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(B) Let $L$ be the length of the string and $m$ be its total mass. The linear mass density is $\mu = m/L$.Consider a point $P$ at a distance $x$ from

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the speed of the pulse decreases at a constant rate as the pulse moves downward.

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(B) Let $L$ be the length of the string and $m$ be its total mass. The linear mass density is $\mu = m/L$.Consider a point $P$ at a distance $x$ from the bottom end of the string.The tension $T$ at point $P$ is equal to the weight of the string below it,so $T = (\mu x)g = \frac{m}{L}gx$.The speed of a transverse pulse in a string is given by $v = \sqrt{T/\mu}$.Substituting $T$,we get $v = \sqrt{\frac{(\mu x)g}{\mu}} = \sqrt{gx}$.As the pulse moves downward,$x$ decreases,so the speed $v$ decreases.To find the rate of change of speed with respect to time,we use $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}$.Since the pulse moves downward,$\frac{dx}{dt} = -v = -\sqrt{gx}$.Calculating $\frac{dv}{dx} = \frac{d}{dx}(\sqrt{gx}) = \frac{g}{2\sqrt{gx}}$.Thus,$a = \left(\frac{g}{2\sqrt{gx}}\right) \cdot (-\sqrt{gx}) = -g/2$.The negative sign indicates that the speed decreases at a constant rate of $g/2$ as it moves downward.

$A$ uniform string is suspended vertically. $A$ transverse pulse is created at the top of the string. Then:

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