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$$\Rightarrow \frac{{v}_{p}}{\frac{\partial y}{\partial x}}=-\frac{\omega}{k}=-v$$
$$\Rightarrow {v}_{p}=-v\frac{\partial y}{\partial x}$$
At a given position “x” and time “t”, the particle velocity is related to wave speed by this equation. Note that direction of particle velocity is determined by the sign of slope as wave speed is a positive quantity. We can interpret direction of motion of the particles on the string by observing “y-x” plot of a wave form. We know that “y-x” plot is a description of wave form at a particular time instant. It is important to emphasize that a wave like representation does not show the motion of wave. An arrow showing the direction of wave motion gives the sense of motion. The wave form is a snapshot (hence stationary) at a particular instant. We can, however, assess the direction of particle velocity by just assessing the slope at any position x=x. See the plot shown in the figure below :
The slope at “A” is positive and hence particle velocity is negative. It means that particle at this position - at the instant waveform is captured in the figure - is moving towards mean (or equilibrium) position. The slope at “B” is negative and hence particle velocity is positive. It means that particle at this position - at the instant waveform is captured in the figure - is moving towards positive extreme position. The slope at “C” is negative and hence particle velocity is positive. It means that particle at this position - at the instant waveform is captured in the figure - is moving towards mean (or equilibrium) position. The slope at “D” is positive and hence particle velocity is negative. It means that particle at this position - at the instant waveform is captured in the figure - is moving towards negative extreme position.
We can crosscheck or collaborate the deductions drawn as above by drawing wave form at another close instant t = t+∆t. We can visualize the direction of velocity by assessing the direction in which the particle at a position has moved in the small time interval considered.
Different forms give rise to a bit of confusion about the form of wave function. The forms used for describing waves are :
$$y\left(x,t\right)=A\mathrm{sin}\left(kx-\omega t\right)$$
$$y\left(x,t\right)=A\mathrm{sin}\left(\omega t-kx\right)$$
Which of the two forms is correct? In fact, both are correct so long we are in a position to accurately interpret the equation. Starting with the first equation and using trigonometric identity :
$$\mathrm{sin}\theta =\mathrm{sin}\left(\pi -\theta \right)$$
We have,
$$\Rightarrow A\mathrm{sin}\left(kx-\omega t\right)=A\mathrm{sin}\left(\pi -kx+\omega t\right)==A\mathrm{sin}\left(\omega t-kx+\pi \right)$$
Thus we see that two forms represent waves moving at the same speed ( $v=\omega /k$ ). They differ, however, in phase. There is phase difference of “π”. This has implication on the waveform and the manner particle oscillates at any given time instant and position. Let us consider two waveforms at x=0, t=0. The slopes of the waveforms are :
$$\frac{\partial}{\partial x}y\left(x,t\right)=kA\mathrm{cos}\left(kx-\omega t\right)=kA=\text{a positive number}$$
and
$$\frac{\partial}{\partial x}y\left(x,t\right)=-kA\mathrm{cos}\left(\omega t-kx\right)=-kA=\text{a negative number}$$
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