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where the integral is just a number. For the frequency one has wo(E) x 1/T x E(n-2)/(2n) that illustrates the above statement. Consider a particle in a washboard potential U(x) = Uo [1- cos (ax)]. (155) Near the minima at ax = 2πn, n = 0,±1, ±2, etc., one has U(x) ≈ kr²/2 with k = Uoa² and behaves similarly to the harmonic oscillator. The whole problem is mathematically equivalent to that of pendulum. At ax = π + 2πn potential energy has maxima, Umax = 2Uo. Let us calculate the period of oscillations, considering the region around the minimum at x = 0. Eq. (148) becomes Im 2m = 2√/2010 [ Uo 0 T=2₁ T = 2, m Uo where I'm is the right turning point satisfying E= Uo [1 - cos (arm)]. The integral can be transformed into the form similar to those above, Next, one can employ the variable change Im dx ✓cos (ax) - cos (axm) 0 da sin² (axm/2) sin (ax/2) (a/2) cos (ax/2) dr - (154) sin² (ax/2) = sin (axm/2) sin sin (ax./2) cos Ed (156) (157) (158) ...
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