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(D) The particle velocity is given by $v_p = \frac{dy}{dt} = -v \left( \frac{dy}{dx} \right)$,where $v$ is the wave velocity and $\frac{dy}{dx}$ is th

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Points $A$ and $E$ have minimum velocity.

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(D) The particle velocity is given by $v_p = \frac{dy}{dt} = -v \left( \frac{dy}{dx} \right)$,where $v$ is the wave velocity and $\frac{dy}{dx}$ is the slope of the wave at that point.$(1)$ For upward velocity,$v_p$ must be positive. This requires the slope $\frac{dy}{dx}$ to be negative. Looking at the graph,the slope is negative at points $D, E,$ and $F$. Thus,these points move upwards.$(2)$ For downward velocity,$v_p$ must be negative. This requires the slope $\frac{dy}{dx}$ to be positive. Looking at the graph,the slope is positive at points $A, B,$ and $H$. Thus,these points move downwards.$(3)$ For zero velocity,the slope $\frac{dy}{dx}$ must be zero. This occurs at the crest and trough,which are points $C$ and $G$.$(4)$ The magnitude of the particle velocity is maximum where the slope $| \frac{dy}{dx} |$ is maximum. This occurs at points $A, E,$ and the point between $G$ and $H$. Points $A$ and $E$ have maximum velocity,not minimum. Therefore,statement $(D)$ is wrong.

$A$ transverse wave is travelling along a string from left to right. The adjoining figure represents the shape of the string at a given instant. At this instant,among the following,choose the wrong statement.

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