 Solution

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

In terms of

R = ℓ_{2} / ℓ_{1}

λ_{\pm} (R) = \frac{ω _{\pm}^{2}}{ω _{0}^{2}} = \frac{1}{2} + \frac{1}{R} \pm \frac{1}{4} + \frac{1}{R ^{2}} .

Both curves decrease monotonically: with

q = 1/ R

d λ_{\pm} / d q = 1 \pm q / 1/4 + q^{2} > 0

, while

d q / d R < 0

. The limits, including their leading approach, are

R \to 0^{+} : R \to \infty : λ_{-} = \frac{1}{2} - \frac{R}{8} + O (R^{3}) \to \frac{1}{2}, λ_{+} = \frac{2}{R} + \frac{1}{2} + O (R) \to \infty; λ_{-} = \frac{1}{R} - \frac{1}{R ^{2}} + O (R^{- 4}) \to 0, λ_{+} = 1 + \frac{1}{R} + O (R^{- 2}) \to 1.

Thus the limiting squared angular frequencies are

ω_{0}^{2} /2

and infinity for short joining fibre, and zero and

ω_{0}^{2}

for very long joining fibre. For orientation, at

R = 1

the values are

(3 \pm 5) ω_{0}^{2} /2

. The high-frequency curve stays above

ω_{0}^{2}

and the low-frequency curve below

ω_{0}^{2} /2

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