# Energy Conservation of waves at a boundary

Consider a wave traveling on a string with velocity $\upsilon$ and mass density $\rho$ having unit length so that the mass of the string is $\rho$. Considering the string to be a simple harmonic oscillator of maximum amplitude $A$ , we know that it's total energy will be E = (1/2)$\rho$$\omega^2$$A^2$. The impedance of the string is given by $\rho$$c. Now consider this wave to be incident on a boundary where this string is connected smoothly (so that they have the same tension) to other string of different impedance at x=0. Let Z_1 = \rho_1$$c_1$ = $T$/$c_1$ where T is the tension of the string be the impedance of the string to the left of the boundary and $Z_2$ = $\rho_2$$c_2 = T/c_2 be the impedance to the right of the boundary. The incident wave coming from the left side of the boundary is given by y_i = A_1exp(\omega$$t$-$k_1$$x) and the reflected is given by y_r = B_1exp(\omega$$t$+$k_1$$x) and transmitted wave is given by y_t = A_2exp(\omega$$t$-$k_2$$x). Now I have the following Questions. • The wave to the left and to the right of the boundary will have the same frequency. How can one physically or mathematically assure this? • While finding out reflection and transmission co-efficients, we use y_i + y_r = y_t. I don't understand as to why should the amplitude of the transmitted wave be the sum of the incident and the reflected waves. • While using this math to explain the sound produced by the whip, we assume Z_2 = 0. But is this explanation sufficient because what basically Z_2 = 0 means is that there are no particles to the right of the boundary. Then how does the wave travel? According to the math above it will but the physics that this concept was built on tells us that is there are no particles to transmit waves, then the wave shouldn't be traveling. What is the correct explanation? • Now if I try to conserve energy at the boundary, by using (1/2)\rho_1$$\omega^2$$A_1^2 = (1/2)\rho_2$$\omega^2$$A_2^2 + (1/2)\rho_1$$\omega^2$$B_1^2 it turns out that the energy is not constant but energy*velocity is a constant. How can one explain this and why is just energy not constant? Also this question at the end might not be connected to the above question but this is something that I have been wondering from the beginning of waves and vibrations course. Why should a system containing n particles have exactly n normal modes? What is the exact meaning of a normal mode can someone physically or mathematically explain the above question? I know the question contains lots of subquestions and I don't mind people answering any or all of the above questions but I need answers please. ## 1 Answer • The frequencies are the same left and right of the boundary because you assume only sinusoid time variations ∝\exp{(i\omega t)} with angular frequency \omega. • The strings are connected at the junction. Therefore the y-coordinate (elongation) of the left and the right string has to be the same at the junction. • Assuming the junction to be at x=0, and that there is an incident, a reflected and a transmitted wave, the exponential wave expressions yield for the amplitudes$$A_1+B_1=A_2$$• This point is not clear. Why do you talk of the sound of a whip? Do you mean that there is no string to the right (Z_2=0 means T_2=0) and you assume an open end for the left string? The question then is how can you get a wave the when T_1=T_2=0 and thus C_1=0? • The expression$$\rho \omega^2 A^2=\rho v_y^2$$corresponds to the (kinetic+potential) energy per unit length of a wave. At the junction the sum of all waves to the left and to the right have the same velocity$v_y$but different$\rho$(and different$c$). Thus the energy densities cannot be the same. On the other hand, for energy conservation, the sum of energy fluxes$P$of the waves (wave power) has to be conserved.$P$is transported energy per time$P=c\rho \omega^2 A^2/2$, • If the y-coordinates are the same then, shouldn't it be$A_1$=$A_2$only. – Sundesh Oct 30 '16 at 19:03 • Because at$x=0$you have an incident wave with amplitude$A_1$and a reflected wave with amplitude$B_1$, the total amplitude of the oscillation of the left string at$x=0$is$A_1+B_2$. This has to be equal to the amplitude$A_2$of the transmitted wave at$x=0\$. – freecharly Oct 30 '16 at 19:15
• I understood your point. – Sundesh Oct 30 '16 at 19:24
• And for the whip, the right side of the boundary is taken to have zero impedance which implies free end and also that the linear mass density is zero to the right. But the left side of the boundary is still present and when you do the math, it tells us that the amplitude of the transmitted wave is twice that of the incident wave. But my question is, the physics that we have built these concepts on till now tells us that no particles in the medium implies wave cannot travel in the medium (because wave consists of infinitely many simple harmonic oscillators). How can one explain this? – Sundesh Oct 30 '16 at 19:24
• And for the energy conservation, why will they have same velocity? Shouldn't it be different because they have different amplitudes? And my question exactly is why energy flux is conserved and not energy? – Sundesh Oct 30 '16 at 19:32