A uniform string oflength 20 m is suspended f | vedclass

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Question

We know that velocity in string is given by

$\mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mu}}$              $...(I)$

Vedclass · Accepted Answer

$2$$\sqrt 2 s$

Vedclass · Answer

We know that velocity in string is given by

$\mathrm{v}=\sqrt{\frac{\mathrm{T}}{\mu}}$              $...(I)$

where $\mu=\frac{\mathrm{m}}{1}=\frac{	ext { mass of string }}{	ext { length of string }}$

The tension $\mathrm{T}=\frac{\mathrm{m}}{\ell} 	imes \mathrm{x} 	imes \mathrm{g}$                   $...(II)$

From $(a)$ and $(b)$

$\frac{d x}{d t}=\sqrt{g x}$

$\mathrm{x}^{-1 / 2} \mathrm{dx}=\sqrt{\mathrm{g}} \mathrm{dt} \quad 	herefore \int_{0}^{\ell} \mathrm{x}^{-1 / 2} \mathrm{dx}-\sqrt{\mathrm{g}} \int_{0}^{\ell} \mathrm{dt}$

$2 \sqrt{l}=\sqrt{g} 	imes t \quad 	herefore t=2 \sqrt{\frac{\ell}{g}}=2 \sqrt{\frac{20}{10}}=2 \sqrt{2}$

A uniform string oflength $20\ m$ 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 supports is (take $g= 10 $ $ms^{-2}$ )

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