Waves

JEE Main level, single correct, medium

73 of 142

Question 73 of 142

A uniform string of length 20m20m 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=10ms−2g=10m{s}^{-2})

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Answer: (A) 22s2\sqrt{2}s

Velocity in a string, at a distance xx, is given by

v=µgxµ=gxv=\sqrt{\dfrac{µgx}{µ}}=\sqrt{gx}

⇒dxdt=gx\Rightarrow \dfrac{dx}{dt}=\sqrt{gx}

⇒∫020dxx=g∫0tdt\Rightarrow {\int }_{0}^{20}\dfrac{dx}{\sqrt{x}}=\sqrt{g}{\int }_{0}^{t}dt

On integrating, we get,
t=22st=2\sqrt{2} s

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