# Shouldn't reflection at the boundary interfere with the original wave to not give any wave?

So I have been taking a wave course, and one thing that I don't understand is how does reflection even allows normal modes of vibration. I'll try to explain my confusion:

Suppose we have a string which is fixed at one of its ends, and I set up a travelling wave of some frequency on the string. This wave travels along the string in the positive x direction and then it reaches the fixed end. Upon reflection a reflected wave with phase shifted by π travels in the negative x direction and so this means the two destructively interfere and give no wave; in all the books I have read (French, Crawford, Howard...) they say that setting up a travelling wave with a certain specific frequency establishes a standing wave on the string, but shouldn't the two reflected wave no matter what the frequency destructively interfere and give no wave?

So my question is:

1. Does reflection of wave always produce a π phase shifted reflected wave?
2. If it does then how does the reflected wave and the original wave interfere to give a standing wave?

I do understand how mathematically solving the wave equation with the proper boundary conditions gives the normal mode/ stationary wave solution and also know how the method of images is incorporated but yet I don't see why the virtual or image pulse should have the same phase while the reflected wave( hence the virtual wave) must be π shifted in case of rigid fixed ends.

Thank you for your time and help.

## 1 Answer

You're right: the reflected wave, being shifted by $$\pi$$, interferes destructively with the incoming wave and gives a zero total disturbance at the end of the string.

If you think about it, the $$\pi$$ phase shift actually derives from the fact that you want a zero disturbance at the boundary: you indeed derive it by applying the boundary condition to the solution of the wave equation. You can think of it as being an "ad hoc" shift to fit the boundary.

The reason why they only interfere destructively at the boundary is that the incoming and reflected waves travel in different directions, so at every point of the string the total disturbance (the sum of the two waves, therefore their interference effect) is time dependent. The only points in which the total disturbance is constant is at the boundary, because "you wanted it to be that way" upon applying the boundary condition, and in the nodes of the stationary wave, because they are the particular points of the string in which the contribution of each wave to the total disturbance changes with time, but the two balance out at every instant to give a zero overall displacement.

Interestingly enough, if the reflection actually caused destructive interference everywhere, you would only need a mirror to eliminate every electromagnetic field arriving at it at normal incidence (with the other appropriate boundary conditions to obtain standing waves of course).

• I get that boundary conditions require the π phase shift because the fixed ends have to be of course nodes but what I don't get is how then can you have anti nodes or any wave at all in the rest of the string because the π shift means they perfectly cancel out algebraically everywhere so how do you physically get a standing wave. – Lucifer Mar 21 '19 at 18:36
• @Lucifer They would only cancel if they were phase shifted and traveling in the same direction. But because they travel in different directions, there are points and times when they do not cancel. – BowlOfRed Mar 21 '19 at 18:46
• Why is that so could u explain a bit more about it, like as to why the two phase shifted waves travelling in opposite direction would not cancel out for all time and space? – Lucifer Mar 21 '19 at 19:34
• Ohh I finally understood it thanks, I went back played around with the sine graphs and also superimposed two π shifted waves travelling in opposite direction and got the standing wave equation.. thanks everyone for their help. – Lucifer Mar 21 '19 at 20:17