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  4. A light string is tied at one end to a fixed support and to a heavy string of equal length L at the other end A as shown in the figure ( Total length of both strings combined is 2L ). A block of mass M is tied to the free end of heavy string. Mass per unit length of the strings are μ and 16μ and tension is T . Find lowest positive value of frequency such that junction point A is a node. <img class=img-fluid question-image alt=Question src=https://cdn.tardigrade.in/q/nta/p-xlrrr0nvwzda.png />

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Q. A light string is tied at one end to a fixed support and to a heavy string of equal length $L$ at the other end $A$ as shown in the figure ( Total length of both strings combined is $2L$ ). A block of mass $M$ is tied to the free end of heavy string. Mass per unit length of the strings are $\mu $ and $16\mu $ and tension is $T$ . Find lowest positive value of frequency such that junction point $A$ is a node.

Question

NTA AbhyasNTA Abhyas 2020Waves

Solution:

$f_{1}=\frac{n_{1}}{2 L}\sqrt{\frac{T}{\mu }}$ , $f_{2}=\frac{n_{2}}{2 L}\sqrt{\frac{T}{16 \mu }}$
$f_{1}=f_{2}\Rightarrow n_{1}=\frac{n_{2}}{4}$
$n_{1}=1$ , $n_{2}=4$
$f_{min}=\left(\frac{1}{2 L}\right)\sqrt{\frac{T}{\mu }}$