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Past exam of the mathematics course of the University of Cambridge

2019

Paper 4

Taking the <gradient> of the quadratic <potential energy> found in part (ii) resolves the linearized force into identical $x$ and $y$ components:
$$
F_x=-\frac{3q^2}{8\pi\epsilon_0\rho^3}x,
\qquad
F_y=-\frac{3q^2}{8\pi\epsilon_0\rho^3}y.
$$
Each component therefore performs <simple harmonic motion> with
$$
\omega^2=\frac{3q^2}{8\pi\epsilon_0m\rho^3}
=\frac{9\sqrt3}{8\pi}\frac{q^2}{m\epsilon_0a^3}.
$$
Thus the <angular frequency> is
$$
\boxed{\omega=\frac{3^{5/4}}{\sqrt{8\pi}}\frac{|q|}{\sqrt{m\epsilon_0a^3}},
\qquad A=\frac{3^{5/4}}{\sqrt{8\pi}}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 4 / 10A / b / iii / Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / ia / Paper 4 / 10A / b / iii / Solution

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