= Period and small-angle limit of a conical pendulum
{title2=$P/P_0=\sqrt{\cos\theta},\quad P_0=2\pi\sqrt{\ell/g}$}
\Image[/conical-pendulum-period.png]
{title=Ideal conical pendulum period decreases with cone angle and its square is proportional to the vertical drop}
For fixed $\ell$ and $g$, the exact <conical pendulum> <oscillation period> satisfies $P/P_0=\sqrt{\cos\theta}$. In the <small-angle approximation>, with $\theta$ in <radians>,
$$
\frac{P}{P_0}=1-\frac{\theta^2}{4}+O(\theta^4),\qquad P_0=2\pi\sqrt{\frac\ell g}.
$$
Thus the limiting revolution period equals the small-amplitude <oscillation period> of a planar <simple pendulum>. This equality has a geometric explanation: to first order, $x(t)=r\cos(\omega t)$ and $y(t)=r\sin(\omega t)$ are perpendicular <simple harmonic motions> of equal <amplitude>, a quarter period apart, with $\omega\to\sqrt{g/\ell}$. At finite <angle>, however, the <conical pendulum>'s <oscillation period> decreases with $\theta$, whereas a planar <simple pendulum>'s <oscillation period> increases with its angular <amplitude>.
The endpoint $\theta=0$ itself is a stationary bob with $r=0$ and no observable revolution; $P_0$ is a limit of nonzero circular motions, not a measured revolution period at rest. For fixed $\ell$, nonzero <conical pendulum> motion also requires $\omega>\sqrt{g/\ell}$, because $\cos\theta=g/(\ell\omega^2)<1$. At the other endpoint, $\theta\to\pi/2$ gives $h\to0$, $P\to0$, and unbounded <tension> and <speed>. These mathematical limits explain why the ideal string cannot sustain a horizontal orbit at finite <tension>; a real experiment ceases to satisfy the ideal assumptions before that limit.
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