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(A) Let the mass of the string be $m$ and its length be $l$. The linear mass density is $\mu = \frac{m}{l}$.The tension at any point in the string is

Vedclass · Accepted Answer

$\sqrt{3} v_A$

Vedclass · Answer

(A) Let the mass of the string be $m$ and its length be $l$. The linear mass density is $\mu = \frac{m}{l}$.The tension at any point in the string is equal to the weight of the portion of the string below that point.For point $A$,the length of the string below it is $\frac{l}{4}$. Thus,the mass below $A$ is $m_A = \mu 	imes \frac{l}{4} = \frac{m}{4}$.The tension at $A$ is $T_A = m_A g = \frac{mg}{4}$.The speed of the wave at $A$ is $v_A = \sqrt{\frac{T_A}{\mu}} = \sqrt{\frac{mg/4}{m/l}} = \sqrt{\frac{gl}{4}}$.For point $B$,the length of the string below it is $\frac{3l}{4}$. Thus,the mass below $B$ is $m_B = \mu 	imes \frac{3l}{4} = \frac{3m}{4}$.The tension at $B$ is $T_B = m_B g = \frac{3mg}{4}$.The speed of the wave at $B$ is $v_B = \sqrt{\frac{T_B}{\mu}} = \sqrt{\frac{3mg/4}{m/l}} = \sqrt{\frac{3gl}{4}}$.Comparing $v_B$ and $v_A$:$v_B = \sqrt{3} 	imes \sqrt{\frac{gl}{4}} = \sqrt{3} v_A$.

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

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