Question

A massless rod of length $l$ is hung from the ceiling with the help of two identical wires attached at its ends. A block is hung on the rod at a distance $x$ from the left end. In the case, the frequency of the $1st$ harmonic of the wire on the left end is equal to the frequency of the $2nd$ harmonic of the wire on the right. The value of $x$ is

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

Answer 1

Answer: (D)

$\frac{l}{5}$

Since, the frequency of the first harmonic from the left is equal to that of second harmonic from right,
${\nu }_1=2{\nu }_2$
Hence, $T_1=T_2$
Thus, according to the question,
$T_1\left(x\right)=T_2(l-x)$
Solving this equation for $T_1$ and $T_2$ we get the value of x = $\frac{l}{5}$