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14.Waves and Sound
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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

A
$\sqrt 3v_A$
B
$v_A$
C
$< v_A$
D
$\sqrt 2v_A$
Solution

String mass $=m$
length $=L$
$\mu=m / L$
Tension at $A=T_{A}=\frac{m}{4} \times g=m g / 4$
velocity of wave at $A=V_{A}=\sqrt{\frac{T_{A}}{\mu}}=\sqrt{\frac{m g / 4}{m / L}}=\sqrt{\frac{g L}{4}}$
Tension at $B=T_{B}=\frac{3}{4} m \times g$
speed of wave at $B=V_{B}=\sqrt{\frac{T_{B}}{\mu}}=\sqrt{\frac{3 m g}{4 \mu}}=\sqrt{\frac{3 g L}{4}}$
$\Rightarrow \quad\left[V_{B}=\sqrt{3} V_{A}\right]$
Standard 11
Physics