Question

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.

• A. $\frac{1}{L}\sqrt{\frac{T}{\mu }}$
• B. $\frac{5}{2L}\sqrt{\frac{T}{\mu }}$
• C. $\frac{3}{2L}\sqrt{\frac{T}{\mu }}$
• D. $\frac{1}{2L}\sqrt{\frac{T}{\mu }}$

Answer 1

Answer: (D)

$f_1=\frac{n_1}{2L}\sqrt{\frac{T}{\mu }}$ , $f_2=\frac{n_2}{2L}\sqrt{\frac{T}{16\mu }}$
$f_1=f_2\Rightarrow n_1=\frac{n_2}{4}$
$n_1=1$ , $n_2=4$
$f_{min}=\left(\frac{1}{2L}\right)\sqrt{\frac{T}{\mu }}$