Question

A massless rod of length $L$ is suspended horizontally with the help of two identical vertical strings $AB$ and $CD$ as shown in the figure. A block of mass $m$ of suspended from the rod at a point $E$. The two strings are made to vibrate by a tuning fork. It is found that string $AB$ vibrates in its $1^{st}$ harmonic and string $CD$ vibrates in its $2^{nd}$ harmonic. Then $BE$ is

• A. $\frac{L}{5}$
• B. $\frac{4L}{5}$
• C. $\frac{3L}{5}$
• D. $\frac{2L}{5}$

Answer 1

The correct option is A $\frac{L}{5}$

Frequency of $1^{st}$ harmonic in $AB$ $=2^{nd}$ harmonic frequency in $CD$

$\frac{1}{2\lambda }\sqrt{\frac{T_1}{\mu }}=\frac{1}{\lambda }\sqrt{\frac{T_2}{\mu }}$

$\frac{1}{4}\frac{T_1}{\mu }=\frac{1}{1}\frac{T_2}{\mu }$

$\Rightarrow T_2=T_1/4$

For rotational equilibrium

$T_{1}X=T_2(L-X)$

$T_{1}X=\frac{T_1}{4}(L-X)$

$4T_{1}X=T_{1}L-T_{1}X$

$5T_{1}X=T_{1}L\Rightarrow X=L/5$

Hence Option (D) is correct.