A sonometer wire has a length 114 cm between two fixed ends. Where should two bridges be placed, so as to…

A sonometer wire has a length 114 cm between 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 $1 : 3 : 4$?
  1. $l_1 = 72\text{ cm}, l_2 = 24\text{ cm}, l_3 = 18\text{ cm}$
  2. $l_1 = 60\text{ cm}, l_2 = 40\text{ cm}, l_3 = 14\text{ cm}$
  3. $l_1 = 52\text{ cm}, l_2 = 30\text{ cm}, l_3 = 32\text{ cm}$
  4. $l_1 = 65\text{ cm}, l_2 = 30\text{ cm}, l_3 = 19\text{ cm}$

Solution

Frequency of wire, $f \propto \frac{1}{l}$ Diagram showing a wire mounted between two fixed ends with two movable bridges dividing it into segments of lengths $72\text{ cm}$, $24\text{ cm}$, and $18\text{ cm}$. $\therefore l_1 : l_2 : l_3 = \frac{1}{f_1} : \frac{1}{f_2} : \frac{1}{f_3} = \frac{1}{1} : \frac{1}{3} : \frac{1}{4}$ or $l_1 : l_2 : l_3 = 12 : 4 : 3$ $\therefore l_1 = \frac{12}{19} \times 114 = 72\text{ cm}$ $l_2 = \frac{4}{19} \times 114 = 24\text{ cm}$ and $l_3 = \frac{3}{19} \times 114 = 18\text{ cm}$

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