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A question about standing wave equation.

We send a harmonic wave that travels down a rope that is fixed at the end with the equation(like in the picture):

$$y = A\sin(kx-\omega t)$$

enter image description here

The wave that travels down a rope gets reflected at the rope’s end and has the equation:

$$ y = A \sin[k(2l-x) - \omega t + π]$$ where $l$ is the length of the rope.

I don't understand this equation. We add $π$ because the waves get inverted when it is reflected, but I don't understand where the $(2l-x)$ part comes from.

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  • $\begingroup$ You need to define $x$. Is it a position relative to some origin or is it a distance travelled? $\endgroup$
    – Bill N
    Apr 11 at 23:12
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In this problem, fixed end is at $x = l$. Corresponding boundary condition has a form $$ y(x = l, t) = 0. $$ Solutions to the wave equation with this boundary condition have the following form $$ y(x,t) = f(kx-\omega t) - f(k(2l-x)-\omega t). \quad (1) $$ In this problem, $f(kx-\omega t) = A\sin(kx-\omega t)$. For $t$ large enough, in the $x<l$ area, only the second term in (1) is nonzero and you have reflected wave packet $$ y(x,t) = -A\sin(k(2l-x)-\omega t) = A\sin(k(2l-x)-\omega t +\pi). $$

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