 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2016/9/e/solution

For

M = 2 m

v_{0} = (1, 1, 1)

v_{a} = (1, 0, - 1)

and

v_{s} = (1, - 1, 1)

0

ω = k / m

and

2 ω

. The initial displacement decomposes as

(0, d, - d) = \frac{d}{4} v_{0} + \frac{d}{2} v_{a} - \frac{3 d}{4} v_{s} .

All initial velocities vanish, so the zero-mode coefficient stays constant and the two oscillatory coefficients use cosines. Thus

x_{1} (t) x_{2} (t) x_{3} (t) = \frac{d}{4} + \frac{d}{2} cos (ω t) - \frac{3 d}{4} cos (2 ω t), = \frac{d}{4} + \frac{3 d}{4} cos (2 ω t), = \frac{d}{4} - \frac{d}{2} cos (ω t) - \frac{3 d}{4} cos (2 ω t) .

The conserved centre of mass is

(x_{1} + 2 x_{2} + x_{3}) /4 = d /4

. Omitting the zero mode would incorrectly force this to zero. The two vibrational frequencies are in the ratio

2

, so there is no common nonzero period.

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