# 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.

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.