 Solution

ID: past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2021/paper-1/section-b/b7/c/solution

q = ℓ_{1} / ℓ_{2}

ϕ_{j} = A_{j} e^{iω t}

, with real physical displacements obtained at the end. In units of

ω_{0}^{2}

K = (1 + q - q - q q) .

λ = ω^{2} / ω_{0}^{2}

, the eigenvalue condition is

det (K - λ 1) = (1 + q - λ) (q - λ) - q^{2} = λ^{2} - (1 + 2 q) λ + q = 0.

Therefore

ω_{\pm}^{2} = ω_{0}^{2} \frac{1}{2} + \frac{ℓ _{1}}{ℓ _{2}} \pm \frac{1}{4} + (\frac{ℓ _{1}}{ℓ _{2}})^{2} .

For finite positive fibre lengths both eigenvalues are positive. The corresponding amplitude ratio is

\frac{A _{2}}{A _{1}} = \frac{q}{q - λ} = \frac{1}{1 - R λ}, R = \frac{ℓ _{2}}{ℓ _{1}} .

The lower normal mode has the spheres twisting in the same sense, and the higher one has opposite signs of angular displacement.

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