Question

A knife-edge divides a sonometer wire into two parts. The fundamental frequencies of the two parts are $v_1$ and $v_2$. The fundamental frequency of the sonometer wire when the knife-edge is removed will be

• A. $v_1+v_2$
• B. $\frac{1}{2}(v_1+v_2)$
• C. $\sqrt{v_1{v}_2}$
• D. $\frac{v_1{v}_2}{v_1+v_2}$

Answer 1

The correct option is D $\frac{v_1{v}_2}{v_1+v_2}$

Let $L_1$ and $L_2$ be the lengths of the two parts of sonometer wire.
Given
$v_1=\frac{1}{2L_1}\sqrt{\frac{T}{m}}$ and $v_2=\frac{1}{2L_2}\sqrt{\frac{T}{m}}$
$v_1{L}_1=v_2{L}_2=\frac{1}{2}\sqrt{\frac{T}{m}}=k\text{(constant)}$,
Thus $L_1=\frac{k}{v_1}$ and $L_2=\frac{k}{v_2}$
The fundamental frequency of the complete sonometer wire without knife edge is
$v=\frac{1}{2(L_1+L_2)}\sqrt{\frac{T}{m}}=\frac{k}{(L_1+L_2)}$
$\frac{1}{v}=\frac{L_1}{k}+\frac{L_2}{k}=\frac{1}{v_1}+\frac{1}{v_2}$
$v=\frac{v_1{v}_2}{v_1+v_2}$,
which is choice (d).