A string of mass m and length l hangs from ce | vedclass

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Question

String mass $=m$

length $=L$

$\mu=m / L$

Tension at $A=T_{A}=\frac{m}{4} 	imes g=m g / 4$

velocity of wave at $A=V_{A}=\sqrt{\frac{T_{A}}

Vedclass · Accepted Answer

$\sqrt 3v_A$

Vedclass · Answer

String mass $=m$

length $=L$

$\mu=m / L$

Tension at $A=T_{A}=\frac{m}{4} 	imes g=m g / 4$

velocity of wave at $A=V_{A}=\sqrt{\frac{T_{A}}{\mu}}=\sqrt{\frac{m g / 4}{m / L}}=\sqrt{\frac{g L}{4}}$

Tension at $B=T_{B}=\frac{3}{4} m 	imes g$

speed of wave at $B=V_{B}=\sqrt{\frac{T_{B}}{\mu}}=\sqrt{\frac{3 m g}{4 \mu}}=\sqrt{\frac{3 g L}{4}}$

$\Rightarrow \quad\left[V_{B}=\sqrt{3} V_{A}ight]$

A string of mass $m$ and length $l$ hangs from ceiling as shown in the figure. Wave in string moves upward. $v_A$ and $v_B$ are the speeds of wave at $A$ and $B$ respectively. Then $v_B$ is

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