A wire of variable mass per unit length mu = | vedclass

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Question

(A) Let $x$ be the distance from the lower end of the wire. The tension $T$ at a distance $x$ from the lower end is equal to the weight of the wire be

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$\sqrt{\frac{8l_0}{g}}$

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(A) Let $x$ be the distance from the lower end of the wire. The tension $T$ at a distance $x$ from the lower end is equal to the weight of the wire below that point.The mass of the wire of length $x$ is $m = \int_0^x \mu(x') dx' = \int_0^x \mu_0 x' dx' = \frac{\mu_0 x^2}{2}$.Thus,the tension $T$ at distance $x$ is $T = mg = \frac{\mu_0 x^2 g}{2}$.The speed of the transverse wave at distance $x$ is $v = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{\mu_0 x^2 g / 2}{\mu_0 x}} = \sqrt{\frac{gx}{2}}$.Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \sqrt{\frac{g}{2}} \sqrt{x}$.Separating variables,$\int_0^{l_0} x^{-1/2} dx = \int_0^t \sqrt{\frac{g}{2}} dt$.$[2x^{1/2}]_0^{l_0} = \sqrt{\frac{g}{2}} t$.$2\sqrt{l_0} = \sqrt{\frac{g}{2}} t$.$t = 2\sqrt{l_0} \sqrt{\frac{2}{g}} = 2\sqrt{\frac{2l_0}{g}} = \sqrt{\frac{8l_0}{g}}$.

$A$ wire of variable mass per unit length $\mu = \mu_0 x$ is hanging from the ceiling as shown in the figure. The length of the wire is $l_0$. $A$ small transverse disturbance is produced at its lower end. Find the time after which the disturbance will reach the other end.

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