For ds2=(1+f(x))2dx2/(1+x2)−(1+x2)dt2, with smooth odd f and ∣f∣<1, the conserved energy gives dt/dτ=E/(1+x2) and (dx/dτ)2=(E2−1−x2)/(1+f)2. Both half-cycle integrals over [−E2−1,E2−1] contain an even weighttimes 1+f. The odd contribution vanishes, leaving the unperturbed half-period π in each clock. The waveforms need not be sinusoidal, but both periods are unchanged for E>1.