I need to understand one seemingly simple thing in wave mechanics, so any help is much appreciated!
When a pulse on a string travels to the right toward an open end(like a massless ring that is free to oscillate only in the vertical direction), then when the wave reaches the end it gets reflected and it becomes a positive pulse traveling to the left. So, my first question is, why do we even have reflection in this case(as far as energy and forces go)? What causes it in physical terms(intuitively)? And why for an open end the reflected wave is a positive wave and why for a closed end the reflected wave is a negative wave? An explanation using the tension of the string and energies will be much appreciated.
Also, for the open end case, the massless ring will have an amplitude of two time the amplitude of the incident wave while the wave gets reflected. Why is this the case? (or to state it differently why does the ratio of the amplitude of the transmitted wave divided by the amplitude of the incident wave equal to +2?).
Thank you!
ΝΟΤΕ: I want an explanation using forces or energy and not "image waves"
Best Answer
Background
Let $\tau$ be the tension and $\mu$ be a linear mass density (i.e., mass per unit length), then the wave equation for a string is given by: $$ \partial_{tt} \psi \left(x,t\right) - \frac{ \tau }{ \mu } \partial_{xx} \psi \left(x,t\right) = 0 \tag{0} $$ where $\partial_{jj} \equiv \partial^{2}/\partial j^{2}$ and $\psi \left(x,t\right)$ is a general solution to this equation, called the wave equation.
This has a simple solution of the form: $$ \psi \left(x,t\right) = A \ e^{i \left( \pm \mathbf{k} \cdot \mathbf{x} \pm \omega t \right)} \tag{1} $$ where $A$ is some amplitude and the phase speed of the wave is given by: $$ \frac{\omega}{k} = \sqrt{\frac{ \tau }{ \mu }} \equiv C \tag{2} $$
We want to find solutions of the form $f\left( x - C \ t \right)$, but this only works for non-dispersive waves and does not work for nonlinear waves. In other words, the solution applies when the wave's phase speed is $C$ = constant.
Reflection and Transmission
First, assume $\tau$ is uniform throughout the string to avoid any unwanted acceleration. Next, let us define a general form: $$ \psi_{j} \left(x,t\right) = f_{j} \left(x - v_{j} t\right) = f_{j} \left(t - \frac{x}{v_{j}} \right) \tag{3} $$ where the subcript $j$ = $i$ for incident, $r$ for reflected, and $t$ for transmitted waves. Now let us assume there is some boundary at $x$ = 0 and that our string has different mass densities on either side. Let's define $\mu_{1}$ for Region 1 (-$\infty < x < 0$) and $\mu_{2}$ for Region 2 ($0 < x < \infty$). Then we have: $$ \begin{align} v_{1} & = \sqrt{\frac{ \tau }{ \mu_{1} }} \tag{4a} \\ v_{2} & = \sqrt{\frac{ \tau }{ \mu_{2} }} \tag{4b} \end{align} $$
Note that the reflected wave, $\psi_{r} \left(x,t\right)$, will have a negative $v_{r}$ and thus a positive sign in the expression for $f$. Since the waves are linear, we can just write them a linear superposition of two waves for Region 1. Then we have: $$ \begin{align} \psi_{1} \left(x,t\right) = \psi_{i} \left(x,t\right) + \psi_{r} \left(x,t\right) = f_{i} \left(t - \frac{x}{v_{1}} \right) + f_{r} \left(t + \frac{x}{v_{1}} \right) \tag{5a} \\ \psi_{2} \left(x,t\right) = \psi_{t} \left(x,t\right) = f_{t} \left(t - \frac{x}{v_{2}} \right) \tag{5b} \end{align} $$
Boundary Conditions
There are two boundary conditions (BCs) that must be met:
