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

$f\alpha \frac{1}{l}\therefore l_1:l_2:l_3=\frac{1}{1}:\frac{1}{3}:\frac{1}{4}=12:4:3$

$l_1=72\text{cm}$      $l_2=24\text{cm}$         $l_3=18\text{cm}$

First bridge is to be placed 72cm from one end. Second bridge to be placed = 72 + 24 = 96 cm from one end.