Exact linear Foucault pendulum trajectory ($z=A{e}^{-i\alpha t}(\cos \beta t+i\alpha \sin \beta t/\beta )$)

For release from rest relative to the building, take $x(0)=A>0$, $y(0)=\dot{x}(0)=\dot{y}(0)=0$. The exact solution of the stated coupled linear differential equations isIndeed $u(0)=A$ and $\dot{u}(0)=i\alpha A$, so $\dot{z}(0)=0$. In the slowly rotating coordinates, the ellipse has semiaxes $A$ and $A|\alpha |/\beta$. A bob released from rest therefore does not describe a perfectly straight line even in this ideal linear model. Replacing the bracket by $\cos {\omega }_{0}t$ gives the familiar leading approximation but introduces a spurious initial transverse velocity $-i\alpha A$. The exact expression keeps the initial condition and separates that tiny orbit width from the slow rotation of its long axis.

For the illustrative $\ell =20\mathrm{m}$ and $g=9.81m/{\mathrm{s}}^2$, the fast period $2\pi /\beta$ is approximately $8.9714\mathrm{s}$. At $45^{\circ }$ north, $|\alpha |/\beta \simeq 7.3624\times 10^{-5}$: the transverse semiaxis is too small to distinguish at the figure's scale. Nevertheless the long-axis orientation changes by about $-63.8138^{\circ }$ in six hours. Changing the sign of geographical latitude conjugates the entire solution, reflecting the trajectory across the eastward axis; setting $\alpha =0$ gives $z=A\cos {\omega }_{0}t$ exactly. These are stronger checks than merely drawing a rotating line.

The coupled equations give an additional independent constraint. Multiplying them by $\dot{x}$ and $\dot{y}$ and adding cancels the Coriolis force's contributions:This is mechanical energy per unit mass in the reduced model; precession does not require a loss of mechanical energy. Also, direct differentiation givesFor this release the constant is $\alpha A^2$. The exact trajectory satisfies both identities. They test the coupled dynamics separately from the visible slow rotation and express the fact that the Coriolis force does no work.