A pulse on a string is shown in the figure. P | vedclass

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(B) The velocity of a particle $v_p$ in a wave is given by $v_p = -v \cdot (	ext{slope})$, where $v$ is the wave speed.If the pulse travels in the po

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If $P$ is moving upwards, then the pulse is travelling in the positive direction.

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(B) The velocity of a particle $v_p$ in a wave is given by $v_p = -v \cdot (	ext{slope})$, where $v$ is the wave speed.If the pulse travels in the positive direction $(v > 0)$, then $v_p = -v \cdot (	ext{slope})$.$1$. At point $P$ (on the trailing edge of the crest), the slope is positive. Thus, $v_p = -v \cdot (	ext{positive}) = 	ext{negative}$. So, $P$ moves downwards if the pulse moves in the positive direction.$2$. Conversely, if the pulse moves in the negative direction $(v $3$. Option $A$: If $P$ is stationary, it implies the superposition of two waves moving in opposite directions (standing wave condition), which is correct.$4$. Option $B$: If $P$ moves upwards, the pulse must be moving in the negative direction (based on the slope analysis). Thus, option $B$ is incorrect.$5$. Option $C$: If $P$ moves downwards, the pulse must be moving in the positive direction. Thus, option $C$ is incorrect.Since both $B$ and $C$ are incorrect statements based on the standard wave equation, and the question asks for the incorrect statement, there is an ambiguity. However, typically in such problems, if $P$ is on the trailing edge, it moves downwards for a positive-moving pulse. Thus, $B$ is clearly incorrect.

$A$ pulse on a string is shown in the figure. $P$ is a particle of the string. State which of the following is incorrect.

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