# Standing waves: Analytical treatment

In the analytical treatment of standing wave on a string (of length $$L$$) which is fixed at both ends as given below, incident wave is taken as $$y_{i} = A \sin(kx-wt)$$ and reflected wave is taken as $$y_{r} = A \sin (kx+wt).$$

My question is that we know reflected ray (from rigid end) is out of phase with incident ray by 180 degrees. Thus the relfected ray should rather be $$y_{r} = -A \sin (kx+wt).$$

You are right about the reflected wave having a minus sign. There is a phase shift of $$180^{\text o}$$ when a wave is reflecting from a fixed boundary. However the sign hardly matters for the standing wave apart from a phase shift as can be seen below.

This is because the identity in this case still gives us stationary solution, except for the arguments of sin and cos interchanged.

• Fellow Traveller, but the wavelengths of harmonics will also change if I take reflected wave as y = -A sin(kx-wt). Commented Feb 20, 2020 at 23:23
• @ArunArora why do you say so? Commented Feb 21, 2020 at 4:43
• If I take y(i) = A sin(kx-wt) and y(r) = A sin(kx+wt) then resultant displacement is y = 2A sin(kx) cos(wt) , as y=0 at x=L (since both ends are rigidly fixed) we get, sin(kL) = 0 or kL = n Pi and If I take y(i) = A sin(kx-wt) and y(r) = A sin(kx+wt) then resultant displacement is y = 2A cos(kx) sin(wt) , as y=0 at x=L (since both ends are rigidly fixed) we get, cos(kL) = 0 or kL = (2n+1) Pi/2, here both harmonics are different Commented Feb 21, 2020 at 8:43
• The physical wave will still be the same as can be seen in the gifs above. In both cases $n$ goes over the same values. The difference between the two is just the constant phase shift. Commented Feb 21, 2020 at 9:05

Saying that on the boundaries the two solutions should be out of phase by $$180^\circ$$ is the same as imposing the following boundary conditions

$$y(0, t) = y(L, t) = 0$$

You can easily see that the given solution respects this boundary conditions, in fact

$$y(0,t) = A\sin(-\omega t)+A\sin(\omega t) = -A\sin(\omega t)+A\sin(\omega t) = 0 \\ y(L,t) = A\sin(2\pi -\omega t)+A\sin(2\pi+\omega t) = A\sin(-\omega t)+A\sin(\omega t) = 0$$ If you use your solution you can see that there wouldn't be the opposite sign and so the only way to vanish at the boundary would be if $$A=0$$, so the trivial solution.

In general you should think the out of phase ad the boundaries as

The solution should vanish at the boundaries because it has fixed ends