 Solution

ID: past-exam-of-the-short-options-paper-shared-by-prelims-and-parts-a-and-b-of-the-physics-course-of-the-university-of-oxford/2012/s19/3/a/i/solution

The quoted proton energies are explicitly kinetic. For mean ring radius

\overset{ˉ}{R} = 215

C = 2 π \overset{ˉ}{R}, γ = 1 + \frac{K}{0.9383 GeV}, β = 1 - γ^{- 2}, f_{rev} = \frac{β c}{C} .

Using these expressions rather than setting

β = 1

at injection gives

f_{rev} (4 GeV) ≃ 217.880 kHz, f_{rev} (20 GeV) ≃ 221.700 kHz, f_{rev} (50 GeV) ≃ 221.885 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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