A uniform string of length 20 m is suspended | vedclass

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(A) The velocity of a wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit length.For a str

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$2\sqrt{2} \ s$

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(A) The velocity of a wave on a string is given by $v = \sqrt{\frac{T}{\mu}}$,where $T$ is the tension and $\mu$ is the mass per unit length.For a string of length $l$ and mass $m$,$\mu = \frac{m}{l}$.At a distance $x$ from the lower end,the tension $T$ is due to the weight of the string below that point: $T = \mu x g$.Substituting this into the velocity formula: $v = \sqrt{\frac{\mu x g}{\mu}} = \sqrt{gx}$.Since $v = \frac{dx}{dt}$,we have $\frac{dx}{dt} = \sqrt{gx}$.Separating variables: $x^{-1/2} dx = \sqrt{g} dt$.Integrating from $x=0$ to $x=l$ and $t=0$ to $t=T_{total}$:$\int_{0}^{l} x^{-1/2} dx = \int_{0}^{T_{total}} \sqrt{g} dt$.$[2x^{1/2}]_{0}^{l} = \sqrt{g} T_{total}$.$2\sqrt{l} = \sqrt{g} T_{total} \implies T_{total} = 2\sqrt{\frac{l}{g}}$.Given $l = 20 \ m$ and $g = 10 \ ms^{-2}$,we get $T_{total} = 2\sqrt{\frac{20}{10}} = 2\sqrt{2} \ s$.

$A$ uniform string of length $20 \ m$ is suspended from a rigid support. $A$ short wave pulse is introduced at its lowest end. It starts moving up the string. The time taken to reach the support is (take $g = 10 \ ms^{-2}$):

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