Some confusions regarding wave motion on a string

Question

Here is the list:

1. While deriving the formula of fundamental frequency for a string fixed at both ends, the two waves coming from both ends are taken as $Asin(kx-wt)$ and $Asin(kx+wt+\phi).$ After applying the boundary condition that $y=0$ at $x=0$, we get that $\phi=0$. but according to my book, standing waves are obtained when waves are produced at one fixed end and then they interfere with the reflected wave. But the reflected wave is inverted, right? Doesn't this mean that the phase difference between the interferering waves if $\pi$? How can it be 0?

2. If I take the interfering waves as $Asin(wt-kx)$ and $Asin(wt+kx+\phi)$ instead, it shouldn't make any differece, right? I've just cahnged the intitial phase. But now, after applying the boundary condition, I get $\phi=\pi$.

3. The incoming wave is taken as Asin(kx-wt) and the reflected wave is taken as Asin(kx+wt+$\phi$). But how can we use the same 'x' in both the equations? The second equation gives the displacement at a distance x from the end and the x in first equations is measured from the origin. I'm guessing that the equatuon of the displacement produced by reflected wave should be Asin(k(L-x)+wt+$\phi$)

There's one more but it's related to sound waves:

The formula for the apparent wavelength of sound waves when the source is moving is $\lambda^{'}=\frac{v-u}{v}\lambda$ which is constant as $v$, $u$ and $\lambda$ are constants. Bu there is a diagram in my book showing the spherical wavefronts originating from a moving point source. In the diagram, the distance between successisive wavefronts decreases with time and hence wavelength varies.

Answer 1

If you add your two sine functions together you get a term which is position dependent and a term which is time dependent.

$$y=\sin(kx-\omega t)+ \sin(kx+\omega t + \phi) = 2\sin\left(kx+\frac \phi 2\right)\cos\left(\omega t-\frac \phi 2\right)$$

You will see from that equation you can satisfy your boundary condition by choosing an appropriate value of $\phi$.

If you choose that $y=0$ at $x=0$ then $\phi=0$ satisfies the boundary condition and $y = 2 \sin(kx)\cos(\omega t)$.
Indeed $\phi = n\pi$ where $n$ is an integer will also satisfy that boundary.

If you make $\phi = \frac \pi 2$ then $y = 2 \cos(kx)\sin(\omega t)$ and you obviously satisfy a different boundary condition.