Past exam of the mathematics course of the University of Cambridge / 2022 / ia / Paper 4 / 9C / b

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- Solution b

Solution

For the proposed circular motion,

\ddot{x}_{1} = - a ω^{2} (cos ω t, sin ω t, 0) .

The separation is

x_{1} - x_{2} = 2 a (cos ω t, sin ω t, 0)

, so the inverse-square force on particle 1 is

F_{12} = - \frac{k m ^{2}}{4 a ^{2}} (cos ω t, sin ω t, 0) .

m \ddot{x}_{1} = F_{12}

holds exactly when

ω^{2} = \frac{km}{4 a ^{3}} .

The equation for particle 2 follows by symmetry.

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