title | permalink |
---|---|
Resogramm |
/Resogramm |
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 $
It is of course possible to solve this 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' $
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, \ \dot e = \dot x\cdot(\underbrace{\ddot x}{=-2\beta\dot x-\omega^2(x-y)} + \omega^2 x) = \dot x\cdot(-2\beta\dot x + \omega^2 y ) \ = \underbrace{-2\beta\dot x^2}{\le 0} + \omega^2\dot x y = -4\beta e - \omega^2(2\beta x^2 - \dot x y) $
Even without the analytical solution it would be clear that a free oscillator (
$ y = 2\frac{\beta}{\omega^2}\dot x $
There is one solution independent of
$ \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 $
So, by swinging at exactly the free frequency
Another question is how much change of energy does the external force contribute. Without the analytical solution there might be some handwaving about how
$ \bar e(t) := 2\Omega\int_0^{\frac1{2\Omega}} e(t-t')e^{+2\beta t'},dt' $