= Solution
Place the junction at $x=0$ and use the <complex amplitude> convention $e^{-i\omega t}$. With incident displacement <wave amplitude> $A_i$, write
$$
y_L=A_i(e^{ik_Lx}+r e^{-ik_Lx})e^{-i\omega t},\qquad
y_R=A_i t e^{ik_Rx}e^{-i\omega t}.
$$
Here $r$ and $t$ are displacement <wave reflection> and <transmission coefficients>, and $k_j=\omega/c_j$. The junction is assumed massless, with no externally applied transverse <force>, and both strings have the common equilibrium <tension> $T$. Continuity of displacement and transverse <force> gives
$$
1+r=t,\qquad Tk_L(1-r)=Tk_Rt.
$$
Since $Tk_j=\omega Z_j$, these become $1+r=t$ and $Z_L(1-r)=Z_Rt$. Solving,
$$
\boxed{\frac{A_r}{A_i}=r=\frac{Z_L-Z_R}{Z_L+Z_R},\qquad
\frac{A_t}{A_i}=t=\frac{2Z_L}{Z_L+Z_R}.}
$$
A larger right-hand <mechanical impedance> produces a reflected displacement with reversed sign, as expected in the fixed-end limit $Z_R\to\infty$. Equal <mechanical impedances> give $r=0$ and $t=1$.
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