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Math Help - incident and reflected waves

  1. #1
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    incident and reflected waves

    let
    y(x, t) := f(x − ct) − f(−x − ct).
    Show that y is a solution of the wave equation c2yxx = ytt and that
    y(0, t) = 0 for all t. Describe what is happening here in terms of
    incident and reflected waves.
    (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.

    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
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  2. #2
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    Quote Originally Posted by James0502 View Post
    let
    y(x, t) := f(x − ct) − f(−x − ct).
    Show that y is a solution of the wave equation c2yxx = ytt and that
    y(0, t) = 0 for all t. Describe what is happening here in terms of
    incident and reflected waves.
    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)= x_0. Then at any t, you are at x= ct+ x_0. 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 x= x_0- ct and y(x_0-ct, t)= f(-x_0+ct-ct)= f(-x_0) 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 the sum of 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
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  3. #3
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    many thanks for the reply, but I'm still not sure exactly what this means in terms of the reflected and incident waves.. I really struggle with these tail bits.. cant really place what is happening physically
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