= Solution
At fixed frequency $k_j=\omega/v_j$, so the displacement reflection and transmission coefficients are
$$
\boxed{r=\frac{v_2-v_1}{v_2+v_1},\qquad t=\frac{2v_2}{v_2+v_1}.}
$$
The heavy-string speed is $v_h=20\,\mathrm{m\,s^{-1}}$ and the light-string speed is $v_l=40\,\mathrm{m\,s^{-1}}$. Therefore
$$
\boxed{r_{h\to l}=\frac13,\quad t_{h\to l}=\frac43,\qquad
r_{l\to h}=-\frac13,\quad t_{l\to h}=\frac23.}
$$
A displacement amplitude greater than the incident amplitude does not mean energy is created: the transmitted pulse also has a different width and propagates in a different medium.
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