Kinda part of my Theory of Everything - Some Novel Approach Including Love ❤️ – Math on Demand Edition

Consider a simple mass on a spring system, yielding a driven harmonic oscillator

\[\ddot x + 2\beta \dot x + \omega^2 (x-y) = 0 \veq{eom}\assumption\]

It is of course possible to solve $\eqref{eom}$ with some analysis:

\[x(t) = A\cos\Big(\underbrace{\sqrt{\omega^2-\beta^2}}_{=:\Omega}t+\phi\Big)e^{-\beta t} + \int_{-\infty}^t\frac{\omega^2}{\Omega}\sin(\Omega (t-t'))e^{-\beta(t-t')} y(t')\, dt' \veq{sol}\sorry\]

But that’s too verbose for now. Let’s instead consider the time evolution of the specific energy:

\[e = \frac12\dot x^2 + \frac{\omega^2}2 x^2 \veq{e}\definition\] \[\begin{aligned} \dot e &= \dot x\cdot(\underbrace{\ddot x}_{=-2\beta\dot x-\omega^2(x-y)} + \omega^2 x) \veqs{edot_deriv}\lean \\ &= \dot x\cdot(-2\beta\dot x + \omega^2 y) %\ltag{edot.2} \\ &= \underbrace{-2\beta\dot x^2}_{\le 0} + \omega^2\dot x y %\ltag{edot.3} \\ &= -4\beta e + \omega^2(2\beta x^2 + \dot x y) \veqs{edot}\sympylean \end{aligned} \tag{edot}\]

Even without the analytical solution it would be clear from (edot.3) that a free oscillator ($y=0$) would permanently loose energy for $\beta\neq0$. Now one interesting question is what kind of drive $y$ is needed in order to keep the energy constant, i.e. $\dot e=0$. Since $\dot x\neq 0$ that means

\[y = 2\frac{\beta}{\omega^2}\dot x \veq{ymaint}\sympy\]

There is one solution independent of $\dot x$:

\[\begin{aligned} \dot y &= 2\frac{\beta}{\omega^2}\ddot x = -4\frac{\beta^2}{\omega^2}\dot x - 2\beta(x-y) = -2\beta y -2\beta(x-y) = -2\beta x, \\ \ddot y &= -2\beta\dot x = -\omega^2 y \end{aligned} \veq{yfree}\sympy\]

So, by swinging at exactly the free frequency $\omega$ (and not the eigenfrequency $\Omega$) the oscillator will maintain its energy. What if instead we want to increase the energy of the system? As seen above,

\[\dot e > 0 \iff |y| > 2\frac{\beta}{\omega^2}|\dot x| \ \text{with}\ \operatorname{sign} y = \operatorname{sign}\dot x \veq{eincr}\sympy\]

i.e. one has to pull in movement direction, but stop at the apex to start pushing and vice versa – which is precisely what intuition causes us to do.

🚧 Located rigor-debt (flagged via /meeting 2026-06-15; tracked in REVIEW_ME.md) — two items in the energy-method chain above:

  1. The opening claim that a free oscillator “would permanently loose energy” is asserted as obvious, but the cited closed form edot (ė = −4βe + ω²(2βx² + ẋy)) is not manifestly ≤ 0. Finding (owner to confirm): the first form ė = −2βẋ² (at y=0) is manifestly ≤ 0 — the wording should lean on that form.
    re okay, explicitly mentioned (edot.3) though we need to fix the ParseError: KaTeX parse error: Multiple \tag in an aligned env, and automated dot-numbering of subequation should be on the wishlist…
  2. The ymaint / yfree derivation skips steps. Their results are machine-verified ✓ by verify/resogram_drive.py and are unaffected by the edot sign fix (which changed only edot’s second equality, not the first form they derive from) — so this is an exposition gap, not a correctness one.
    re same

AI flags only — the owner resolves; the math/prose stays untouched.

Another question is how much change of energy does the external force contribute. Without the analytical solution there might be some handwaving about how $\dot e$ varies between zero and $-4\beta e$ in the absence of external forces, thus averaging to $\dot e\approx -2\beta e$. But I’ll just use the analytical solution now, which boils down to:

\[e = \frac{A^2\omega}{2}e^{-2\beta t}\Big(\omega + \beta\cos\big(2(\Omega t+\phi)-\delta\big)\Big),\qquad \delta=\operatorname{atan2}(\Omega,\beta) \veq{esol}\numeric\]

This hints at a sensible sliding average:

\[\bar e(t) := \frac\Omega\pi \int_0^{\frac\pi\Omega} e(t-t')e^{+2\beta t'}\,dt'\]