These can be written mathematically as: $$ \begin{align} \psi_{1} \left(0,t\right) & = \psi_{2} \left(0,t\right) \tag{6a} \\ \partial_{x} \psi_{1} \left(x,t\right) \vert_{x = 0} & = \partial_{x} \psi_{2} \left(x,t\right) \vert_{x = 0} \tag{6b} \end{align} $$ where these equations can be rewritten in terms of $f_{j}$ (and integrating the second) to find: $$ \begin{align} f_{i} \left(t - \frac{x}{v_{1}} \right) + f_{r} \left(t + \frac{x}{v_{1}} \right) & = f_{t} \left(t - \frac{x}{v_{2}} \right) \tag{7a} \\ v_{2} \left[ \ f_{i} \left(t\right) - f_{r} \left(t\right) \right] & = v_{1} f_{t} \left(t\right) \tag{7b} \end{align} $$ We can solve these two equations for $f_{r}$ and $f_{t}$ in terms of $f_{i}$ to find: $$ \begin{align} f_{r} & = \left( \frac{ v_{2} - v_{1} }{ v_{1} + v_{2} } \right) \ f_{i} \tag{8a} \\ f_{t} & = \left( \frac{ 2 \ v_{2} }{ v_{1} + v_{2} } \right) \ f_{i} \tag{8b} \end{align} $$
Coefficients/Amplitudes
We can see from the last two equations that the amplitudes of the reflected ($R$) and transmitted ($T$) wave are given by: $$ \begin{align} R & = \left( \frac{ v_{2} - v_{1} }{ v_{1} + v_{2} } \right) = \left( \frac{ \sqrt{ \mu_{1} } - \sqrt{ \mu_{2} } }{ \sqrt{ \mu_{1} } + \sqrt{ \mu_{2} } } \right) \tag{9a} \\ T & = \left( \frac{ 2 \ v_{2} }{ v_{1} + v_{2} } \right) = \left( \frac{ 2 \ \sqrt{ \mu_{1} } }{ \sqrt{ \mu_{1} } + \sqrt{ \mu_{2} } } \right) \tag{9b} \end{align} $$
Massless Ring
A massless ring1 at one end of a string2 is treated as a form of impedence. Because the ring is massless, we require that the net transverse (i.e., orthogonal to the direction of wave propagation, say, along the x/horizontal direction) force be zero. A finite transverse force would result in an infinite acceleration. The only difference in this case is that we need to use a non-uniform tension. So we just follow the same steps as above but use $\tau_{j}$ for Region $j$ and so we have: $$ \begin{align} \tau_{1} \ \sin{\theta_{1}} & = \tau_{2} \ \sin{\theta_{2}} \tag{10a} \\ \tau_{1} \ \partial_{x} \psi_{1} \left(x,t\right) \vert_{x = 0} & = \tau_{2} \ \partial_{x} \psi_{2} \left(x,t\right) \vert_{x = 0} \tag{10b} \end{align} $$ where the angles, $\theta_{j}$, are relative to the x/horizontal direction. We can define the impedence as $Z_{j} = \tau_{j}/v_{j} = \sqrt{ \tau_{j} \ \mu_{j} }$, which allows us to redefine the reflection ($R$) and transmission ($T$) coefficients as: $$ \begin{align} R & = \left( \frac{ Z_{1} - Z_{2} }{ Z_{1} + Z_{2} } \right) \tag{11a} \\ T & = \left( \frac{ 2 \ Z_{1} }{ Z_{1} + Z_{2} } \right) \tag{11b} \end{align} $$
Massive Ring
In contrast to a massless ring, a massive ring requires an alteration of the BCs since we now need to include Newton's laws. We can assume the string applies a force and the massive ring undergoes an acceleration, allowing us to write: $$ \begin{align} F & = \tau \partial_{x} \psi \left(x,t\right) \tag{12a} \\ m \ a & = m \ \partial_{tt} \psi \left(x,t\right) \tag{12b} \end{align} $$ Note that $F$ in Equation 12a is the vertical force on the ring due to the tension in the string, $m$ in Equation 12b is the mass of the ring, and $a$ in Equation 12b is the acceleration of the ring3.
We can see that in the limit as $m \rightarrow 0$ we have $\partial_{x} \psi \rightarrow 0$, thus the force is null as is necessary for a massless system. We also see that as $m \rightarrow \infty$ we have $\partial_{tt} \psi \rightarrow 0$, which implies a constant velocity for the massive ring (i.e., it would be zero here as the initial condition is that it starts from rest).
Boundary Examples
Now we can provide a few useful examples:
Footnotes
References