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

Hint: A sonometer is a device which is used to determine the relationship between the frequency of the sound and the tension in the string when a string is plucked. The frequency in the string is inversely proportional to the length of the string.

Formula used:
The formula for the frequency in the string is given by,
$f = \dfrac{1}{{2l}}\sqrt {\dfrac{T}{m}} $
Where f is the fundamental frequency, l is the length of the string, T is the tension in the string and m is the mass per unit length.

Complete step by step solution:
It is given in the problem that the sonometer wire has a length of 114 cm between two fixed ends and we need to find where the two bridges should be placed so that the wire is divided into three segments having fundamental frequencies in the ratio $1:3:4$.
As the relationship between the frequency and the length is equal to,
$ \Rightarrow f\alpha \dfrac{1}{l}$
Where f is the frequency and l is the length.
The ratio of the frequency given as,
$ \Rightarrow {f_1}:{f_2}:{f_3} = 1:2:3$
Since $f\alpha \dfrac{1}{l}$ we get,
$ \Rightarrow {f_1}:{f_2}:{f_3} = 1:2:3$
$ \Rightarrow {l_1}:{l_2}:{l_3} = \dfrac{1}{1}:\dfrac{1}{2}:\dfrac{1}{3}$
$ \Rightarrow {l_1}:{l_2}:{l_3} = 6:3:2$
The length of the three segments is equal to,
For length of first segment,
$ \Rightarrow {l_1} = \dfrac{6}{{11}} \times 110$
$ \Rightarrow {l_1} = 60cm$
For length of second segment
$ \Rightarrow {l_2} = \dfrac{3}{{11}} \times 110$
$ \Rightarrow {l_2} = 30cm$
For length of third segment
$ \Rightarrow {l_3} = \dfrac{2}{{11}} \times 110$
$ \Rightarrow {l_3} = 20cm$.

The length of segments is given by${l_1} = 60cm$,${l_2} = 30cm$ and ${l_3} = 20cm$.

Note: The sonometer works on the principle of resonance. The sonometer can also verify the laws of vibration on stretched strings and help in finding the frequency of the turning fork. If the frequency of the applied force becomes equal to the natural frequency then the resultant frequency is of very high amplitude.