrm{P}^{-1}\mathrm{S} \mathrm{P} = \rm{diag}\left(\lambda_{1},\lambda_{2},\cdots,\lambda_{n}\right) \tag{10} \end{equation} Defining \begin{equation} \mathbf{y}\equiv\mathrm{P}^{-1}\mathbf{x} \tag{11} \end{equation} and multiplying (09) by $\:\mathrm{P}^{-1}\:$, we have \begin{equation} \ddot{\mathbf{y}}+\left(\mathrm{P}^{-1}\mathrm{S} \mathrm{P}\right)\mathbf{y}=0 \tag{12} \end{equation}

that is $\:n \:$ independent differential equations

\begin{equation} \ddot{y}_{\rho}+\lambda_{\rho}y_{\rho}=0, \quad \rho=1,2,\cdots, n \tag{13} \end{equation} Note that taking the inner product of (03) with the "velocity" $\:n-$vector $\:\dot{\mathbf{x}} \:$ \begin{equation} \dot{\mathbf{x}}= \begin{bmatrix} \dot{x}_{1}\\ \dot{x}_{2}\\ \dot{x}_{3}\\ \vdots\\ \dot{x}_{n-1}\\ \dot{x}_{n} \end{bmatrix} \in \mathbb{R}^{n} \tag{14} \end{equation}

we have

\begin{equation} \langle\mathrm{M}\ddot{\mathbf{x}},\dot{\mathbf{x}}\rangle+\langle\mathrm{K}\mathbf{x},\dot{\mathbf{x}}\rangle=0 \tag{15} \end{equation}

that is the equation of conservation of energy

\begin{equation} \dfrac{ \mathrm{d} }{\mathrm{d}t }\left[\frac12\langle\mathrm{M}\dot{\mathbf{x}},\dot{\mathbf{x}}\rangle+\frac12\langle\mathrm{K}\mathbf{x},\mathbf{x}\rangle\right]=0 \tag{16} \end{equation}

related to the question.

For the special case of a common spring constant $\:k_{\rho}=k \: (\rho=1,2,\cdots, n+1) \:$ and common particle mass $\:m_{\rho}=m \: (\rho=1,2,\cdots, n) \:$, equation (08) gives

\begin{equation} \ddot{\mathbf{x}}+\omega_{o}^{2}\,\mathrm{\Xi}\,\mathbf{x}=0 \tag{17} \end{equation}

where \begin{equation} \omega_{o}\equiv \sqrt{\dfrac{k}{m}}= \textrm{fundamental frequency} \tag{18} \end{equation}

and $\:\mathrm{\Xi}\:$ the following $\:n \times n\:$ tridiagonal symmetric matrix (a special case of the so-called Toeplitz matrices)

\begin{equation} \mathrm{\Xi}= \begin{bmatrix} 2& -1 & 0 & \cdots & 0 & 0 \\ -1 & 2 & -1 & \cdots & 0 & 0 \\ 0 & -1 & 2& \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \cdots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & 2 & -1 \\ 0 & 0 & 0 & \cdots & -1& 2 \end{bmatrix} \tag{19} \end{equation} with real positive eigenvalues

\begin{equation} \xi_{\rho}= 4\sin^{2}\left[ \rho\dfrac{\pi}{2(n+1)} \right]=2\Bigg(1-\cos\left[ \rho\dfrac{\pi}{(n+1)} \right]\Bigg), \quad \rho=1,2,\cdots, n \tag{20} \end{equation}

and eigenvectors(1) $\:\mathbf{e}_{\rho}\:$ with $\:\sigma-$component

\begin{equation} \left(\mathbf{e}_{\rho} \right)_{\sigma}=\sqrt{\dfrac{2}{n+1}}\sin\left( \rho \sigma \dfrac{\pi}{n+1} \right) , \quad \rho,\sigma=1,2,\cdots, n \tag{21} \end{equation} In this special case the system of independent equations (13) is

\begin{equation} \ddot{y}_{\rho}+\left(\xi_{\rho}\omega_{o}^{2}\right)y_{\rho}=0, \quad \rho=1,2,\cdots, n \tag{22} \end{equation}

that is :

The motion of a system of $\:n\:$ particles of the same mass $\:m\:$ connected by $\:n+1\:$ ideal springs of the same constant $\:k\:$, see Figure above, is the superposition of $\:n\:$ independent harmonic oscillations with frequences \begin{equation} \omega_\rho=\sqrt{\xi_{\rho}}\omega_o=2\omega_o \sin\left[\rho\dfrac{\pi}{2(n+1)} \right],\quad \omega_{o}\equiv \sqrt{\dfrac{k}{m}} , \quad \rho=1,2,\cdots,n-1,n \tag{23} \end{equation} as shown in Figure below.

enter image description here

(1) Any $\:n \times n\:$ tridiagonal symmetric Toeplitz matrix has the same eigenvectors !!!

EDIT

For other more general cases a useful theorem from "Matrix Theory" by Joel N.Franklin, is given below unchanged :


Theorem Let $\:\mathrm{M}\:$ and $\:\mathrm{K}\:$ be $\:n \times n\:$ Hermitian matrices. If $\:\mathrm{M}\:$ positive definite, then there is a $\:n \times n\:$ matrix $\:\mathrm{C}\:$ for which \begin{equation} \mathrm{C}^{*}\mathrm{M}\mathrm{C}=\mathrm{I} \quad \textrm{and} \quad \mathrm{C}^{*}\mathrm{K}\mathrm{C}=\Lambda= \rm{diag}\left(\lambda_{1},\lambda_{2},\cdots,\lambda_{n}\right) \tag{t-17} \end{equation}

The numbers $\:\lambda_{j}\:$ are real. If $\:\mathrm{K}\:$ is positive definite, the $\:\lambda_{j}\:$ are positive. The $\:\lambda_{j}\:$ are generalized eigenvalues satisfying

\begin{equation} \mathrm{K}\,c^{j}=\lambda_{j}\,\mathrm{M}\,c^{j}, \quad c^{j}\ne 0 \quad (j=1,\cdots, n) \tag{t-18} \end{equation}

If $\:\mathrm{K}\:$ and $\:\mathrm{M}\:$ are real, then a real matrix $\:\mathrm{C}\:$, with columns $\:c^{j}\:$, may be found satisfying (t-17) and (t-18).


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    $\begingroup$ I must say I really am impressed with your figures and formatting; one of the best here. +1. $\endgroup$
    – user36790
    Nov 22 '16 at 8:04
  • $\begingroup$ @MAFIA36790 : I have always in my mind your kindness and your friendly to me comments in the past : I joined Physics SE as diracpaul in June'15 and I quit the site in Sep'15 for personal reasons. I came back as Frobenius in Mar'16. $\endgroup$
    – Frobenius
    Nov 22 '16 at 8:21
  • $\begingroup$ That's great! I was about to mention that yours were similar to someone I knew who unfortunately left Phys.SE a year ago; but that would make the above comment chatty. Nevertheless, welcome again @Frobenius. $\endgroup$
    – user36790
    Nov 22 '16 at 8:35

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