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(C) Let $x$ be the distance from the bottom end of the string. The linear mass density is $\mu(x) = (\lambda/L)x$.The tension $T(x)$ at a distance $x$

Vedclass · Accepted Answer

The time taken by a pulse to reach from bottom to top will be $\sqrt{8L/3g}$.

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(C) Let $x$ be the distance from the bottom end of the string. The linear mass density is $\mu(x) = (\lambda/L)x$.The tension $T(x)$ at a distance $x$ from the bottom is due to the weight of the segment of length $x$ in the accelerating frame. The effective acceleration is $g_{eff} = g + 2g = 3g$.The mass of the segment of length $x$ is $m(x) = \int_0^x \mu(x') dx' = \int_0^x (\lambda/L)x' dx' = \frac{\lambda x^2}{2L}$.The tension $T(x) = m(x) \cdot g_{eff} = \frac{3g\lambda x^2}{2L}$.The wave velocity $v(x) = \sqrt{T(x)/\mu(x)} = \sqrt{\frac{3g\lambda x^2 / 2L}{\lambda x / L}} = \sqrt{\frac{3gx}{2}}$.Since $v = dx/dt$,we have $dt = dx / \sqrt{3gx/2} = \sqrt{2/3g} \cdot x^{-1/2} dx$.Integrating from $x=0$ to $x=L$,the time $t = \int_0^L \sqrt{2/3g} \cdot x^{-1/2} dx = \sqrt{2/3g} \cdot [2x^{1/2}]_0^L = \sqrt{2/3g} \cdot 2\sqrt{L} = \sqrt{8L/3g}$.

One end of a string of length $L$ is tied to the ceiling of a lift accelerating upwards with an acceleration $2g$. The other end of the string is free. The linear mass density of the string varies linearly from $0$ to $\lambda$ from bottom to top.

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