A sonometer wire of length 114 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 : 4?

A sonometer wire of length 114 cm is fixed at both ends, and we need to place two bridges to divide it into three segments with fundamental frequencies in the ratio 1 : 3 : 4. For a string fixed at both ends, the fundamental frequency $$f$$ is given by $$f = \frac{1}{2L} \sqrt{\frac{T}{\mu}}$$, where $$L$$ is the length of the segment, $$T$$ is the tension (constant throughout the wire), and $$\mu$$ is the linear mass density (also constant). Since $$T$$ and $$\mu$$ are constant, the frequency $$f$$ is inversely proportional to the length $$L$$, so $$f \propto \frac{1}{L}$$.

Given the frequency ratio $$f_1 : f_2 : f_3 = 1 : 3 : 4$$, the lengths will be inversely proportional to these frequencies. Therefore, the ratio of the lengths is $$L_1 : L_2 : L_3 = \frac{1}{1} : \frac{1}{3} : \frac{1}{4}$$. To simplify, multiply each term by the least common multiple (LCM) of the denominators 1, 3, and 4, which is 12:

$$L_1 : L_2 : L_3 = \frac{1}{1} \times 12 : \frac{1}{3} \times 12 : \frac{1}{4} \times 12 = 12 : 4 : 3$$

So, $$L_1 : L_2 : L_3 = 12k : 4k : 3k$$ for some constant $$k$$. The total length is the sum of the segments:

$$12k + 4k + 3k = 19k = 114 \text{ cm}$$

Solving for $$k$$:

$$19k = 114 \implies k = \frac{114}{19} = 6$$

Now, find the lengths:

$$L_1 = 12k = 12 \times 6 = 72 \text{ cm}$$

$$L_2 = 4k = 4 \times 6 = 24 \text{ cm}$$

$$L_3 = 3k = 3 \times 6 = 18 \text{ cm}$$

The segments must be adjacent and cover the entire wire. The bridges are placed such that the segments are formed consecutively. Starting from one end (say end A), the first bridge is at a distance $$L_1 = 72$$ cm from A. The second bridge is at $$L_1 + L_2 = 72 + 24 = 96$$ cm from A. The third segment from 96 cm to 114 cm is $$L_3 = 18$$ cm.

To verify, the fundamental frequencies are proportional to the reciprocals of the lengths:

$$f_1 \propto \frac{1}{72}, \quad f_2 \propto \frac{1}{24}, \quad f_3 \propto \frac{1}{18}$$

The ratio is:

$$f_1 : f_2 : f_3 = \frac{1}{72} : \frac{1}{24} : \frac{1}{18}$$

Simplify by multiplying each term by the LCM of 72, 24, and 18, which is 72:

$$\frac{1}{72} \times 72 = 1, \quad \frac{1}{24} \times 72 = 3, \quad \frac{1}{18} \times 72 = 4$$

So, $$f_1 : f_2 : f_3 = 1 : 3 : 4$$, which matches the given ratio.

Now, check the options:

• Option A: Bridges at 36 cm and 84 cm. Segments: 36 cm, 48 cm (84 - 36), 30 cm (114 - 84). Frequencies proportional to $$\frac{1}{36}$$, $$\frac{1}{48}$$, $$\frac{1}{30}$$. Ratios: $$\frac{1}{36} : \frac{1}{48} : \frac{1}{30} = 20 : 15 : 24$$ (after multiplying by 720, the LCM), which is not 1:3:4.
• Option B: Bridges at 24 cm and 72 cm. Segments: 24 cm, 48 cm (72 - 24), 42 cm (114 - 72). Frequencies proportional to $$\frac{1}{24}$$, $$\frac{1}{48}$$, $$\frac{1}{42}$$. Ratios: $$\frac{1}{24} : \frac{1}{48} : \frac{1}{42} = 14 : 7 : 8$$ (after multiplying by 336, the LCM), which is not 1:3:4.
• Option C: Bridges at 48 cm and 96 cm. Segments: 48 cm, 48 cm (96 - 48), 18 cm (114 - 96). Frequencies proportional to $$\frac{1}{48}$$, $$\frac{1}{48}$$, $$\frac{1}{18}$$. Ratios: $$\frac{1}{48} : \frac{1}{48} : \frac{1}{18} = 3 : 3 : 8$$ (after multiplying by 144, the LCM), which is not 1:3:4.
• Option D: Bridges at 72 cm and 96 cm. Segments: 72 cm, 24 cm (96 - 72), 18 cm (114 - 96). Frequencies proportional to $$\frac{1}{72}$$, $$\frac{1}{24}$$, $$\frac{1}{18}$$. Ratios: $$\frac{1}{72} : \frac{1}{24} : \frac{1}{18} = 1 : 3 : 4$$ (as shown above).

Hence, the correct answer is Option D.