Chapter 15, Problem 40P

(III) A cord stretched to a tension F_T consists of two sections (as in Fig. 15–19) whose linear densities are in μ₁ and μ₂. Take x = 0 to be the point (a knot) where they are joined, with μ₁ referring to that section of cord to the left and μ₂ that to the right. A sinusoidal wave, D = A sin[k₁(x – v₁t)], starts at the left end of the cord. When it reaches the knot, part of it is reflected and part is transmitted. Let the equation of the reflected wave be D_R = A_R sin[k₁(x + v₁t)] and that for the transmitted wave be D_T = A_T sin[k₂(x – v₂t)]. Since the frequency must be the same in both sections, we have ω₁ = ω₂ or k₁v₁ = k₁v₂. (a) Because the cord is continuous, a point an infinitesimal distance to the left of the knot has the same displacement at any moment (due to incident plus reflected waves) as a point just to the right of the knot (due to the transmitted wave). Thus show that A = A_T + A_R. (b) Assuming that the slope (∂D/∂x) of the cord just to the left of the knot is the same as the slope just to the right of the knot, show that the amplitude of the reflected wave is given by

$A_{\text{R}}=\left(\frac{{\upsilon }_1-{\upsilon }_2}{{\upsilon }_1+{\upsilon }_2}\right)A=\left(\frac{k_2-k_1}{k_2+k_1}\right)A$.

(c) What is A_T in terms of A?