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:
- The string is continuous
- The slope of the string is continuous
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:
- Uniform String: $\mu_{1} = \mu_{2}$ or $v_{1} = v_{2}$
- Solid(inifinite?) Wall at $x = 0$: $\mu_{2} \rightarrow \infty$ or $v_{2} = 0$
- Zero mass string for $x > 0$: $\mu_{2} \rightarrow 0$ or $v_{2} = \infty$
- Massless ring on vertical, frictionless pole at $x = 0$: $\tau_{2} = 0$ $\rightarrow$ $Z_{2} = 0$
- Massive ring on vertical, frictionless pole at $x = 0$:
- $\lim_{m \rightarrow 0} \ \partial_{x} \psi = 0$ $\Rightarrow$ $R$ = 1 and $T$ = 2
- $\lim_{m \rightarrow \infty} \ \partial_{tt} \psi = 0$ $\Rightarrow$ $R$ = -1 and $T$ = 0
Footnotes
- The ring must be massless to maintain the boundary conditions without requiring an infinite force to do so. This results because we require continuity in slope and tension at the junction. A finite mass would also result in $Z_{2} \neq 0$, as it would act like an second tension.
- Assume the ring is on a frictionless vertical rod.
- As an aside, one should note that BCs and differential equations are the primary constituents of a problem. This is relevant because the superposition rule was not used in Equations 12a and 12b in contrast to the approach used in earlier sections. The use of a superposition is just one of many possible methods one can use to solve the differential equations but is not required and may not apply in some circumstances. That is, the BCs and differential equations exist independent of whether one can apply the superposition rule.
References
- French, A.P. (1971), Vibrations and Waves, New York, NY: W. W. Norton & Company, Inc.; ISBN:0-393-09936-9.
- Whitham, G.B. (1999), Linear and Nonlinear Waves, New York, NY: John Wiley & Sons, Inc.; ISBN:0-471-35942-4.
Best Answer
As with light, radio waves are electromagnetic waves. As a result, they would obey the same behaviour. The only issue is the wavelength. Radio waves have longer wavelengths (lower frequencies) than light. So if the particular phenomenon depends on the wavelength, then that would affect how the radio waves would behave. Keep in mind that some material properties, such as the refractive index (dielectric constant) depends on wavelength.