If the right string is a ``brick wall" then (i.e. $\mu_r \to \infty$ ) then the transmitted amplitude is 0. If the right string is identical to the left string (i.e. $\mu_r = \mu_l$ ) then the transmitted amplitude is 0. Both situations are possible, so in fact answers (A) and (E) are both wrong.

Now, consider the case where $\mu_r \to \infty$ again. In this limit, answers (B) and (D) give 2 and -1 respectively. Therefore, these answers must both be wrong. This eliminates all answers except answer (C), therefore answer (C) is the correct one. And indeed, answer (C) goes to 0 as $\mu_r \to \infty$ .

\section*{alternative solution:}

Let
$\begin{align*}\tilde{f}_I &= \bar{A}_I e^{i(k_1 z - \omega t)} \\\tilde{f}_R &= \bar{A}_R e^{i(-k_1 z - \omega t)} \\...$
be the incident, reflected, and transmitted waves respectively. Actually, the real parts of these functions are the incident, reflected, and transmitted waves respectively.
The wave must be continuous at $z = 0$ . Also, the derivative of the wave must be continuous at $z = 0$ because the problem statement does not gives us the mass of the knot. Therefore,
$\begin{align*}\bar{A}_I + \bar{A}_R &= \bar{A}_T \\k_1 \bar{A}_I - k_1 \bar{A}_R &= k_2 \bar{A}_T\end{align*}$
$\bar{A}_T$ can be easily found from these equations:
$\begin{align}\bar{A}_T = \bar{A}_I \frac{2 k_1}{k_1 + k_2} = \bar{A}_I \frac{2 k_1/ k_2}{k_1/k_2 + 1} \label{eqn81:1}\end{ali...$
Recall the following relationships:
$\begin{align*}\frac{\omega}{k_1} &= v_1 = \sqrt{T/\mu_l} \\\frac{\omega}{k_2} &= v_2 = \sqrt{T/\mu_r}\end{align*}$
These imply that
$\begin{align*}k_1/k_2 = \sqrt{\mu_l/\mu_r}\end{align*}$
Plugging this into equation 1 gives
$\setcounter{equation}{1}\begin{align}\bar{A}_T = \bar{A}_I \frac{2 \sqrt{\mu_l/\mu_r}}{\sqrt{\mu_l/\mu_r} + 1} \label{eqn81:2...$
Presumably, the amplitudes referred to in the problem statement are the real amplitudes $A_T = \left|\bar{A}_T \right|$ and $A_I = \left|\bar{A}_I \right|$ . By taking the modulus of both sides of equation 2, we find that
$\begin{align*}A_T = A_I \frac{2 \sqrt{\mu_l/\mu_r}}{\sqrt{\mu_l/\mu_r} + 1}\end{align*}$
We are given that $A_I = 1$ . Therefore,
$\begin{align*}A_T = \boxed{ \frac{2 \sqrt{\mu_l/\mu_r}}{\sqrt{\mu_l/\mu_r} + 1}}\end{align*}$
Therefore, answer (C) is correct.

Solution to 1996 Problem 81