Pendulum with a prescribed translating support ($l\ddot{\theta }+{\ddot{a}}_x\cos \theta +(g+{\ddot{a}}_y)\sin \theta =0$)

For support displacement $\mathbf{a}(t)$ and angle measured from the downward vertical, the mass position is $(a_x+l\sin \theta ,a_y-l\cos \theta )$. Its Euler-Lagrange equation is $l\ddot{\theta }+{\ddot{a}}_x\cos \theta +(g+{\ddot{a}}_y)\sin \theta =0$. In the support frame the effective acceleration is $\mathbf{g}-\ddot{\mathbf{a}}$; taking its torque about the hinge gives $l^2\ddot{\theta }\hat{\mathbf{z}}=\mathbf{l}\times (\mathbf{g}-\ddot{\mathbf{a}})$. The prescribed support can do work, so pendulum energy need not be conserved.

For small angles about the downward vertical and prescribed pivot position $x=A\cos \omega t$, a simple pendulum obeys the displayed forced-oscillator equation. Off resonance, its harmonic particular solution is $[A{\omega }^2/(l({\omega }_0^2-{\omega }^2))]\cos \omega t$, with ${\omega }_0^2=g/l$. At exact resonance, a pendulum initially vertical with zero angular velocity has $\theta =(A{\omega }_0/(2l))t\sin {\omega }_{0}t$ in the undamped linear model. The growing envelope eventually violates the small-angle assumption; damping and nonlinear frequency shifts matter in a real pendulum.

For a small, rapid horizontal displacement $a\cos \omega t$, the leading fast angular excursion is $\xi =-(a/l)\cos \Theta \cos \omega t$. Correlating that excursion with the driving torque gives the averaged force $\ddot{\Theta }=-(g/l)\sin \Theta +(a^2{\omega }^2/2l^2)\sin \Theta \cos \Theta$. Its effective potential is $-mgl\cos \Theta +m{a}^2{\omega }^2{\cos }^2\Theta /4$. The downward state loses linear stability above $a\omega =\sqrt{2gl}$ and two tilted minima appear. Small $a/l$ and slow dynamics compared with $\omega$ are required; horizontal and vertical support vibration produce different effective potentials.