A rope of length L and mass M hangs freely fr | vedclass

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(C) The speed of a transverse wave on a string is given by $v = \sqrt{\frac{T_s}{\mu}}$,where $T_s$ is the tension and $\mu = \frac{M}{L}$ is the mass

Vedclass · Accepted Answer

$\frac{T}{\sqrt{2}}$

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(C) The speed of a transverse wave on a string is given by $v = \sqrt{\frac{T_s}{\mu}}$,where $T_s$ is the tension and $\mu = \frac{M}{L}$ is the mass per unit length.At a distance $x$ from the bottom,the tension $T_s$ is due to the mass of the rope below it,which is $m = \frac{M}{L} x$. Thus,$T_s = \left(\frac{M}{L} x\right)g$.The speed at distance $x$ is $v(x) = \sqrt{\frac{(M/L)xg}{M/L}} = \sqrt{gx}$.Since $v = \frac{dx}{dt}$,we have $dt = \frac{dx}{\sqrt{gx}}$.Integrating from $x=0$ to $x=L$ to find the total time $T$:$T = \int_0^L \frac{dx}{\sqrt{gx}} = \frac{1}{\sqrt{g}} [2\sqrt{x}]_0^L = 2\sqrt{\frac{L}{g}}$.Now,to find the time $T_1$ to cover the first half length ($x=0$ to $x=L/2$):$T_1 = \int_0^{L/2} \frac{dx}{\sqrt{gx}} = \frac{1}{\sqrt{g}} [2\sqrt{x}]_0^{L/2} = 2\sqrt{\frac{L}{2g}} = \frac{2}{\sqrt{2}}\sqrt{\frac{L}{g}} = \frac{T}{\sqrt{2}}$.

$A$ rope of length $L$ and mass $M$ hangs freely from the ceiling. If the time taken by a transverse wave to travel from the bottom to the top of the rope is $T$,then the time taken to cover the first half length is:

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