Standing waves: Analytical treatment

Question

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

Davide Morgante · Answer

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

$$ y(0,t) = Asin(-omega t)+Asin(omega t) = -Asin(omega t)+Asin(omega t) = 0 
 y(L,t) = Asin(2pi -omega t)+Asin(2pi+omega t) = Asin(-omega t)+Asin(omega t) = 0
$$
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

Superfast Jellyfish · Answer

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.

Standing waves: Analytical treatment

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