Question

A wire has a length of 114 cm between two fixed ends. Where should two bridges be placed to divide the wire into three segments whose fundamental frequencies are in the ratio 1 : 3 : 4?

Answer 1

In case of a given wire under constant tension, fundamental frequency of vibration $f\propto \left(\frac{1}{l}\right)$
$\therefore l_1:l_2:l_3=\frac{1}{1}:\frac{1}{3}:\frac{1}{4}=12:4:3$
Let the lengths be $12x,4x$ and $3x$.
Total length = 114 cm
$\Rightarrow 12x+4x+3x=114cm\Rightarrow 19x=114cm\Rightarrow x=\frac{114}{19}=6cm$
$\therefore l_1=72cm$, $l_2=24cm$ and $l_3=18cm$
So the first bridge is to be placed 72 cm from one end. The second bridge is to be placed 24 cm from the first bridge.