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(D) The particle velocity $v_p$ of a point on a string is given by the relation $v_p = -v 	imes (	ext{slope of the graph at that point})$, where $v$

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$\downarrow$            $\uparrow$             $\downarrow$

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(D) The particle velocity $v_p$ of a point on a string is given by the relation $v_p = -v 	imes (	ext{slope of the graph at that point})$, where $v$ is the wave speed.Since the wave is traveling from left to right, the slope determines the direction of the particle's motion.At point $1$: The slope of the curve is positive. Therefore, $v_p = -v 	imes (	ext{positive}) = 	ext{negative}$, which means the velocity is downward $(\downarrow)$.At point $2$: The slope of the curve is negative. Therefore, $v_p = -v 	imes (	ext{negative}) = 	ext{positive}$, which means the velocity is upward $(\uparrow)$.At point $3$: The slope of the curve is positive. Therefore, $v_p = -v 	imes (	ext{positive}) = 	ext{negative}$, which means the velocity is downward $(\downarrow)$.Thus, the directions are $\downarrow, \uparrow, \downarrow$ respectively.

The diagram below shows an instantaneous position of a string as a transverse progressive wave travels along it from left to right. Which one of the following correctly shows the direction of the velocity of the points $1$, $2$ and $3$ on the string?

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