A rope of length L and uniform linear density | vedclass

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Question

$(b)$ Tension at a section $x$ distance from free end is

$T=m g=\mu \cdot x \cdot g$

where, $\mu=$ mass per unit length of rope.

Wave speed o

Vedclass · Accepted Answer

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

Vedclass · Answer

$(b)$ Tension at a section $x$ distance from free end is

$T=m g=\mu \cdot x \cdot g$

where, $\mu=$ mass per unit length of rope.

Wave speed over the rope is given by

$v=\sqrt{\frac{T}{\mu}} =\sqrt{\frac{\mu \cdot x \cdot g}{\mu}}=\sqrt{g \cdot x}$

$\Rightarrow \quad v^{2} =g x$

Comparing above equation with

$v^{2}-u^{2}=2 a s, 	ext { we get }$

$\Rightarrow v^{2}=2 a x=g x$

$\Rightarrow 	ext { Acceleration, } a=\frac{g}{2}$

Now, if $t=$ time for wave pulse to cover a distance $x$,

then from

$s=u t+\frac{1}{2} a t^{2},$

we have

$x=\frac{1}{2} 	imes \frac{g}{2} 	imes t^{2}$

$	ext { or } t =2 \sqrt{\frac{x}{g}}$

$\Rightarrow t \propto \sqrt{x}$

or time to travel complete length of rope $t \propto \sqrt{L} .$

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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