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(A) The wave pulse moves at a speed $v_w = 2 \ cm/s$. At $t = 0$,the trailing edge of the pulse is at a distance of $2 \ cm$ from the free end. In $t

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(A) The wave pulse moves at a speed $v_w = 2 \ cm/s$. At $t = 0$,the trailing edge of the pulse is at a distance of $2 \ cm$ from the free end. In $t = \frac{3}{4} \ s$,the pulse travels a distance $d = v_w 	imes t = 2 	imes \frac{3}{4} = 1.5 \ cm$. Since the trailing edge was $2 \ cm$ away,after $0.75 \ s$,the trailing edge is at $2 - 1.5 = 0.5 \ cm$ from the free end. This means the free end is currently located on the trailing slope of the triangular pulse. The slope of the trailing edge is $m = \frac{	ext{height}}{	ext{base}} = \frac{1 \ cm}{1 \ cm} = 1$. For a wave pulse,the particle velocity $v_p$ is given by $v_p = -v_w 	imes 	ext{slope}$. Since the trailing edge has a negative slope (relative to the direction of motion),the particle velocity at the free end is $v_p = -2 	imes (-1) = 2 \ cm/s$.

$A$ triangular pulse moving at $2 \ cm/s$ on a rope approaches an end at which it is free to slide on a vertical pole. What is the particle speed at the free end at $\frac{3}{4} \ s$ from the instant shown? (in $cm/s$)

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