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Two-mass spring eigenmodes at explicit parameters: eigenrelations, a mode solution, and the zero-frequency case

Example

Take $m = 1$ , $k = 1$ , $c = 3$ in Two equal masses with two outer springs and a coupling spring: eigenmodes, decoupling and the explicit normal frequencies, so that $K = (\begin{array}{cc} 4 & - 3 \\ - 3 & 4 \end{array})$ and $u_{\pm} = \frac{1}{\sqrt{2}} (1, \pm 1)^{T}$ ; the proposition gives $K u_{+} = u_{+}$ and $K u_{-} = 7 u_{-}$ , hence squared frequencies $k / m = 1$ and $(k + 2 c) / m = 7$ . The curve

$x (t) = \cos (t) u_{+} + \cos (\sqrt{7} t) u_{-}$

solves $x^{''} = - K x$ with $x (0) = (\sqrt{2}, 0)$ and $x^{'} (0) = 0$ . In the degenerate case $k = 0$ , $c = 1$ one has $K = (\begin{array}{cc} 1 & - 1 \\ - 1 & 1 \end{array})$ and $K u_{+} = 0$ , so $x (t) = (1 + t) u_{+}$ solves $x^{''} = - K x$ and is unbounded on $[0, \infty)$ : the zero-frequency mode is affine, not oscillatory. All entries are exact; no numerical integration or measurement is involved.

Facts & Assumptions

Given: The parameters of the two cases, the matrix $K$ and vectors $u_{\pm}$ of each case, and the system $x^{''} = - K x$ of Two equal masses with two outer springs and a coupling spring: eigenmodes, decoupling and the explicit normal frequencies with $m = 1$ .

[L1]

The proposition (Two equal masses with two outer springs and a coupling spring: eigenmodes, decoupling and the explicit normal frequencies): for $m > 0$ , $k \geq 0$ , $c \geq 0$ one has $K u_{+} = k u_{+}$ and $K u_{-} = (k + 2 c) u_{-}$ , the system $m x^{''} = - K x$ is equivalent to the two scalar oscillator equations $m y_{+}^{''} + k y_{+} = 0$ , $m y_{-}^{''} + (k + 2 c) y_{-} = 0$ via $x = y_{+} u_{+} + y_{-} u_{-}$ , and for $k = 0$ the family $x (t) = (A + B t) u_{+}$ consists of solutions with $K u_{+} = 0$ (The damped driven oscillator model: second-order equation, phase-plane system and characteristic data, The homogeneous damped oscillator: solution classification by the discriminant and uniqueness for initial data, The degenerate and repulsive oscillator: solutions for k=0 and the explicit growing solution for k<0 for the scalar solution families).

[L2]

Matrix and derivative data (Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes, The vector space $M_{m \times n} (F) : = F^{m \times n}$ of $m$ by $n$ matrices over a field, with entrywise operations, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The chain rule for total derivatives: $D (g \circ f) (a) = D g (f (a)) \circ D f (a)$ , The derivatives of sine and cosine are cosine and minus sine, Sine and cosine defined by their real power series, The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length): matrix-vector multiplication is the displayed sum; $\frac{d}{d t} \cos t = - \sin t$ and $\frac{d}{d t} \cos (\sqrt{7} t) = - \sqrt{7} \sin (\sqrt{7} t)$ , so $\frac{d^{2}}{d t^{2}} \cos (\sqrt{7} t) = - 7 \cos (\sqrt{7} t)$ by the chain rule; and $∥ (A + B t) u_{+} ∥_{2} = ∣ A + B t ∣$ because $u_{+}$ is a unit vector.

Verification

Proof technique: direct; multiply the two explicit matrices into the displayed vectors and substitute the resulting eigenrelations into the two solutions.

1.1givenL1L2algebra

Eigenrelations in the first case: direct multiplication gives $K u_{+} = \frac{1}{\sqrt{2}} (\begin{array}{c} 4 - 3 \\ - 3 + 4 \end{array}) = \frac{1}{\sqrt{2}} (\begin{array}{c} 1 \\ 1 \end{array}) = u_{+}, K u_{-} = \frac{1}{\sqrt{2}} (\begin{array}{c} 4 + 3 \\ - 3 - 4 \end{array}) = \frac{7}{\sqrt{2}} (\begin{array}{c} 1 \\ - 1 \end{array}) = 7 u_{-},$ so $k = 1$ and $k + 2 c = 7$ are the eigenvalues and the squared frequencies are $1$ and $7$ ; the second case has $K u_{+} = \frac{1}{\sqrt{2}} (1 - 1, - 1 + 1)^{T} = 0$ .

2.1step 1.1givenL2algebra

The mode solution: with $x (t) = \cos (t) u_{+} + \cos (\sqrt{7} t) u_{-}$ one has $x^{''} (t) = - \cos (t) u_{+} - 7 \cos (\sqrt{7} t) u_{-} = - (\cos (t) K u_{+} + \cos (\sqrt{7} t) K u_{-}) = - K x (t)$ by step 1.1 and [L2], so $x^{''} = - K x$ ; also $x (0) = u_{+} + u_{-} = (\sqrt{2}, 0)$ and $x^{'} (0) = 0$ because both sine factors vanish at $0$ .

3.1step 1.1givenL1L2algebra∎

The zero-frequency case: for $k = 0$ , $c = 1$ and $x (t) = (1 + t) u_{+}$ one has $x^{''} = 0$ and $- K x = - (1 + t) K u_{+} = 0$ by step 1.1, so $x$ solves the system; moreover $∥ x (t) ∥_{2} = 1 + t$ tends to $+ \infty$ , so the motion is unbounded and affine rather than oscillatory, as clause 4 of [L1] predicts.

Two-mass spring eigenmodes at explicit parameters: eigenrelations, a mode solution, and the zero-frequency case

Example

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

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