💼 Case: damped pendulum with a moving vertical support
We define up to be positive
The acceleration of the pivot is
$$ \ddot \zeta=a\cos(2\omega t) $$
End of setup start of math
when the pivot goes up the pendulum feels a downwards acceleration so $g$ increases
$$ g(t)=g+\ddot\zeta=g(1+h\cos(2\omega t)) $$
where $h=a/g$
The equation of motion is
$$ \ddot \theta+\gamma \dot \theta + \omega^2_0(1+h\cos(2\omega t))\sin\theta=0 $$
🔎 Lets explore the solution at $\theta =0$, so lets consider small perturbations around this point
Lets set $\gamma =0$ (so much for using a damped pendulum) we get the Mathieu equation
$$ \ddot \theta + \omega ^2 _0 [1+h\cos(2\omega t)]\theta =0 $$
We seek a solution of the form $\theta (t)=e^{\lambda t}f(t)$ which is Floquet theory
Since we know for $h=0$ the solution is
$$ \theta(t)=\theta_0 \cos(\omega_0 t+\phi) $$
where $\theta_0$ and $\phi$ are determined by the initial conditions
Then for small $h$ we seek a general solution of the form
$$ \theta(t)=\theta_0 e^{\lambda t} \cos(\omega t+\phi) $$
which is a sub-harmonic frequency
🗒️ Note: Stable if $\lambda <0$ and unstable if $\lambda >0$
Lets sub our solution in assuming $\theta_0\sim 1$ for convenance
$$ \begin{aligned} \left [ \frac{\text d^2}{\text d t^2} + \omega_0 ^2 (1+h\cos(2\omega t)) \right ] (e^{\lambda t} \cos(\omega t +\phi))&=0 \\ \footnotesize{e^{\lambda t}[(\omega_0^2-\omega^2 +\lambda^2)\cos(\omega t+\phi)-2\lambda \omega \sin(\omega t+\phi)+\frac{h}{2}\omega_0^2 \cos(\omega t-\phi)+\frac h2 \omega^2_0 \cos(3\omega t+\phi)]}&=0
\end{aligned} $$
Expanding $\cos(\omega t+\phi)$, $\cos(\omega t-\phi)$ and $\sin(\omega t+\phi)$ while disregarding $\cos(3\omega t+\phi)$ because it is negligible
Since the initial conditions require both the $e^{\lambda t} \sin(\omega t)$ terms and $e^{\lambda t} \cos(\omega t)$ to be zero we can split the equation into 2 to write
$$ \begin{aligned} \left [ \omega^2 _0 - \omega ^2 + \lambda ^2 +\frac h2 \omega ^2 _0 \right ]\cos \phi - 2\lambda \omega \sin \phi &=0 \\ 2\lambda \omega \cos \phi+\left [ \omega^2 _0 - \omega ^2 + \lambda ^2 -\frac h2 \omega ^2 _0 \right ]\sin \phi &=0 \end{aligned} $$
This is two homogenous equations where solutions only exist if the determinant is zero
$$ \lambda ^4 + 2 (\omega_0^2 + \omega ^2 ) \lambda ^2 + (\omega_0 ^2 -\omega^2)^2 - \frac {h^2}4 \omega ^4_0=0 $$
Discriminent is $16 \omega_0^2 \omega ^2 + h ^2 \omega_0^4>0$ this means the equations has the following real solution
$$ \lambda ^2 = -(\omega_0^2 +\omega^2 )\pm \frac 12 \sqrt{16 \omega_0^2 \omega ^2 + h ^2 \omega_0^4} $$
🗒️ Note: in the case of - $\lambda$ is imaginary which is stable as it is only unstable for $\lambda >0$
to have $\lambda >0$ we need $\lambda ^2 >0$
$$ 16 \omega_0^2 \omega ^2 + h ^2 \omega_0^4>4(\omega^2_0 + \omega^2 )^2 $$
Which is possible if
$$ h>2\left | 1-\frac{\omega^2}{\omega_0^2} \right | $$
we get an exact resonance when $\frac{\omega^2}{\omega_0^2}=1$
General non-dimensional case we have
$$ \ddot \theta+(\alpha + \beta \cos(2\tau ))\theta=0 $$
🌕 Unshaded region: stable hanging down equilibrium
🌑 Shaded region: unstable hanging down equilibrium (swinging motion)
🗒️ Note: resonances at $\alpha =1,4,9,16,\ldots$
This time $\gamma\ne 0$ and we follow the same steps expanding for small $h$ with a Floquet theory
$$ (\omega_0^2 -\omega ^2 +\lambda ^2 +\gamma \lambda )^2 - \frac{h^2}{4}\omega^4_0 + ( 2\omega \lambda +\omega \gamma )^2 =0 $$
Marginal stability when $\lambda =0$ which requires
$$ \frac{h^2}{4}=\left ( 1-\frac{\omega^2_0}{\omega^2} \right )+\frac{\omega^2}{\omega_0^4 }\gamma^2 $$
At resonance $\omega=\omega_0$ the hanging down solution $(\theta=0)$ is linearly unstable if $h>\frac{2\gamma}{\omega_0}$
With damping we now have a minimum amplitude required to drive instability
The path $A\to B$ goes thought the solid line where instability is exponential
To describe the states inside the resonance region we need higher Taylor expansions
