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 $\mathrm{g}=10{\mathrm{ms}}^{–2}$ )

A uniform string of length 20m is suspended from a rigid support

As $\mu$ is mass per unit length of the rope, then $\mu =m/L$

As $\upsilon =\sqrt{\frac{T}{\mu }}$

$\frac{dx}{dt}=\sqrt{\frac{mgx/L}{m/L}}=\sqrt{g}\sqrt{x}$

$\frac{dx}{\sqrt{x}}=\sqrt{g}dt$

Integrating on both sides, we get ${\int }_0^L{x}^{-1/2}dx=\sqrt{g}{\int }_0^{t}dt$

$t=2\sqrt{L}/\sqrt{g}=2\times \sqrt{\frac{20}{10}}=2\sqrt{2}s$