Look at y(x,t)= f(x- ct) alone. Suppose you were to move along the x-axis with velocity c, starting, say, at x(0)= . Then at any t, you are at . Looking at the wave, you would see [tex]y(x,t)= y(ct+ x_0,t)= f(ct-ct)= f(0)[\math] which does not depend on t: you keep seeing the same thing, you are always on the same point on the wave. That is because, as you are moving along the x-axis, the wave is moving with you. y= f(x- ct) describes a wave that is moving along the x-axis with velocity c.

Now look at y(x,t)= f(-x- ct). If you were to move along the x-axis with velocity c, as before, f(x0+ ct,t)= f(-x0-ct-ct)= f(-x0-2ct) which varies with t. You are NOT moving with the wave. But if you move with velocity -c, then and which does not depend on t. This represents a wave moving with velocity -c.

y(x,t)= f(x- ct)- f(-x-ct) represents y as the difference of two identical waves, which is why f(x,0)=0 for all x, moving at the same speed but in opposite directions.

(ii) If, by contrast,

y(x, t) := f(x − ct) + f(−x − ct)

show that y is once again a solution of the wave equation. What

boundary condition does y now satisfy at x = 0? Describe what

is happening here in terms of incident and reflected waves.[/quote]

Now this is thesumof two waves moving at the same speed but opposite directions. Now y(x, 0)= 2f(x).

I can do it all except for the explanation bit.. I'm guessing that in the first one, the reflected wave is coming back along the line of the incident wave, and in the second one, its waves are reversed.. but this is really a guess

any help much appreciated