= Solution
Write the complex displacements as $y_1=(A e^{ik_1x}+B e^{-ik_1x})e^{-i\omega t}$ and $y_2=C e^{ik_2x-i\omega t}$. At a massless join, continuity of displacement and transverse <force> gives
$$
A+B=C,\qquad T k_1(A-B)=T k_2C.
$$
Eliminating $B$ gives the displacement <transmission coefficient> and <reflection coefficient>,
$$
\boxed{t=\frac CA=\frac{2k_1}{k_1+k_2},\qquad
r=\frac BA=\frac{k_1-k_2}{k_1+k_2}.}
$$
The <angular frequency> is shared across a stationary interface, whereas the <wavelength> changes. The equal-tension assumption is used in the slope condition.
\Image[/past-exam-of-the-mathematical-methods-ii-course-of-the-university-of-oxford/2016-resit-string-transmission.png]
{title=Displacement continuity and energy sharing at a light-to-heavy string junction}
Back to article page
