Question

A massless rod $BD$ is suspended by two identical massless strings $AB$ and $CD$ of equal lengths. A block of mass $m$ is suspended from point $P$ such that $BP$ is equal to $x$. If the fundamental frequency of the left wire is twice the fundamental frequency of right wire, then the value of $x$ is

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

Answer 1

The correct option is C $\frac{l}{5}$

We know that,
$f\propto v\propto \sqrt{T}$

From the data given in the question,
$f_{AB}=2f_{CD}\Rightarrow f_{AB}^2=f_{CD}^2\Rightarrow T_{AB}=4T_{CD}$

Since, rod $BD$ is in equilibrium, Net torque about $P$ will be zero.
${\tau }_{net}=0$
$\Rightarrow T_{AB}\times x=T_{CD}\times (l-x)$
$\Rightarrow 4T_{CD}\times x=T_{CD}\times (l-x)\Rightarrow 4x=l-x$
$\Rightarrow x=\frac{l}{5}$

Thus, option (c) is the correct answer.