If the right string is a ``brick wall" then (i.e. ) then the transmitted amplitude is 0. If the right string is identical to the left string (i.e. ) 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 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 . \section*{alternative solution:} Let 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 . Also, the derivative of the wave must be continuous at because the problem statement does not gives us the mass of the knot. Therefore, can be easily found from these equations: Recall the following relationships: These imply that Plugging this into equation 1 gives Presumably, the amplitudes referred to in the problem statement are the real amplitudes and . By taking the modulus of both sides of equation 2, we find that We are given that . Therefore, Therefore, answer (C) is correct. |

Please send questions or comments to X@gmail.com where X = physgre.