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

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$1 - \downarrow, 2 - \uparrow, 3 - \downarrow$

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

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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