# 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 $$y_{r} = -A \sin (kx+wt).$$ 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

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). Feb 20 '20 at 23:23
• @ArunArora why do you say so? Feb 21 '20 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 Feb 21 '20 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. Feb 21 '20 at 9:05