= Solution
Place the massless junction at $x=0$, with common <tension> $T$ on both sides. Using a common time factor $e^{-i\omega t}$, write the left-hand wave as $A_i e^{ik_1x}+A_r e^{-ik_1x}$ and the right-hand wave as $A_t e^{ik_2x}$. A stationary junction does not change the <frequency>, while $k_j=\omega\sqrt{\rho_j/T}$.
<displacement> continuity requires $A_i+A_r=A_t$. Transverse <force> balance on a massless junction requires equal tension-weighted slopes, giving $k_1(A_i-A_r)=k_2A_t$. Hence the signed <displacement> coefficients are
$$
\boxed{r=\frac{A_r}{A_i}=\frac{k_1-k_2}{k_1+k_2},\qquad t=\frac{A_t}{A_i}=\frac{2k_1}{k_1+k_2}}.
$$
Here $\rho_2=4\rho_1$, so $k_2=2k_1$ and $\boxed{A_r=-A_i/3,\quad A_t=2A_i/3}$. The negative reflected coefficient signifies inversion. If “reflection and transmission coefficients” instead denotes <power> fractions, their values are $\mathcal R=1/9$ and $\mathcal T=8/9$, as derived separately below; <power> and <displacement> coefficients must not be conflated.
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