 Solution

ID: past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2016-resit/10/c/i/solution

y_{1} = (A e^{i k_{1} x} + B e^{- i k_{1} x}) e^{- iω t}

and

y_{2} = C e^{i k_{2} x - iω t}

. At a massless join, continuity of displacement and transverse force gives

A + B = C, T k_{1} (A - B) = T k_{2} C .

Eliminating

B

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

The angular frequency is shared across a stationary interface, whereas the wavelength changes. The equal-tension assumption is used in the slope condition.

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