The length of a sonometer wire A B is 110 cm. Where should the two bridges be placed from A to

step 1 In this question we have given the length of sonometer wire that is 110 semi and we have to find

The length of a sonometer wire A B is 110 cm. Where should...

The length of a sonometer wire $A B$ is $110 \mathrm{~cm}$. Where should the two bridges be placed from $A$ to divide the wire in three segments whose fundamental frequencies are in the ratio of $1: 2: 3$ ? (A) $30 \mathrm{~cm}, 90 \mathrm{~cm}$ (B) $60 \mathrm{~cm}, 90 \mathrm{~cm}$ (C) $40 \mathrm{~cm}, 70 \mathrm{~cm}$ (D) None of these

The length of a sonometer wire $A B$ is $110 \mathrm{~cm}$. Where should the two bridges be placed from $A$ to divide the wire in three segments whose fundamental frequencies are in the ratio of $1: 2: 3$ ?
(A) $30 \mathrm{~cm}, 90 \mathrm{~cm}$
(B) $60 \mathrm{~cm}, 90 \mathrm{~cm}$
(C) $40 \mathrm{~cm}, 70 \mathrm{~cm}$
(D) None of these

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Key Concepts

Fundamental Frequency of a Vibrating String

This concept explains that the lowest (fundamental) frequency at which a string vibrates depends on its tension, mass per unit length, and length. Under fixed tension and density, the frequency is given by an equation showing it is inversely related to the length of the vibrating segment.

Inverse Proportionality of Frequency to Length

When the tension and linear mass density of a string remain constant, the frequency of vibration is inversely proportional to the effective length of the string. Reducing the length increases the frequency, making it a critical idea when aiming to achieve a specific ratio of frequencies by dividing the string into segments.

Harmonic Frequency Ratios

In the study of vibrating strings, harmonic frequencies are integer multiples of the fundamental frequency. Understanding how to manipulate the lengths of various segments to obtain frequencies in a specific ratio, such as 1:2:3, is essential for solving problems related to standing waves on a string.

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