Question

A sonometer wire has a length 1144 cm between two fixed ends. Where should two bridges be placed so as to divide the wire in to three segments whose fundamental frequencies are in the ratio 1 : 3 : 4

• A. $l_1=72cm,l_2=24cm,l_3=18cm$
• B. $l_1=60cm,l_2=40cm,l_3=14cm$
• C. $l_1=52cm,l_2=30cm,l_3=32cm$
• D. $l_1=65cm,l_2=30cm,l_3=19cm$

Answer 1

The correct option is A $l_1=72cm,l_2=24cm,l_3=18cm$

Let the lengths of the three segments formed be $l_1$ , $l_2$ and $l_3$ and their respective fundamental frequencies be ${\nu }_1$, ${\nu }_2$ and ${\nu }_3$.

Fundamental frequency in a string fixed at both ends $\nu =\frac{v}{2l}$

$⟹$ $\nu \propto \frac{1}{l}$ for $v=constant$

$⟹$ ${\nu }_1:{\nu }_2:{\nu }_3=\frac{1}{l_1}:\frac{1}{l_2}:\frac{1}{l_3}$

$1:3:4=\frac{1}{l_1}:\frac{1}{l_2}:\frac{1}{l_3}$ $⟹l_1:l_2:l_3=12:4:3$ ..........(1)

As $l_1=72cm$, $l_2=24cm$ and $l_3=18cm$ satisfies the equation (1).

Hence option A is correct.