Question

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

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

Answer 1

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

Frequency of first harmonic in AB = second harmonic frequency in CD

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

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

$⟹T_2=\frac{T_1}{4}$

For rotational equilibrium

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

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

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

$=5T_{1}X=T_{1}L$

$=X=\frac{L}{5}$