# Wave reflection and open end boundary condition intuition

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"

• I am not entirely sure what you are looking for beyond what people have given as answers (myself included). I discussed why the massless ring responds to an incident pulse using both forces and boundary conditions. The transmission/reflection coefficients I discussed are proxies for energy (since the energy of a wave is often proportional to its amplitude squared). The most intuitive way to look at this is from the boundary conditions. So could you explain what is missing in the present answers? – honeste_vivere Oct 8 '15 at 22:36
• I was seeking something along the lines of my own answer (see it below). You have explained it by forces but not in the detailed way as to give an understanding of why there is reflection, but just by which reflection it is. Also, I was seeking an explanation of why the ring moves 2A (see at my answer, maybe yoy can extend it) – TheQuantumMan Oct 9 '15 at 6:46

# 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:

1. The string is continuous
2. 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 $v_{1} = v_{2}$
• $R$ = 0 and $T$ = 1
• Solid(inifinite?) Wall at $x = 0$: $\mu_{2} \rightarrow \infty$ or $v_{2} = 0$
• $R$ = -1 and $T$ = 0
• Zero mass string for $x > 0$: $\mu_{2} \rightarrow 0$ or $v_{2} = \infty$
• $R$ = 1 and $T$ = 2
• Massless ring on vertical, frictionless pole at $x = 0$: $\tau_{2} = 0$ $\rightarrow$ $Z_{2} = 0$
• $R$ = 1 and $T$ = 2
• 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

1. 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.
2. Assume the ring is on a frictionless vertical rod.
3. 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.
• Hi honeste_vivere. Could you please help me understand why the masslessless of the ring is so crucial? What do we lose if we have a massive ring. After all, the string supporting transverse oscillations is not massless either. It has a fixed mass density. – 299792458 Apr 17 '18 at 14:18
• @TheDarkSide - It need not be massless in principle, but giving the ring a finite mass adds several complications to the analysis. It would be similar to joining two strings of different mass densities but the ring would have a much more rigid response. I guess the easy answer is that making the ring massless greatly simplifies everything and it's not impossible to make a real system with a very small mass ring compared to the string/rope. – honeste_vivere Apr 17 '18 at 15:04
• Thanks. The reason why I asked is that, the masslessness of the ring is a very crucial requirement from the point of view of the Neumann boundary conditions for the free ends of string problem. With a massive ring, the argument collapses. See here for example, or here for a reference. I find the argument very loose, as explained in the comment discussion below the top answer, and would greatly appreciate if you could help me in understanding this nuance. – 299792458 Apr 17 '18 at 15:18
• @TheDarkSide - I think it's not that everything collapses completely, rather that the solutions become piecewise continuous instead of continuous everywhere. I am inclined to think a massive ring would also have the implication of imposing something akin to damping, like evanescence in the "ring region" rather than the $Z_{2} = 0$ result for a massless ring. It becomes extremely complicated very quickly if the ring does not perfectly fit the rod because then you need to worry about torques and moments of inertia applied to rotate the ring. – honeste_vivere Apr 17 '18 at 15:25
• But is it correct to say that a string of finite length always supports only a discrete spectrum of frequencies, even with one end/ both ends free? So, with both ends rigidly clamped, we get discrete freq's, with one end fixed and one end free, still discrete freq's supported (though they are different kind of modes), and with both ends free, still different freq's but all discrete. So, this system has no continuous spectrum at all? – 299792458 Apr 17 '18 at 15:31

I have found a good explanation in the net to my question, so I am sharing it just in case somebody else wants an answer too. Note that the question about "why it reaches double the amplitude" remains , as well as a new problem on why does the answer say that the ring has momentum(because its massless):

