Question

When a string of length 'l' is divided into three segments of length $l_{1}$, $l_{2}$ and $l_{3}$, the fundamental frequencies of three segments are $n_{1}$, $n_{2}$ and $n_{3}$ respectively. The original fundamental frequency 'n' of the string is

• A. $n=n_{1}+n_{2}+n_{3}$
• B. $\sqrt{n}=\sqrt{n_{1+\sqrt{n_{2+\sqrt{n_{3$
• C. $\frac{1}{n}=\frac{1}{n_{1+\frac{1}{n_{2+\frac{1}{n_{3$
• D. $\frac{1}{\sqrt{n=\frac{1}{\sqrt{n_{1}+\frac{1}{\sqrt{n_{2}+\frac{1}{\sqrt{n_{3}$

Hint:

Logic Tip: This is similar to calculating equivalent resistance for resistors in parallel, but applied to frequency and length.

Answer 1

The Correct Option is C

Concept: For a stretched string vibrating in its fundamental mode: \[ n \propto \frac{1}{l} \] or, \[ n = \frac{k}{l} \] where $k$ is a constant (depends on tension and linear density).
Step 1: Express length in terms of frequency \[ l = \frac{k}{n} \] For three parts: \[ l_1 = \frac{k}{n_1}, \quad l_2 = \frac{k}{n_2}, \quad l_3 = \frac{k}{n_3} \]
Step 2: Use total length relation \[ l = l_1 + l_2 + l_3 \] Substituting: \[ \frac{k}{n} = \frac{k}{n_1} + \frac{k}{n_2} + \frac{k}{n_3} \]
Step 3: Simplify Dividing throughout by $k$: \[ \frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3} \]
Step 4: Final Result \[ \boxed{\frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3 \]