Reflection of wave at open end (boundary condition)

[Joker93]

Hello!
I need to understand one seemingly simple thing in wave mechanics, so any help is much appreciated!
When a pulse 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 this think called reflection? 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?

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!

 

[PietKuip]

The ring goes twice as high because the inertia is only half of the inertia of points not near the boundary.

I once made this animation to help with intuition:
https://commons.wikimedia.org/wiki/File:Reflection_at_fixed_and_free_string_ends_single_pulse.gif

The red line is used for constructing an "image wave". Superposition with a pulse of opposite sign keeps the endpoint fixed. To get the ring go up twice as high, the image wave needs to be of the same sign.

 

[Joker93]

The ring goes twice as high because the inertia is only half of the inertia of points not near the boundary.

I once made this animation to help with intuition:
https://commons.wikimedia.org/wiki/File:Reflection_at_fixed_and_free_string_ends_single_pulse.gif

The red line is used for constructing an "image wave". Superposition with a pulse of opposite sign keeps the endpoint fixed. To get the ring go up twice as high, the image wave needs to be of the same sign.

The image wave is something that we constructed to help us visualize what is happening during the reflection process.. But, what is causing the point at the boundary to behave like this? And I am only talking about the actual wave. Can you explain it to me without using the image wave but using forces or energy?

 

[PietKuip]

The restoring force on the ring comes only from one side, not from both sides

Can you explain it to me ... using forces or energy?

The restoring force on the ring comes only from one side. So it is half as large as for the pulse when it as away from the end.

 

[bcrowell]

Let I be the amplitude of the incident wave, R the amplitude of the reflected one, and T the amplitude of the transmitted one.

In the case you're talking about, T=0. Therefore by conservation of energy, the reflected wave, once it has moved past the entire incident wave and is no longer superposing with it, must have the same squared amplitude that the incident wave had, $R^2=I^2$. The two possibilities are $R=I$ and $R=-I$.

Which of these two possibilities we actually get depend on the type of boundary conditions. Let the wave be given by the function y(x), and let the boundary be at x=0. The two most common types of boundary conditions are y(0)=0 and y'(0)=0. In the former case, we must have $R=-I$ so that the reflected wave cancels the incident one at x=0. In the latter case, we need $R=-I$ so that the incident and reflected waves have opposite derivatives.

The energy argument is generic for all waves that can be specified by a single scalar function y(x). An argument about forces is only going to apply to one type of wave (e.g., waves on a string), and will not be as interesting.

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?).

In the situation you're talking about, there is no transmitted wave. The ratio of the amplitude of the *reflected* wave to that of the incident wave, R/I, is 1. It's their sum I+R that is twice the amplitude of I, I+R=2I.

 

[Joker93]

Let I be the amplitude of the incident wave, R the amplitude of the reflected one, and T the amplitude of the transmitted one.

In the case you're talking about, T=0. Therefore by conservation of energy, the reflected wave, once it has moved past the entire incident wave and is no longer superposing with it, must have the same squared amplitude that the incident wave had, $R^2=I^2$. The two possibilities are $R=I$ and $R=-I$.

Which of these two possibilities we actually get depend on the type of boundary conditions. Let the wave be given by the function y(x), and let the boundary be at x=0. The two most common types of boundary conditions are y(0)=0 and y'(0)=0. In the former case, we must have $R=-I$ so that the reflected wave cancels the incident one at x=0. In the latter case, we need $R=-I$ so that the incident and reflected waves have opposite derivatives.

The energy argument is generic for all waves that can be specified by a single scalar function y(x). An argument about forces is only going to apply to one type of wave (e.g., waves on a string), and will not be as interesting.
In the situation you're talking about, there is no transmitted wave. The ratio of the amplitude of the *reflected* wave to that of the incident wave, R/I, is 1. It's their sum I+R that is twice the amplitude of I, I+R=2I.

yes, but how can we explain these with forces on the string?