A string of mass $m$ and length $l$ is hanging from ceiling as shown in the figure. Wave in string move upward. $v_{A}$ and $v_{B}$ are the speed of wave at $A$ and $B$ respectively. Then $v_{B}$ is

\sqrt{3} v_{A}

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v_{A}

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< v_{A}

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\sqrt{2} v_{A}

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Solution

The correct option is A $\sqrt{3} v_{A}$

String mass $= m$
length $= L$
$μ = m / L$
Tension at $A = T_{A} = \frac{m}{4} \times g = m g / 4$
Velocity of wave at $A = V_{A} = \sqrt{\frac{T_{A}}{μ}} = \sqrt{\frac{m g / 4}{m / L}} = \sqrt{\frac{g L}{4}}$
Tension at $B = T_{B} = \frac{3}{4} m \times g$
speed of wave at $B = V_{B} = \sqrt{\frac{T_{B}}{μ}} = \sqrt{\frac{3 m g}{4 μ}} = \sqrt{\frac{3 g L}{4}}$
$\Rightarrow [V_{B} = \sqrt{3} V_{A}]$

$∴$ Option (A) is correct.

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