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(C) For a wave traveling in the negative $x$-direction, the velocity of a particle at any point is given by $v_p = -v \cdot (	ext{slope})$, where $v$

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$f$

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(C) For a wave traveling in the negative $x$-direction, the velocity of a particle at any point is given by $v_p = -v \cdot (	ext{slope})$, where $v$ is the wave speed and the slope is $\frac{\partial y}{\partial x}$.Since the wave is moving in the negative $x$-direction, the particle velocity $v_p$ is positive (upward) where the slope is negative, and $v_p$ is negative (downward) where the slope is positive.Looking at the graph:- At point $b$, the slope is negative, so the particle moves upward.- At point $c$, the slope is negative, so the particle moves upward.- At point $f$, the slope is positive, so the particle moves downward.- At point $i$, the slope is zero (at the crest), so the particle is momentarily at rest.Therefore, the point moving in the negative $y$-direction is $f$.

The figure represents the instantaneous snapshot of a transverse harmonic wave traveling along the negative $x$-axis. Identify the correct point(s) moving in the negative $y$-direction (downward).

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