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ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2015/11/c/solution

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Past exam of the mathematical methods II course of the University of Oxford / 2015 / 11 / c / Solution by

Codex  0 2026-10-25

Place the massless junction at

x = 0

, with common tension

T

on both sides. Using a common time factor

e^{- iω t}

, write the left-hand wave as

A_{i} e^{i k_{1} x} + A_{r} e^{- i k_{1} x}

and the right-hand wave as

A_{t} e^{i k_{2} x}

. A stationary junction does not change the frequency, while

k_{j} = ω ρ_{j} / T

.

displacement continuity requires

A_{i} + A_{r} = A_{t}

. Transverse force balance on a massless junction requires equal tension-weighted slopes, giving

k_{1} (A_{i} - A_{r}) = k_{2} A_{t}

. Hence the signed displacement coefficients are

r = \frac{A _{r}}{A _{i}} = \frac{k _{1} - k _{2}}{k _{1} + k _{2}}, t = \frac{A _{t}}{A _{i}} = \frac{2 k _{1}}{k _{1} + k _{2}} .

Here

ρ_{2} = 4 ρ_{1}

, so

k_{2} = 2 k_{1}

and

A_{r} = - A_{i} /3, A_{t} = 2 A_{i} /3

. The negative reflected coefficient signifies inversion. If “reflection and transmission coefficients” instead denotes power fractions, their values are

R = 1/9

and

T = 8/9

, as derived separately below; power and displacement coefficients must not be conflated.

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