Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

If $n_1, n_2$ and $n_3$ are the fundamental frequencies of three segments into which a string is divided, then the original fundamental frequency $n$ of the string is given by

$\frac{1}{n}=\frac{1}{n_1}+\frac{1}{n_2}+\frac{1}{n_3}$

$\frac{1}{\sqrt{n}}=\frac{1}{\sqrt{n_1}}+\frac{1}{\sqrt{n_2}}+\frac{1}{\sqrt{n_3}}$

$\sqrt{n}=\sqrt{n_1}+\sqrt{n_2}+\sqrt{n_3}$

$n=n_1+n_2+n_3$

Step-by-Step Solution

Identify the formula for fundamental frequency: The fundamental frequency $n$ of a stretched string of length $l$ , tension $T$ , and linear mass density $\mu$ is given by $n = \frac{1}{2l}\sqrt{\frac{T}{\mu}}$ .
Relate length to frequency: Assuming the tension $T$ and linear mass density $\mu$ remain constant for all segments of the string, the length $l$ is inversely proportional to the frequency $n$ . Thus, $l = \frac{C}{n}$ , where $C = \frac{1}{2}\sqrt{\frac{T}{\mu}}$ is a constant.
Use the property of total length: The total length of the string is equal to the sum of the lengths of the individual segments into which it is divided: $l = l_1 + l_2 + l_3$
Substitute the length-frequency relationship: Substituting $l = \frac{C}{n}$ into the length equation gives: $\frac{C}{n} = \frac{C}{n_1} + \frac{C}{n_2} + \frac{C}{n_3}$ Dividing both sides by the constant $C$ , we obtain the relationship between the fundamental frequencies: $\frac{1}{n} = \frac{1}{n_1} + \frac{1}{n_2} + \frac{1}{n_3}$

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started