 Solution

ID: past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2014/paper-1/3/e/solution

Separating

G = W (q) - βt

gives

[W^{'} (q)]^{2} / (2 m) + m ω^{2} q^{2} /2 = β

. On the positive-momentum branch choose

G = \int_{0}^{q} 2 m (β - \frac{1}{2} m ω^{2} s^{2}) d s - βt .

With

Q = β

and

A = 2 β / (m ω^{2})

Q

gives

P = - G_{β} = t - \int_{0}^{q} \frac{m d s}{2 m ( β - m ω ^{2} s ^{2} /2 )} = t - \frac{1}{ω} arcsin (q / A) .

Since

P = γ

is constant,

q = A sin [ω (t - γ)] .

The displayed integral is a local positive-

p

branch. Continuing the trajectory through turning points switches momentum branches and extends the sine solution over the complete orbit.

β

γ

fixes its time phase.

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