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 $\frac{v_1{v}_2}{v_1+v_2}$

$V_1\propto \frac{1}{l_1}$
$V_1=\frac{K}{l_1}$ k is proportionality constant
$V_2=\frac{k}{l_2}$
$V=\frac{1}{l}$
$V=\frac{1}{l_1+l_2}$
Putting the values of $l_1$ and $l_2$
$V=\frac{v_1{v}_2}{v_1+v_2}$