A heavy uniform rope is suspended vertically | vedclass

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Question

(A) Let the total mass of the rope be $M$ and its length be $L$. Let $\mu = \frac{M}{L}$ be the mass per unit length.Consider the pulse at a distance

Vedclass · Accepted Answer

Constant and equal to $\frac{g}{2}$

Vedclass · Answer

(A) Let the total mass of the rope be $M$ and its length be $L$. Let $\mu = \frac{M}{L}$ be the mass per unit length.Consider the pulse at a distance $x$ from the bottom end of the rope.The tension $T$ at a distance $x$ from the bottom is equal to the weight of the rope segment of length $x$ below that point: $T = \mu x g$.The speed of the pulse $v$ is given by $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{\mu x g}{\mu}} = \sqrt{xg}$.The acceleration of the pulse is $a = \frac{dv}{dt} = \frac{dv}{dx} \cdot \frac{dx}{dt}$.Since $v = \sqrt{xg}$,we have $\frac{dv}{dx} = \frac{1}{2\sqrt{x}} \sqrt{g}$.Also,$\frac{dx}{dt} = v = \sqrt{xg}$.Therefore,$a = \left( \frac{1}{2\sqrt{x}} \sqrt{g} \right) \cdot \sqrt{xg} = \frac{1}{2} \sqrt{g} \cdot \sqrt{g} = \frac{g}{2}$.Since the acceleration $a = \frac{g}{2}$ is independent of $x$,the acceleration of the pulse is constant throughout its motion.

$A$ heavy uniform rope is suspended vertically from a ceiling and is in equilibrium. $A$ pulse is generated at the bottom end of the rope as shown. As the pulse travels up the rope,its acceleration at any instant is ($g$ is acceleration due to gravity).

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