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Two coupled equal masses: mode frequencies sqrt(k/m) and sqrt((k+2c)/m), and the free zero mode when k=0

Example

Let $m > 0$ , $k \geq 0$ and $c \geq 0$ , and consider the constant-coefficient system $M q^{''} + K q = 0$ on $R^{2}$ with

$M = (\begin{array}{cc} m & 0 \\ 0 & m \end{array}), K = (\begin{array}{cc} k + c & - c \\ - c & k + c \end{array}) .$

Then $K$ is symmetric and non-negative, $S = M^{- 1 / 2} K M^{- 1 / 2} = K / m$ has eigenvalues $λ_{1} = k / m$ with eigenvector $\frac{1}{\sqrt{2}} (1, 1)$ and $λ_{2} = (k + 2 c) / m$ with eigenvector $\frac{1}{\sqrt{2}} (1, - 1)$ , and the general solution is the sum of the modal solutions of Decoupling of the constant-coefficient quadratic system $M q^{''} + K q = 0$ into normal modes, with the zero-mode and energy identities:

if $k > 0$ the first mode oscillates with angular frequency $ω_{1} = \sqrt{k / m}$ , while if $k = 0$ it is the free affine mode $a_{1} + b_{1} t$ along $\frac{1}{\sqrt{2}} (1, 1)$ ;
the second mode oscillates with angular frequency $ω_{2} = \sqrt{(k + 2 c) / m}$ whenever $k + 2 c > 0$ , and if $k = c = 0$ it is also a free affine mode $a_{2} + b_{2} t$ along $\frac{1}{\sqrt{2}} (1, - 1)$ .

In particular a zero eigenvalue is not an oscillator but a free mode.

Facts & Assumptions

Given: The numbers $m > 0$ , $k \geq 0$ , $c \geq 0$ , the matrices $M = m I_{2}$ and $K$ displayed above, and the vectors $w_{1} = \frac{1}{\sqrt{2}} (1, 1)$ , $w_{2} = \frac{1}{\sqrt{2}} (1, - 1)$ .

[L1]

The quadratic-mode theorem applies to $M$ symmetric positive definite and $K$ symmetric non-negative, produces the eigenvalues and modes of $S = M^{- 1 / 2} K M^{- 1 / 2}$ , and gives the general solution and the modal energies (Decoupling of the constant-coefficient quadratic system $M q^{''} + K q = 0$ into normal modes, with the zero-mode and energy identities).

[L2]

$M^{1 / 2} = \sqrt{m} I_{2}$ and $M^{- 1 / 2} = \frac{1}{\sqrt{m}} I_{2}$ (Positive definite symmetric matrices have positive definite square roots, and the congruence M^{-1/2} K M^{-1/2} preserves symmetry and positivity, Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes).

Verification

technique · direct; multiply the explicit matrices and apply the theorem

1.1givenL2algebra

$S = M^{- 1 / 2} K M^{- 1 / 2} = K / m$ , and $S w_{1} = \frac{1}{m} (\begin{array}{cc} k + c & - c \\ - c & k + c \end{array}) \frac{1}{\sqrt{2}} (\begin{array}{c} 1 \\ 1 \end{array}) = \frac{1}{\sqrt{2}} \frac{1}{m} (\begin{array}{c} k \\ k \end{array}) = \frac{k}{m} w_{1},$ $S w_{2} = \frac{1}{m} (\begin{array}{cc} k + c & - c \\ - c & k + c \end{array}) \frac{1}{\sqrt{2}} (\begin{array}{c} 1 \\ - 1 \end{array}) = \frac{1}{\sqrt{2}} \frac{1}{m} (\begin{array}{c} k + 2 c \\ - (k + 2 c) \end{array}) = \frac{k + 2 c}{m} w_{2} .$ Since $w_{1}, w_{2}$ are orthonormal, $O = (w_{1} w_{2})$ is orthogonal and diagonalises $S$ with the displayed eigenvalues; the eigenvalues are non-negative, $λ_{1} = k / m > 0$ exactly when $k > 0$ , and $λ_{2} = (k + 2 c) / m > 0$ exactly when $k + 2 c > 0$ .

2.1step 1.1L1

By [L1] the general solution is $q = M^{- 1 / 2} O z$ , where each modal coordinate is given by the theorem according to the sign of its eigenvalue: if $k > 0$ then $z_{1} = a_{1} \cos (ω_{1} t) + b_{1} \sin (ω_{1} t)$ with $ω_{1} = \sqrt{λ_{1}}$ , and if $k = 0$ then $z_{1} = a_{1} + b_{1} t$ ; if $k + 2 c > 0$ then $z_{2} = a_{2} \cos (ω_{2} t) + b_{2} \sin (ω_{2} t)$ with $ω_{2} = \sqrt{λ_{2}}$ , and if $k = c = 0$ then $z_{2} = a_{2} + b_{2} t$ . The modal energies are constant. The verification of the scalar solution formulas is the one given by the theorem via The derivatives of sine and cosine are cosine and minus sine.

3.1step 1.1L1algebra∎

The zero-mode branch $k = 0$ : then $λ_{1} = 0$ and the first modal solution is the affine motion $a_{1} + b_{1} t$ along $w_{1}$ by [L1], which is unbounded unless $b_{1} = 0$ ; the second eigenvalue is $2 c / m \geq 0$ , with $2 c / m > 0$ when $c > 0$ , so the second mode oscillates with frequency $\sqrt{2 c / m}$ unless also $c = 0$ , in which case both modes are affine. Thus when $k = 0 < c$ the only free direction is $\frac{1}{\sqrt{2}} (1, 1)$ , while when $k = c = 0$ both independent directions are free; in either case a zero mode is not an oscillator.

Remarks

The example is the two-body spring model rewritten in normal coordinates. The potential $U (q) = \frac{k}{2} (q_{1}^{2} + q_{2}^{2}) + \frac{c}{2} (q_{1} - q_{2})^{2}$ has Hessian $K$ , so this is also the linearisation of the mechanical system at the equilibrium $q = 0$ by Linearisation of the mechanical system at an equilibrium: M(q_0) xi'' + D²U(q_0) xi = 0, and the small-oscillation quadratic model; the example verifies the modal computation on explicit matrices and claims nothing about the nonlinear system.

Two coupled equal masses: mode frequencies sqrt(k/m) and sqrt((k+2c)/m), and the free zero mode when k=0

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources