A rope of length L and uniform linear density | vedclass

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(B) The tension $T$ at a section at a distance $x$ from the free end is given by $T = m g = \mu x g$,where $\mu$ is the mass per unit length of the ro

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The time taken by the pulse to travel the length of the rope is proportional to $\sqrt{L}$.

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(B) The tension $T$ at a section at a distance $x$ from the free end is given by $T = m g = \mu x g$,where $\mu$ is the mass per unit length of the rope.The wave speed $v$ on the rope is given by $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{\mu x g}{\mu}} = \sqrt{g x}$.Since $v = \sqrt{g x}$,the speed increases as the pulse moves up (as $x$ increases).To find the time $t$ taken to travel a distance $x$,we use $v = \frac{dx}{dt} = \sqrt{g x}$.Rearranging gives $dx / \sqrt{x} = \sqrt{g} dt$.Integrating both sides from $0$ to $x$ and $0$ to $t$,we get $\int_{0}^{x} x^{-1/2} dx = \int_{0}^{t} \sqrt{g} dt$.$2\sqrt{x} = \sqrt{g} t$,which implies $t = 2\sqrt{\frac{x}{g}}$.For the total length $L$,the time taken is $t = 2\sqrt{\frac{L}{g}}$,so $t \propto \sqrt{L}$.

$A$ rope of length $L$ and uniform linear density is hanging from the ceiling. $A$ transverse wave pulse,generated close to the free end of the rope,travels upwards through the rope. Select the correct option.

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