Question

a massless rod of length $L$ is suspended by two identical strings $AB$ and $CD$ of equal length. A block of mass $m$ is suspended from point $O$ such that $BO$ is equal to ${}^{\prime }{x}^{\prime }$. Further it is observed that the frequency of $1^{st}$ harmonic in $AB$ is equal to $2^{nd}$ harmonic frequency in $CD$, ${}^{\prime }{x}^{\prime }$ is.

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

Answer 1

Answer: (A)

$\frac{L}{5}$

Frequency of the first harmonic of $AB$
$=\frac{1}{2l}\sqrt{\frac{T_{AB}}{m}}$, when $l$ is the length of the two strings.
Frequency of the $2^{nd}$ harmonic of $CD$
$=\frac{1}{l}\sqrt{\frac{T_{CD}}{m}}$
Given that the two frequencies are equal.
$=\frac{1}{2l}\sqrt{\frac{T_{AB}}{m}}=\frac{1}{l}\sqrt{\frac{T_{CD}}{m}}⟹\frac{T_{AB}}{4}=T_{CD}$
$⟹T_{AB}=4T_{CD}$ ......(i)
For rotational equilibrium of massless rod taking torque about point $O$.
$T_{AB}\times x=T_{CD}(L-x)$ ......(ii)
Solve to get, $x=\frac{L}{5}$