= Solution
The quoted proton energies are explicitly kinetic. For mean ring radius $\bar R=215$ m,
$$
C=2\pi\bar R,\qquad\gamma=1+\frac{K}{0.9383\,\mathrm{GeV}},\qquad\beta=\sqrt{1-\gamma^{-2}},\qquad f_{\rm rev}=\frac{\beta c}{C}.
$$
Using these expressions rather than setting $\beta=1$ at injection gives
$$
\boxed{f_{\rm rev}(4\,\mathrm{GeV})\simeq217.880\,\mathrm{kHz},\quad f_{\rm rev}(20\,\mathrm{GeV})\simeq221.700\,\mathrm{kHz},\quad f_{\rm rev}(50\,\mathrm{GeV})\simeq221.885\,\mathrm{kHz}.}
$$
The frequency approaches $c/C$ as energy increases; it is the velocity increase, not a change in the prescribed mean ring circumference, that produces the frequency sweep.
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