The end of the stationary string, figure (a), is attached to a ring that is free to move in the vertical direction on a frictionless pole. A pulse is now sent down the string in figure (b). Let us consider the forces on the string that come from the pulse, that is, we will ignore the gravitational forces on the string. The arrows up and down on the pulse represent the forces upward and downward, respectively, on the particles of the string. The pulse propagates to the right in figure (c) and reaches the ring in figure (d). The force upward that has been propagating to the right causes the ring on the end of the string to move upward. The ring now rises to the height of the pulse as the center of the pulse arrives at the ring, figure (e). Although there is no additional force upward on the ring, the ring continues to move upward because of its momentum. We can also consider this from an energy standpoint. At the location of the top of the pulse, the ring has a kinetic energy upward. The ring continues upward until this kinetic energy is lost. As the ring moves upward it now pulls the string up with it, eventually overcoming the forces downward at the rear of the pulse, until the string to the left of the ring has a net force upward acting on it, figure (f). This upward force is now propagated along the string to the left by pulling each adjacent particle to its left upward. Because the ring pulled upward on the string, by Newton’s third law the string also pulls downward on the ring, and the ring eventually starts downward, figure (g). As the ring moves downward it exerts a force downward on the string, as shown by the arrows in figure (h). The forces upward and downward propagate to the left as the pulse shown in figure (i). The net result of the interaction of the pulse to the right with the movable ring is a reflected pulse of the same size and shape that now moves to the left with the same speed of propagation. The incoming pulse was right side up, and the reflected pulse is also right side up. The movable ring at the end of the string acts like the hand, moving up and down to create the pulse.

Note the best way to understand this behavior is to consider the simplest case of just a “half pulse”. Generate a pulse by just moving the string upward at some position and let this propagate to the free end (see its behavior), then at some later time bring the string down to its initial position. This will be like a very long pulse, however you can see what happens by adding the forces together in each part of the pulse following their respective reflections from the free end.

Also, here is a simulation so you can all do the very long pulse and see all these stuff in action.

• Hey can you kindly tell me the reference from which you got that beautiful image? – claws May 2 '17 at 13:24
• Aww, sorry I can't even remember since I wrote the answer a relatively long time ago. – TheQuantumMan May 2 '17 at 14:04

This question is very broad so I will try to answer only its parts.

Why do we even have reflection? Impedance of the string is the key. Wave on the string reaches the point with impedance discontinuity and therefore reflection occurs. In your case there is reflection from "zero impedance boundary". The opposite textbook example is the reflection from "infinite impedance boundary". There are of course cases in between, e.g. (highly theoretical) ending of the string on a piston with the same impedance as a string. Wave on the string would therefore still be "travelling".

Intuition on that case: There is an amount of energy carried on the string divided between displacement (i.e. kinetic energy) and tension (i.e. potential energy). At the proposed unbound end there is no factor which could cause tension increase and therefore (from conservation of energy) results a larger displacement.

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? Maybe this counterquestion might help you get it intuitively: What would be the factor or agent turning the amplitude to the opposite direction? For "infinite impedance ending" is something like that present: the impedance. But for "zero impedance ending"...?

Why the double amplitude during reflection? Try to model it (in your mind or numerically). What is happening. The wave is incidenting but in the same time reflecting. It's therefore the sum of both waves and in the infititely small time interval when the maximum hits the boundary, the amplitudes of incidenting and reflecting waves is the same and sum therefore is maximum amplitude times two.

Source recommendation: I strongly recommend the Ian G. Main's book Vibrations and Waves in Physics for more intuition and a further discussion.

• Thank you for your answer.I know about these but the wave traveling the other way is just a trick. I want to have an explanation with forces or energy.. – TheQuantumMan Sep 30 '15 at 13:05
• OK, I'll try to edit the answer to add more intuition. – Victor Pira Oct 2 '15 at 10:22

Let's start with the intuition you already have, for a wave travelling along a continuous string. The transversal motion is a result of the tension from both sides: One, where the excitation (or wave packet) is coming from, is pulling a piece of string at the front of the wave outwards, whilst the other, where the wave is only just about to arrive, is resisting this motion by still being at rest. Similar but opposite forces balance in such a way that the excitation continues to travel forward, leaving any piece of the string at rest once it passes that piece.

If you remove the side the wave is travelling to, you get a larger amplitude because this half, otherwise resisting the transverse motion, is missing. The other half, from where the wave packet arrived, will hence get pulled further, more than what would suffice to simply end its motion after the wave packet passes it if it were in the middle of the string. So you have a new excitation in which the very end of your string is acting on adjacent bits of the same string: There is a new wave that travels back.

If you realize that a perfect string for such waves will be perfectly elastic and hence conserve mechanical energy, you can also argue that the energy carried by your wave packet has to continue to exist in some form. The momentarily stretched, open ended string will naturally recoil, which constitutes a new wave (or, if you like, the old one reflected).