Question

The length of a sonometer wire 'AB' is 110 cm where should the two bridges be places from end 'A' to divide the wire in three segments whose fundamental frequencies are in the ratio 1:2:3?

• A. 60 cm and 90 cm
• B. 90 cm and 100 cm
• C. 40 cm and 80 cm
• D. 50 cm and 90 cm

Answer 1

Answer: (A)

60 cm and 90 cm

Answer 2

Let the three segments be of lengths $l_1$, $l_2$, and $l_3$ and their fundamental frequencies be $n_1$, $n_2$, and $n_3$ respectively. Since the frequency is inversely proportional to the length, we have:

$$n_i \propto \frac{1}{l_i}$$

Given that

$$n_1 : n_2 : n_3 = 1 : 2 : 3,$$

we can write

$$\frac{1}{l_1} : \frac{1}{l_2} : \frac{1}{l_3} = 1 : 2 : 3.$$

Taking reciprocals, the segment lengths are in the ratio:

$$l_1 : l_2 : l_3 = \frac{1}{1} : \frac{1}{2} : \frac{1}{3} = 6 : 3 : 2.$$

Since the total length is 110 cm, compute:

$$l_1 = \frac{6}{6+3+2} \times 110 = \frac{6}{11} \times 110 = 60~\text{cm},$$ $$l_2 = \frac{3}{11} \times 110 = 30~\text{cm},$$ $$l_3 = \frac{2}{11} \times 110 = 20~\text{cm}.$$

The first bridge is placed at the end of $l_1$ (60 cm from A), and the second bridge is placed after $l_1 + l_2$ (i.e., 60 cm + 30 cm = 90 cm from A).

Core Explanation:

• Frequency ∝ 1/length.
• Ratio of lengths $= 6:3:2$.
• $l_1 = 60$ cm, $l_2 = 30$ cm, $l_3 = 20$ cm.
• Bridges at 60 cm and 90 cm from A.