Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 2 / 12

\frac{d E}{d t} = m \dot{r} \cdot \ddot{r} + k r \cdot \dot{r} = \dot{r} \cdot (m \ddot{r} + k r) = 0.

The equations of motion make the bracket zero. Hence the mechanical energy is constant in time. The central restoring force can exchange kinetic and potential energy but supplies no change in their sum.

b

ω = k / m

x = A cos ω t + B sin ω t, y = C cos ω t + D sin ω t .

The four initial data give

A = X_{0}

B = C = 0

, and

D = V_{0} / ω

. Thus

x (t) = X_{0} cos ω t, y (t) = \frac{V _{0}}{ω} sin ω t .

This is the position component of the isotropic harmonic oscillator flow. Substitution also gives the constant energy

E = (k X_{0}^{2} + m V_{0}^{2}) /2

The stated condition is equivalent to

(V_{0} / ω)^{2} = X_{0}^{2}

. Therefore the solution from (b) obeys

x (t)^{2} + y (t)^{2} = X_{0}^{2} (cos^{2} ω t + sin^{2} ω t) = X_{0}^{2} .

For nonzero initial data this is a circle centred at the origin with radius $∣ X_{0} ∣$ . The sign of

X_{0} V_{0}

chooses its direction of traversal. If

X_{0} = V_{0} = 0

, the same algebra gives only the stationary origin, a degenerate radius-zero case rather than a genuine circular orbit.

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