Consider the Kapitz pendulum problem for the case of horizontal oscillations of a suspension point.
1. Find the equilibrium positions
2. Find the maximal deviation of the pendulum from the equilibrium position at which it will still return to this position
My attempt for the first assignment:
Equilibrium positions are defined as points at which the sum of all acting forces is zero. In the case of a horizontal oscillation of a suspension point, we can assume that it moves along the horizontal axis with amplitude $A$ and angular velocity $\omega $: $x = A \sin \omega t$. The pendulum itself also moves in the horizontal plane, so the equilibrium positions will be at points where the acceleration of the pendulum and the acceleration of the suspension point are equal. Let $\theta $-angle between the vertical axis and the pendulum, then the acceleration of the pendulum $-g\sin \theta $. The acceleration of the suspension point is $-A\omega ^2\sin \omega t$. Then the equilibrium equation is
$$-g\sin \theta =-A\omega ^2\sin \omega t$$
My attempt for the second assignment:
Suppose that the Kapitsa pendulum behaves like a simple harmonic oscillator at small deviations. Then the maximum deflection of the pendulum from its equilibrium position (i.e. the amplitude of the oscillations of the pendulum), at which it will still return to this position, will be determined by the point where the force acting on the pendulum is zero. In our case, the forces on the pendulum include the gravitational force and the force due to the oscillation of the suspension. Consider the system of equations $$\begin{cases} mg\sin \theta =-ml\theta'' \\ mA\omega \cos \omega t=ml\theta '' \end{cases}$$
As we considered the case above, for sufficiently small fluctuations $$i\theta ''+g\theta =A\omega ^2\cos \omega t$$
Suppose there is a solution $\theta (t)=\Re \left ( C\exp\left \{ i \omega t \right \} \right )$, then $$\theta' (t)=\Re \left ( i\omega C\exp \left \{ i\omega t \right \} \right ), \; \theta ''(t)=\Re \left ( -\omega ^2C\exp\left \{ i\omega t \right \} \right )$$ $$-g\Re \left ( C\exp \left \{ i\omega t \right \} \right )=\Re \left ( -i\omega ^2C\exp \left \{ i\omega t \right \}+A\omega ^2\cos \omega t \right )$$
Equate the real parts, then $$C=\frac{A\omega ^2}{g+i\omega ^2}\Rightarrow \theta (t)=\Re \left ( \frac{A\omega ^2\exp \left \{ i\omega t \right \}}{g+i\omega ^2} \right )$$
The amplitude of this oscillation will be the maximum deflection of the pendulum from its equilibrium position $$\left | \frac{A\omega ^2}{g+i\omega ^2} \right |$$
Am I reasoning correctly and have I solved two problems?