# Standing wave equation

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):

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

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.

• You need to define $x$. Is it a position relative to some origin or is it a distance travelled? Apr 11 at 23:12

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 $$x area, only the second term in (1) is nonzero and you have reflected wave packet $$y(x,t) = -A\sin(k(2l-x)-\omega t) = A\sin(k(2l-x)-\omega t +\pi).$$