= Exact linear Foucault pendulum trajectory
{c}
{title2=$z=Ae^{-i\alpha t}(\cos\beta t+i\alpha\sin\beta t/\beta)$}
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> is
$$
z(t)=Ae^{-i\alpha t}\left(\cos\beta t+i\frac\alpha\beta\sin\beta t\right),\qquad
\beta=\sqrt{\omega_0^2+\alpha^2}.
$$
Indeed $u(0)=A$ 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 $\cos\omega_0t$ 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.
\Image[/foucault-pendulum-exact-slow-trajectories.png]
{title=Exact trajectories of the linear Foucault pendulum equations at the actual Earth rotation rate, showing one swing near zero, three, and six hours at northern, equatorial, and southern latitudes}
For the illustrative $\ell=20\,\mathrm m$ and $g=9.81\,\mathrm{m/s^2}$, the fast period $2\pi/\beta$ is approximately $8.9714\,\mathrm s$. At $45^\circ$ north, $|\alpha|/\beta\simeq7.3624\times10^{-5}$: 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 $z=A\cos\omega_0t$ 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:
$$
\frac d{dt}\left[\frac12(\dot x^2+\dot y^2)+\frac12\omega_0^2(x^2+y^2)\right]=0.
$$
This is <mechanical energy> per unit <mass> in the reduced model; <precession> does not require a loss of <mechanical energy>. Also, direct <differentiation> gives
$$
\frac d{dt}\left[x\dot y-y\dot x+\alpha(x^2+y^2)\right]=0.
$$
For 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>.
Back to article page
