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(C) In a transverse harmonic wave, the velocity of a particle is given by $v_p = -v \cdot (	ext{slope})$, where $v$ is the wave speed and the slope i

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$d$

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(C) In a transverse harmonic wave, the velocity of a particle is given by $v_p = -v \cdot (	ext{slope})$, where $v$ is the wave speed and the slope is $\frac{\partial y}{\partial x}$.For a wave traveling in the negative $x$-direction, the particle velocity is $v_p = v \cdot (	ext{slope})$.The particle velocity is maximum at the mean position (where the displacement $y = 0$) because the slope is maximum there.Looking at the figure, the points $o$, $d$, and $h$ are at the mean position $(y = 0)$.Among the given options, point $d$ is at the mean position.Points $b$ and $f$ are at the extreme positions where the velocity is zero.Therefore, the point moving with maximum velocity is $d$.

The figure represents the instantaneous picture of a transverse harmonic wave traveling along the negative $x$-axis. Choose the correct alternative(s) related to the movement of the points shown in the figure. The points moving with maximum velocity is/are:

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