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The equation for standing wave with one of its end tied is $2A \cos (\omega t) \sin (kx)$ and the amplitude of the standing wave is $ 2A \sin(kx)$

What is $x$? Is it the distance from the source of the wave or the distance from where the wave was reflected?

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  • $\begingroup$ Amplitude of a standing wave is not the same at every point on the wave. It is a function of the distance of the point you want to find the amplitude of from the source, given by that equation. So, x gives the location of the point you want to find the amplitude of. $\endgroup$ – Shreyansh Darshan Aug 27 '17 at 10:14
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You have fixed one end of the string, that means that one end of the string must have zero amplitude for all time. You can enforce this condition by making the end where the string is fixed $x=0$, since $2A \cos \omega t\sin kx = 0$ for all time if $x=0$.

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Recall the definition of a standing wave: it is some wave, which can be described by a function $f$, that oscillates in time but does not move. In other words, the wave oscillates about its resting point without ever really changing its "shape": extremal points (crests or troughs) remain crests or troughs, zero points remain zero points, midpoints stay as midpoints, and so on.

Consider the standard example of a standing wave, a vibrating string:

basic standing string wave

Since the standing wave is 1-dimensional, we can write it as $f(x) = A\sin(kx)^{[1]}$, just like we would any other waveform. Multiplying by $\cos(\omega t)$ is how we model oscillating about the resting point as we go forward in time. The scaled standing wave is multiplied by a value continuously varying between $-1$ and $1$, so no point can get further from rest than it originally was. That's what it means for the wave to keep its shape.

Whether or not $x=0$ denotes the start of the wave depends on the situation. If you're doing the high school experiment of holding one end of a tied string and flicking it, then yes, $x=0$ is the start. If instead you have a guitar string, you probably wouldn't have plucked it from the end, and the origin can be any point along the string. What matters is that the string is secured there, and thus resonant vibrations will bounce back from that point. $$$$ $[1]:$ The actual standing wave equation has $2A$ instead of $A$ because the standing wave forms from the overlapping of two waves: one moving to the left from the starting position, one to the right. Since both original waves would have amplitude $A$, the combined standing wave has maximum amplitude $2A$ from constructive interference. If the wave really did originate from $x=0$, then you can ignore this as a formalism.

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