Waves

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Question 75 of 142

A sonometer wire of length 114 cm114 \mathrm{~cm} is fixed at both the ends. Where should the two bridges be placed so as to divide the wire into three segments whose fundamental frequencies are in the ratio 1:3:41: 3: 4 ?

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Answer: (D) At 72 cm72 \mathrm{~cm} and 96 cm96 \mathrm{~cm} from one end

Total length of the wire, L=114 cm\mathrm{L}=114 \mathrm{~cm} n1:n2:n3=1:3:4\mathrm{n}_1: \mathrm{n}_2: \mathrm{n}_3=1: 3: 4
Let L1, L2\mathrm{L}_1, \mathrm{~L}_2 and L3\mathrm{L}_3 be the lengths of the three parts
As n∝1 L\mathrm{n} \propto \frac{1}{\mathrm{~L}}
∴L1:L2:L3=11:13:14=12:4:3∴L1=72 cm(1212+4+3×114)L2=24 cm(419×114) and L3=18 cm(319×114)\begin{aligned} & \therefore \quad \mathrm{L}_1: \mathrm{L}_2: \mathrm{L}_3=\frac{1}{1}: \frac{1}{3}: \frac{1}{4}=12: 4: 3 \\ & \therefore \quad \mathrm{L}_1=72 \mathrm{~cm}\left(\frac{12}{12+4+3} \times 114\right) \\ & \mathrm{L}_2=24 \mathrm{~cm}\left(\frac{4}{19} \times 114\right) \\ & \text { and } \quad \mathrm{L}_3=18 \mathrm{~cm}\left(\frac{3}{19} \times 114\right) \end{aligned}
Hence the bridges should be placed at 72 cm72 \mathrm{~cm} and 72+24=96 cm72+24=96 \mathrm{~cm} from one end.

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